Misplaced Pages

Revision as of 06:14, 21 February 2007

	Monty Hall problem/Archive 5 is a featured article; it (or a previous version of it) has been identified as one of the best articles produced by the Misplaced Pages community. Even so, if you can update or improve it, please do so.
	This article appeared on Misplaced Pages's Main Page as Today's featured article on July 23, 2005.
Article milestones Date Process Result May 3, 2005 Peer review Reviewed June 25, 2005 Featured article candidate Promoted January 29, 2007 Featured article review Kept
Current status: Featured article

Mathematics NA‑class

This page is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Misplaced Pages. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.MathematicsWikipedia:WikiProject MathematicsTemplate:WikiProject Mathematicsmathematics NA This page does not require a rating on Misplaced Pages's content assessment scale.

Archives
Archive 1: July 2004 Archive 2: July 2004 – July 2005 Archive 3: August 2005 – February 2006 Archive 4: March 2006 – November 2006

Archiving notes

I've moved the existing talk page to Talk:Monty Hall problem/Archive2, so the edit history is now with the archive page. I've copied back a few recent threads. Older discussions are in Talk:Monty Hall problem/Archive1. Hope this helps, Wile E. Heresiarch 15:28, 28 July 2005 (UTC)

I've done similarly to produce Talk:Monty Hall problem/Archive3. In keeping with Wile E. Heresiarch I moved the page so the edit history is with the archive page, and copied back the current (March 2006) discussions. _└/_talk 13:04, 10 March 2006 (UTC)

Likewise for Talk:Monty Hall problem/Archive4. Gzkn 06:15, 27 December 2006 (UTC)

Classic Riddle Misdirection

Ok, the main problem I see is with the invalid grouping of outcomes, which causes the number of possibilities to be reduced. The following 2 are syntactically identical ways of stating the same thing, both of which illustrate the number of possible outcomes:

• The player originally picked the door hiding goat number 1. The game host has shown the goat number 2.
• The player originally picked the door hiding goat number 2. The game host has shown the goat number 1.
• The player originally picked the door hiding the car. The game host has shown goat 1.
• The player originally picked the door hiding the car. The game host has shown goat 2.

OR

• The player originally picked a door hiding a goat. The game host has shown the other goat.
• The player originally picked the door hiding the car. The game host has shown either goat.

In the first example, if the player chooses to switch, the car is won in the first two cases and lost in the second two. A player choosing to stay with the initial choice loses in the first two and wins in the third and fourth cases. In the second example, if the player stays they lose in one and win in the other.

The original question is a classic case of riddle misdirection, nothing more. The fact that it is as old as it is and people still do not see it astounds me. If one of the incorrect doors is eliminated, and the player is then forced to make a decision, it changes to a straight 50/50 chance of getting the correct door. The initial choice can then be discarded. This article seriously needs to be revised.

Mvandemar 05:30, 8 November 2006 (UTC)

Quote The original question is a classic case of riddle misdirection, nothing more - true, what is interesting is not the question, but the insistence by intelligent people on the wrong answer. Again and again, people with higher educations within the sciences (if not exactly mathematics) have published their incorrect "solutions", claiming that everyone defending the correct solution are missing something, are stupid, are a disgrace, or whatever. That is an interesting story, similar in character to what is dealt with in another featured article, 0.999....--Niels Ø 12:29, 8 November 2006 (UTC)

Actually, I may have to eat my words. I just created a simulator to prove my point, and seem to be wrong. I thought I had done so before, where I randomized the full scenario, at http://www.endlesspoetry.com/goatcar2.php (with the source at http://www.endlesspoetry.com/goatcar2.phps should anyone care to view it) with 20,000 iterations per run. However, a friend then pointed out that I had merely shown that if the player randomizes his choice to stay with the original door or switch to the remaining one, that the odds then change to 50/50. I modified the routine, so that the player always switched, at http://www.endlesspoetry.com/goatcar3.php (append an "s" for the source), and sure enough the player won 2/3rds of the time. Mvandemar 14:54, 8 November 2006 (UTC)

I have great respect for you for testing your own theory and listening to your friend. You might have already spotted where the error came in in the chain of reasoning you listed above, but in case you haven't, the answer is that you listed four possibilities above, and treated them as if all four were equally probable, when actually two of the possibilities are only half as probable, on the average, as the others.

The scenarios "The player originally picked the door hiding the car. The game host has shown goat 1" and "The player originally picked the door hiding the car. The game host has shown goat 2" can only happen when the precondition "The player originally picked the door hiding the car." is true. Since the player originally picks the door hiding the car only 1/3 of the time, those two possibilities combined must have a total probability of no more than 1/3.

Another way to look at it that might remove some of the misdirection is to realize that we are only interested in the car. It doesn't matter if what's behind the other doors is a goat, a boat, a baboon, a spitoon or nothing at all, so it doesn't matter if we change the problem so that behind every door there is either the car or just nothing at all. When we looked at the problem in terms of a car and two goats, it was easy to get misdirected into thinking that it matters which goat we are shown. But when we look at it in terms of just one car, and nothing else, it's fairly obvious that it doesn't matter which of two "nothings" Monty shows us -- all that matters is whether he is showing us a nothing from a set of two nothings (1/3 of the time) or a nothing from a set of a nothing and a car (2/3 of the time). -- Antaeus Feldspar 22:10, 9 November 2006 (UTC)

The solution in two completely intuitive sentences

1) Your first guess is most likely wrong.

It has a 2 in 3 chance of being wrong.

2) If your first guess is wrong, and you switch, you will always win.

--220.237.67.125 15:58, 8 November 2006 (UTC)

The explanation above is only true for external obser who knows the host choice is not random. However in the perspective of the contestant who assumes the choice of the host is random the chance is still 50/50. I suggest you emphasise this fact in the article. Nice and clean explanation though from user above.

Well, not really. The fact is that under the given conditions, a strategy of always switching will succeed 2/3 of the time, whether or not the contestant:

• knows the 'hidden' facts of the game, such as that the host must always offer a choice,
• knows the 'open' facts of the game, such as that there is only one car and all the other doors are goats,
• understands the facts he does know about the game.

The contestant may think a different strategy is optimal, or may even be something like a robot that is capable of performing a simple strategy but incapable of 'thinking' about the strategy it performs. But the success of the strategy is still the same. -- Antaeus Feldspar 19:25, 23 November 2006 (UTC)

Wasn't this explaination originally in the article? I thought it was since I think it was the moment when it clicked in my mind. Or perhaps it just occured to me by myself. It is IMHO, one of the easiest ways to think about the problem. Nil Einne 18:08, 9 December 2006 (UTC)

Actually this explanation seems perfectly correct, whether or not the host choice is random. If the contestant's first choice is incorrect then opening the second door must indicate the correct choice. Either Monty Hall opens the second door and reveals a goat, in which case the contestant should switch to the last remaining door, or else Monty Hall opens a door and reveals the car, in which case the contestant should obviously switch to that door.--Lorenzo Traldi 10:41, 20 December 2006 (UTC)

can someone please explain the whole "what if monty doesn't know"

ok so a lot of people have been saying that if Monty forgets where the car is but luckily picks a door with a goat in to "eliminate" this makes the remaining choice 50/50. i am sorry but i thought i was understanding this thing up until that. how does it? how can odds change based on the host who is surely peripheral to the experiment? maybe i've missed it but i consider this to be unexplained and SEVERLY confusing 28/11

Please see #Aids to understanding, above. -- Rick Block (talk) 15:01, 28 November 2006 (UTC)

I think it is severely confusing, actually, because for such an apparently small change, it has a profound effect on the entire basis of the problem.

The change may become clearer if we make it clearer which parts of the problem are random and which are not. Let's replace all random choices with die rolls. Since we're randomizing the method of choosing the doors, we can safely de-randomize the doors themselves; we'll call them A, B and C and specify that the car is always behind door A.

With these changes in place, let's look at the original problem, the one where the host knows what's behind each door. A die is rolled to choose a door for the player: a result of 1 or 2 gives him door A; 3 or 4 gives him door B; 5 or 6 gives him door C. If the player gets a 1 or 2, the host opens door B or door C (and which one it is doesn't matter, of course); if the player gets a 3, 4, 5, or 6, the host opens whichever one of doors B and C the player didn't get, and offers a chance to switch to door A. Out of six faces the die could have come up, four result in a situation where the player would win by switching.

Now let's look at the "forgotten goat" variant, the one where Nonty has forgotten or never knew where the goat was. Now Monty's choice is random, too, so we'll determine it by the same die roll: on an odd number, he picks the door which comes earlier in the alphabet, and on an even number, he picks the later door.

Now, our problem states that Monty, having forgotten which door contains the goat, guesses at the door and luckily guesses correctly because he opens it to reveal a goat. However, what if the die comes up a 3? According to the rules we've set out, a result of 3 means the player chooses door B and Monty chooses the earlier of the two doors left, A. But we are told this didn't happen; we are told that Monty made his random choice after forgetting which door contained the car and the door he chose contained a goat. If the die came up a 3, then the door he chose would have contained the car; by similar logic, the die coming up a 5 would also result in Monty showing the car.

The answer is that even though these are possible results of the die, they are results that we know did not happen this time around (just as if Monty were to open his door and show the car, we would know with 100% certainty that the player did not pick the car this time around.) There are only four die rolls which could have resulted in Monty opening a door at random and getting a goat: 1, 2, 4, and 6. Out of those die rolls, 1 and 2 represent the player's door having the car, while 4 and 6 represent the car being behind the remaining unpicked door. Thus, the chances in this variation, unlike the original problem, are 1 in 2. -- Antaeus Feldspar 17:41, 28 November 2006 (UTC)

A simple solution to the "Monty doesn't know" problem is that if Monty doesn't know, then his choice is between two doors that each hav a 1 in 3 chance (the other 1 in 3 chance door being already picked by the contestant). Assuming Monty picks the goat door (as we must) this simply removes one of the 1 in 3 chance doors and we are left with two doors that each had a 1 in 3 chance (the same chance)when there were three and so now as equal probabilities have a 1 in 2 chance since there are only two.Davkal 18:09, 28 November 2006 (UTC)

yes but how is the immediate above different from the original problem?? you could use surely use the exact same logic for the original problem. lets forget probabilities here and return to the root of the question "are you more likely to win by switching or not?".

Response: it's not clear you can talk about "more likely" without discussing probabilities, so the answer is no, in the new game you are not more likely to win by switching because the odds are 50/50.Davkal 13:45, 29 November 2006 (UTC)

now i understand that the reason (in the original question) you will win is because you have essentially split the three doors into two sets, then after you pick between the two sets. thus you have a better chance of winning with the set of two (even tho one has been revealed, tho this is meaningless). surely if you were to repeat the experiment with the host not knowing and took the results of the games where he randomly chose the goat (discarding the games where he chose the car as the question states the chances if he picks the goat) you would still have a better chance of switching? mentioning such things as the odds of monty picking the car HAVE NOTHING TO DO WITH THE QUESTION. the question states he has chosen the door with a goat behind it. thus logically you still have a better chance of switching.

Response: if you were to repeat the experiment in the way described then all the cases that are discarded (the cases where Monty picks the car by accident) are cases where you would have won by switching (i.e, you didn't pick the car first). Overall, then, your score drops to 1 in 3 by switching, 1 in 3 by sticking and the "removed" 1 in 3 where Monty picks the car. So your choice to stick or switch is now between the two remaining 1 in 3 chances which equals 50/50 in those remaining cases.Davkal 13:45, 29 November 2006 (UTC)

imagine two of the "monty hall problems* playing side by side on a stage with a screen seperating them. on either side of the screen, both contestants pick their first box. on one side of the screen, the producer tells monty which of the remaining two boxes he must reveal (showing a goat), he does so. on the other side the host cannot here the producer very well and cannot be sure which box to pick. still he manages to get the one with the goat in (which he had a 2/3 chance of anyway). now according to what people have said on this site, even though both players games have *PHYSICALLY AND EMPIRICALLY* proceeded in exactly the same way, the first contestant should switch, while it does not make a difference if the second one does. now i know that *odds* can be made to show that it's equal chance, but odds aren't real life. to put it another way, in the first game, obviously the player has a better chance (2/3) if they switch. now imagine both games and the way they went, and if they have different odds, would the odds change if the producer of the second game confirmed to the host that his random guess was right?

Response: the difference is that in multiple trials one Monty would never pick the car where the other Monty (who can't hear the producer) would pick the car 1 in 3 times. The probabilities, then, for what happens in these cases are different, because what would actually happen in these cases IS different. The point being that your chances of picking the ace of spades out of a pack of cards at random are 1 in 52, where your chances of picking it out after looking through the deck to find it are 1 in 1. This is true even in cases where you have just picked the ace of spades out by random chance, and true even if you wanted to say (which I would not) the in both cases what has happened was physically and empirically identical.Davkal 13:45, 29 November 2006 (UTC)

ok i think i understand what you mean, the discarded ones (i.e the ones where monty picks the car) are scenarios where it would be beneficial to switch, so in a case of multiple games, the time when in would be beneficial to switch are halved. thanks for explaining this to me as it was really starting to get me annoyed. but if it was just one or two games, and this thing happened, it would still make sense to switch, right? i mean, in a real life situation, even if this was happening, it would still make sense to switch, as you still have a lesser chance of getting it right first time? 29/11/06

No, because all the multiple trials are intended to show here is what could be the case. And what they show is that in a one-off case where you pick a random door, and then Monty opens a random door from the other two and reveals a goat it's 50/50 between the remaining doors. This is because the one-off case is one of only two remaining possibilities - you picked the car and Monty picked a goat or you picked a goat and Monty picked a goat - in one you already have the car, in the other you don't. The third possibility, you picked a goat and Monty picks the car can be ruled out in the one-off case because we know it hasn't happened (i.e. Monty didn't pick the car).Davkal 15:27, 29 November 2006 (UTC)

i'm sorry but now you've really confused me. consider this

"This is because the one-off case is one of only two remaining possibilities - you picked the car and Monty picked a goat or you picked a goat and Monty picked a goat "

how is this different from the original case!? monty is always going to pick a goat, so i fail to see how in a one-off one time only case, this changes the odds? say two games on the same stage, side by side one time only. both games proceed exactly the same however monty one is told by the producer to reveal a certain box at the second stage. monty two (who cannot see the other game) does not have a producer, but coincidentally picks the same box. the two contestants must have the same odds right? — Preceding unsigned comment added by 194.168.3.18 (talk • contribs)

Both games DO NOT proceed in exactly the same way because in one Monty's behaviour is guided in a non-car revealing way, whereas in the other it is only by chance that this happens. This means that the probabilities are different. It might be clearer why this affects the odds if you think: what are the chances of Monty revealing a car in the original version (none); and what are the chances he will reveal a car in the new version (1 in 3) - this is what is different and this is what results in different odds after Monty's actions.Davkal 16:09, 29 November 2006 (UTC)

The fact that the second Monty would have to "coincidentally" pick the same box to make the situations come out the same (outwardly) should indicate that in fact the situations aren't the same at all. The first situation involves one random choice out of three (the player picking one of Monty's three doors) and one choice that isn't random (Monty selecting a door that he knows contains a goat), but the second situation involves two random choices (the player picking one of Monty's three doors, and Monty picking one of the two doors the player left behind, which might have the car behind them.)

Here's a breakout of possibilities for the first situation:

1. The player picks the car. Monty opens either of the remaining two goat doors. Winning strategy: STAY
2. The player picks the first goat. Monty opens the door containing the second goat. Winning strategy: SWITCH
3. The player picks the second goat. Monty opens the door containing the first goat. Winning strategy: SWITCH

Now here's the breakout for the second situation:

1. The player picks the car. Monty opens the door containing the first goat. Winning strategy: STAY
2. The player picks the car. Monty opens the door containing the second goat. Winning strategy: STAY
3. The player picks the first goat. Monty opens the door containing the car.
4. The player picks the first goat. Monty opens the door containing the second goat. Winning strategy: SWITCH
5. The player picks the second goat. Monty opens the door containing the car.
6. The player picks the second goat. Monty opens the door containing the second goat. Winning strategy: SWITCH

The player makes a choice between three doors; Monty then makes a choice between two. This means there are six possibilities. However, we are told that when Monty opens the door, it has a goat behind it. This means that any "possibility" that would result in Monty opening a door with the car behind it didn't happen. Out of those which are still possible, half of them have the car behind the player's door. -- Antaeus Feldspar 01:00, 30 November 2006 (UTC)

Hang on, you've just counted your options here and said "there are four valid options and it's half-and-half." You've completely forgotten that your first two options, wherein "the player picks the car," comprise only 1/3 probability out of the entire set of options. Remember that the player chooses first and there is only a 1/3 chance that the player picked the car.

There is no quantitative difference between "Monty knew which door had a goat" and "Monty correctly guessed which door had a goat" as long as he didn't open the player's door. Maelin (Talk | Contribs) 02:23, 30 November 2006 (UTC)

No, because two out of the six intial options is 1/3. The four options that remain to make it 50/50 excludes the two cases where Monty picks the car because were told that hasn't happpened. Davkal 02:37, 30 November 2006 (UTC)

"You've completely forgotten that your first two options, wherein "the player picks the car," comprise only 1/3 probability out of the entire set of options." No, I haven't forgotten that. They do comprise only 1/3 probability out of the entire set of options, but only if by "entire set of options" you mean "all the possible outcomes from the player picking one of three doors and Monty picking one of the two doors left." When you restrict it to "all the possible outcomes from the player picking one of three doors and Monty picking one of the two doors left that result in Monty's door containing a goat," you've eliminated two of the six possible outcomes.

Suppose you roll a die without looking at the result. The chances of a 1 coming up are 1/6, and so are the chances of a 2, so are the chances of a 3, et cetera; you have six possible outcomes and each of them has a 1/6 chance of being what actually happened when you rolled the die. But suppose you ask a friend to look at the die and tell you if it's a 6. "No," he says, "it's not a 6." What does this do to the chance that you rolled a 6? Obviously, that's gone down to nothing, since you know through your friend that it didn't come up as a 6. What does this do to the chances for all the other faces? They can't still each be 1/6, because then they only add up to 5/6 and not to 1. What actually happens is that the chances for the five remaining faces each go up to 1/5. In just the same way, in the variant problem, each possible outcome has a probability of 1/6 until the door Monty picked by chance is shown to contain a goat; this eliminates two of the six "possible" outcomes by showing them not to be possible and leaves each of the remaining outcomes with a probability of 1/4.

Why doesn't the same thing happen in the original problem? Because nowhere in the original problem do we get evidence that eliminates possible outcomes. No matter which door the player chooses, Monty always has a door chosen from the other two which he can open to show a goat. If the player chooses door A and Monty opens door C, it could be because door B contains the car... or door A holds the car and Monty just happened to choose C though he could have as easily chosen B. The fact that he opens a door to reveal a goat doesn't eliminate any outcomes and so it doesn't change any probabilities. -- Antaeus Feldspar 05:44, 30 November 2006 (UTC)

I think the problem here is where we're trying to analyse a probability situation in which we have arbitrarily decided that a random event turned out "just so". I'm not sure the same mathematical analysis is valid here and I'm not convinced about this assumption that when you chop off a branch of a probability tree by saying "it happens to not turn out like that" that you can just redistribute the "lost" probability evenly among the remaining branches. However I'm not so sure of my own reasoning here so I'll wait until someone can clarify this. Maelin (Talk | Contribs) 11:57, 30 November 2006 (UTC)

Well, maybe a visual demonstration will make it clearer why redistributing the probability of a "lost" (disproven) outcome evenly among the remaining possible outcomes is the correct thing to do.

Suppose you want to get a random color from the set of Black and White, but the only randomizer you have produces a random color each time you use it from the set of Black, Red and White. (Let's say it's a special die from a board game, and for some reason you don't have a coin to flip.) You decide that if the die comes up Black or White, you'll accept that as the outcome, and if it comes up Red, you'll simply roll another result.

Now, let's visually depict this. Draw a long, thin box lengthwise across a sheet of paper; this represents the entire probability space. Now divide the box into equal thirds; mark the section on the left as "Black" and the section on the right as "White". These represent the portions of the probability space where "Black" or "White" came up on the first roll.

The section in the middle represents the portion of the probability space where the first roll came up "Red", but it doesn't get marked "Red" because Red is our signal to roll again. So we divide this middle section into three equal portions again, marking the left one "Black" and the right "White". Now we've marked all portions of the probability space where we got an answer on the first or the second roll.

If you keep doing this, I think you'll see the following:

• Each time we color in Black and White sections to represent those colors coming up on the nth roll, the portion of the probability space that is unmarked decreases. Continuing to infinity (to represent rolling, and rolling, and rolling, until we get a non-Red result) would result in the entire probability space being colored in.
• Each time we color in Black and White sections, we add them in precisely equal amounts, so the proportion of Black to White always remains the same: if we continue to infinity, the entire probability space will be split evenly between Black and White.
• This simulates our variant problem: Black represents the cases where the player gets the car on the first try; White represents the cases where the player gets a goat, and Monty opens the door that shows the other goat; Red represents the cases where the player gets a goat and Monty opens the door to show the car. Since our problem specifies that Monty does not open the door to show the car (just as we specified that "Red" is not a valid result) we keep re-randomizing until we get a result that is valid -- which, as we just saw, is equivalent to dividing the probability space among only those results which are valid. -- Antaeus Feldspar 22:13, 30 November 2006 (UTC)

I think I get it now. It's because the contestant who knows that Monty forgot but was lucky can no longer think, "What choice pays off most frequently in this situation?" but instead needs to think. "What are the possible ituations I might be in, and for each one, what is the probability of it being the case in this instance?" Maelin (Talk | Contribs) 08:21, 4 December 2006 (UTC)

i'm sorry but i still don't agree with this. the question is not about options or possible scenarios, but if monty picks a goat, now consider this ;

":# The player picks the car. Monty opens the door containing the first goat. Winning strategy: STAY

1. The player picks the car. Monty opens the door containing the second goat. Winning strategy: STAY"

in the strategy of the player, it doesn't matter which of the goats monty picks. these two scenarios are basically one. in the original scenario, imagine that the player picks the car, even if monty knows that the two remaining boxes are goats he still essentially "decides" which one to reveal, yet this obviously isn't counted in the analysis. the argument of the original solution was that because you were picking only one out of the three, the choice of switching was essentially a choice between staying with your one box out of three or switching to what was essentially two choices out of the original three. it has also been noted that the "reveal" by monty is a diversionary tactic and "red herring" of the puzzle designed to make it more counter-intuitive. the fact that in the "monty doesn't know" scenario, he has still picked the goat, means the choice to switch is still based on the "one choice versus two". the player is still more likely to have chosen the goat at first (how can you dispute this) thus switching will always be a better option (at least in a one off situation, the question is supposed to put us in the situation of the player and ask us what we would/should do. multiple situations, i.e. ones where monty would discard the games where he accidently picked the car would undoubtedly lower the benefits of switching). the reason (in the original game) why switching is beneficial is more based on the primary "split" than it is on monty's reveal. he will always reveal a goat, and in the "monty doesn't know" scenario, regardless of whether it was random or not, he has still picked a goat, thus the benefits of switching remain the same. your mathematics are undoubtedly right, yet i think the reasoning is wrong. ec 30/11 — Preceding unsigned comment added by 194.168.3.18 (talk • contribs)

In the original problem, it doesn't matter which of the goats Monty picks because Monty's knowledge always guarantees it to be a goat. In the variant problem, Monty is picking at random: when the player has picked the car, Monty's random guessing will always pick a goat (since goats are all that are left); when the player has picked a goat, it's only half the time that Monty's random choice will pick a goat. Yes, the chance that the player didn't initially pick the car is twice the chance that he did -- but only half the outcomes where the player didn't pick the car match what we know, that Monty opened a door to reveal a goat. The fact that only half the player-picked-a-goat outcomes are Monty-shows-a-goat outcomes cancels out the fact that they are twice as frequent as player-picked-a-car outcomes. -- Antaeus Feldspar 17:59, 30 November 2006 (UTC)

I think it's time for another experiment. Take three cards including the ace of spades. The ace of spades counts as the car. Step 1: shuffle the cards and then deal them face down so you don't know where the ace of spades is. Step 2: pretend you're the contestant and move one card to the side. Step 3: Pretend you're Monty and turn one of the remaining cards over. What happens is this: in about one-third of the trials the card you turn over as Monty will be the ace and you will have to abandon those trials; in the other two-thirds of the cases (where "Monty" reveals a "goat"), you will see that that the original "contestant's choice" card is the ace half the time and the third card picked by neither the "contestant" nor "Monty" is the ace half the time. That is, in the trials that reflect the new puzzle (Monty picks a goat/non-ace) you win half the time if you always switch and you win half the time if you always stick. Do it, try it, and then once you are satisfied that there is no advantage to be had in the new game from either switching or sticking, try to work out what is wrong with your arguments to the contrary above. Davkal 12:34, 30 November 2006 (UTC)

right gonna do it now at work to pass the time... — Preceding unsigned comment added by 194.168.3.18 (talk • contribs)

The extreme case

Maybe this will help. Consider what happens if Monty has a very specific kind of forgetfulness, so that he attempts to execute the classical problem, but he always gets confused and opens the wrong door. Then if you pick a goat, Monty always shows you the car. So if Monty shows you a goat, that means you picked the car, and you shouldn't switch.

The forgetful problem (in which it doesn't matter if you switch) lies somewhere between the above problem, in which you shouldn't switch, and the classical problem, in which you should.

To be precise, if Monty's memory is such that he opens the wrong door with probability p, the probability that you picked the car, given that he reveals a goat, is given by

${\displaystyle {\frac {1\cdot (1/3)}{1\cdot (1/3)+(1-p)\cdot (2/3)}}={\frac {1}{3-2p}}.}$

You can have fun plugging in values of 0, 1/2, and 1 for p. Melchoir 03:24, 30 November 2006 (UTC)

I was initially confused by this, too, but Marylin is right. The way I've always envisaged the Monty Hall problem is about information: does Monty's choice tell you (as the player) anything new? The Venn diagram representation demonstrates that in the original problem the answer is, NO. Monty might as well ask you to take either your first choice or switch to take whatever is behind both the other doors, for all the good his revealing the goat does. Now in this situation where Monty picks from the remaining doors at random, YES, his picking the goat does tell you something new. He is more likely to pick the goat if you picked the car. The fact that he did pick the goat means the conditional probability that you picked the car is higher. As it turns out, for the case where he picks at random it comes out that your chances of winning are 50:50 whether you switch or not. In the 'extreme' case, if Monty is guaranteed to mess up and reveal the car if you picked a goat, then again, YES, you do learn something new, and in this extreme you can be sure that if Monty picked a goat, you have the car and should not switch, just as Melchoir's formula reveals. Ironically, the reasoning that people incorrectly apply to the original problem produces the correct answer to the 'forgetful Monty' version, and vice versa! BJC 16:45, 14 December 2006 (UTC)

The long discussion above is similar to a decision between two possible rules for the "clueless host" game. Possible Rule 1: if Monty accidentally reveals a car, the contestant is allowed to choose it. Possible Rule 2: If Monty accidentally reveals a car, the game is canceled and the contestant gets to play again, presumably after the car and goats have been rearranged behind the doors.

Under Possible Rule 1, the original 1/3 vs 2/3 analysis applies ... except the bonus for switching is obvious because sometimes the contestant is looking at the car and can decide to take it. So under this rule, switching is still advisable -- when you see a car take it! When you see a goat, you lose nothing by switching to the non-revealed door, but you also gain nothing. The whole advantage of switching involves the car having been revealed.

Under Possible Rule 2, that advantage has been declared illegal. Consequently the advantage of switching has been removed, and there is no effective difference between switching and staying.

If a careless Monty Hall reveals a goat by accident, that's the same as operating under Possible Rule 2. If the real Monty Hall reveals a goat, that's the same as operating under Possible Rule 1, because in the cases in which he might have accidentally revealed the car he would open the other door instead. --Lorenzo Traldi 04:21, 20 December 2006 (UTC)

I believe I follow what you're saying, Lorenzo, but in actuality, your "Possible Rule 1" can never play a role in the problem under discussion. Even in the variant problem, where it is clear that Monty could pick the door that has the car behind it, the problem statement clearly states that he doesn't. Which, I freely concede, is a damn confusing thing to grasp: on the one hand he can because the random choices make that possible but on the other hand he can't because the problem statement says he didn't? But that's actually the case. It's not any different from saying "You flip two coins and at least one of them comes up heads; what is the chance that the other one is tails?" Obviously the randomizer (the two coins) offers the possibility of "both coins come up tails" but our problem statement rules out that possibility by telling us that "at least one of them comes up heads".

Now, "Possible Rule 2" is perhaps an easier way to model the probabilities involved in situations like this of "all these outcomes are possible according to our randomizer but only some of them are possible according to the problem statement". It is not necessary for "do-overs" to actually, literally occur (which makes it somewhat unfortunate that someone added "the game ends prematurely one-third of the time" to discussion of the variant problem, as that seems to miss the point that the problem statement states what the host's door reveals and it is never the car.) However, calling a "do-over" every time you try to simulate the problem and get preconditions contrary to the problem statement makes it possible to do some of the traditional forms of analysis such as, well, simulating the problem with rolling dice and flipping coins. -- Antaeus Feldspar 05:49, 20 December 2006 (UTC)

I agree. The "possible rules" simply give reasonable whole-game contexts. I guess I wouldn't say Possible Rule 1 can never play a role in the problem under discussion, because the never seems to forbid Monty Hall's carelessness, which is after all the point in this variant. I'd rather simply say that once the contestant sees a goat we realize a car has not been revealed this time. (Technically, with the careless host the conditional probability of selecting the door with the car when a goat has been revealed is .50, either by switching to the still-closed door or staying with the door chosen first. The conditional probability of selecting the door with the car when that's the one Monty opened should certainly be 1.)--Lorenzo Traldi 10:34, 20 December 2006 (UTC)

"I guess I wouldn't say Possible Rule 1 can never play a role in the problem under discussion, because the never seems to forbid Monty Hall's carelessness, which is after all the point in this variant." You're confusing two different things. You're confusing what is possible according to the random events and what is possible according to the evidence. The random events do indeed hold the possibility that Monty will open the door and show the car but that does not matter because the evidence says that he did not. If a problem says "You flipped a coin and it came up heads" then anything you say starting with "but if it comes up tails instead" is irrelevant. -- Antaeus Feldspar 17:39, 20 December 2006 (UTC)

I think the use of "never" is what causes the confusion, as I said.--Lorenzo Traldi 19:48, 21 December 2006 (UTC)

If Monty does not know, and picks a goat, it is random which goat he picks, and he will always pick a goat. If monty does know, it is random which goat he picks, and he will always pick a goat. Since nothing happens when he picks a car, well, he can't, because then there would be no strategy. If he knows, then he will not pick the car regardless. The results are completely the same. Saying they are different is like saying that if you have a two thirds chance of picking a pair of socks if you put on gloves, and you forget what you are supposed to do to get the two thirds chance, yet still put on gloves, you will have a fifty percent chance of picking socks. Again, He will pick the car then goat one or two (one third chance), goat number one, and then monty picks goat number two (one third chance), and goat number two then goat number one (one third chance). In fact, whether or not he knows, you should always switch if you get the chance, because if you got the chance, Monty most certainly picked a goat. Again, another analogy- if you pick up a tissue because you know your supposed to, and blow your nose with it, and if you pick up a tissue when you are not sure if your a supposed to, there is no chance of blowing your nose with it.

Your logic resembles the logic involved in the original problem. Your math and reasoning are completely off, because your judgement is impaired by incorrect reason. No offense, but it is kind of funny that you got the problem right, then got the problem wrong.

Given this logic, I think to avoid confusion it would be necessary to adjust the article to say that there is a ⅔s chance, because there is. Squarethecircle 00:07, 6 January 2007 (UTC)

Deal or No Deal vs. Let's Make a Deal

It may be instructive to contrast the situation in the Monty Hall Problem with the situation that occurs in the game show Deal or No Deal when the player in Deal or No Deal is down to two cases and facing the decision to switch the case he or she originally selected with the only other remaining case. Suppose that in Deal or No Deal there are 26 cases containing different prize amounts and the highest prize possible is $1,000,000 as in the original version of the show. If there are only two prize amounts left on the board and one of them is $1,000,000 while the other is $0.01, the probability of winning $1,000,000 does not increase if the player switches the case he or she originally selected with the only other case remaining. However, if the host Howie Mandel eliminated the other 24 cases because he knew they didn't contain $1,000,000 then the probability that the player would win $1,000,000 by switching cases would be 25/26. Now what if Howie helped out in eliminating some of the 24 cases and the player got lucky in eliminating the others?  :-) — Preceding unsigned comment added by 69.141.232.16 (talk • contribs)

Well from my knowledge of Deal or No Deal from the Australian version, it's more complicated/different. In Deal or No Deal, if you end up with two cases, one with $0.01 and one with $1,000,000, your choices are to take the bank offer which would be about $500k or assume your cases is the $1 million. If your case has the $0.01, you end up with $0.01. You can't choose the other case as such, you only choose between whatever is in your case and the bank offer. If the host knowingly opens 24/26 cases without the $1 million, it will be a very, very bad idea for you to think your case has 1$ million since there's a 1/26 chance it does whereas as 25/26 chance it only has $0.01. So the chances are very likely you'd end up with $0.01. Far better to accept the bank deal. Of course, if the host does do that, then it doesn't really make sense for the bank to offer you about $500k either Nil Einne 18:17, 9 December 2006 (UTC)

Yeah, I'd take the $500,000 bank offer. In that situation, a 'fair' bank offer would be $1,000,000 * (1/26) + $0.01 (25/26) ~= $38,461.55. That reminds me of something I've wondered about Deal or No Deal: Suppose, in the normal version of the show where the host doesn't help out, that there are three cases left, $0.01, $500,000 and $1,000,000 and the player turns down a reasonable offer, say $450,000. If the player then opens the case containing $500,000, the banker could force the player to get lucky by offering a really low amount, like $0.02. I wonder if there's anything stated in the rules of the show about the offers or if the player is supposed to just trust that the banker wouldn't do such a thing.  :-) 69.141.232.16 21:30, 9 December 2006 (UTC)

Comment about clueless host

Vos Savant's answer makes it seem that the host's cluelessness makes a difference. In fact it does not. The contestant now knows that the open door hid a goat. As the contestant originally had only a 1/3 chance of choosing the door with the car, there is a 2/3 chance of getting the car by switching, just as in The solution above. --Lorenzo Traldi 20:05, 19 December 2006 (UTC)

In actual fact it does make a difference, please see #can someone please explain the whole "what if monty doesn't know" (above). -- Rick Block (talk) 20:19, 19 December 2006 (UTC)

The host's cluelessness does make a difference because it introduces a chance that the host may ruin the game by accidentally opening a door to show a car and forcing a do-over.* This ruination cannot happen when the player has chosen the car (obviously) but it happens one out of every two times when the player chooses a goat. This means that out of six combinations of doors picked by the player and the host, there are two that result in the winning strategy being to switch, there are two that result in the winning strategy being to stay, and there are two that force do-overs. Assuming that we'll never hit an infinite series of do-overs, this means we really have four combinations: two that result in the winning strategy being to switch, and two that result in the winning strategy being to stay.

(*This is, anyways, the cleanest way to model the problem conditions which state that the forgetful host's random door opens to reveal a goat, even though clearly sometimes random choices would lead to the car being revealed instead.) -- Antaeus Feldspar 22:49, 19 December 2006 (UTC)

The difference in the clueless host stems from the fact that in the original problem, it's a "clean" probability situation - you can consider all the options and draw out a standard probability tree. In the clueless situation, however, you need to arbitrarily lop branches off. Instead of "Any of these options could have happened", it's "any of these options could have happened, except for these two which, just by luck, didn't. You're restricting yourself to analysing a certain subset of all possible states, a subtree of the total probability tree, in which a particular random event went a specific way. Maelin (Talk | Contribs) 01:01, 20 December 2006 (UTC)

It looks like there are now two threads on this -- but anyway under the clueless host assumption, the advantage of switching is only that Monty might sometimes show the contestant the car. (That is, my comment above was wrong -- apologies.) If the contestant is allowed to keep the car under those circumstances, then "always switch" is still good advice -- 1/3 of the time you'll switch to the car because you're looking at it, 1/3 of the time you'll switch to the car because Monty has revealed a goat and your door has a goat, and 1/3 of the time you'll switch to a goat when you have the car and Monty has revealed a goat. In the four cases in which Monty has revealed a goat, switching is 50-50. In the two cases in which Monty has revealed the car switching is 100-0. :-) --Lorenzo Traldi 05:05, 20 December 2006 (UTC)

I'm sorry, Lorenzo, but talking about 'what happens in those situations where Monty opens the door and it shows the car' is only going to confuse people (including yourself) to no benefit, because the problem states that this does not happen. Introducing a case that does not happen and saying "when this happens, the player is offered a chance to switch to the car that has been revealed" has nothing to do with the problem under discussion because it is part of the problem statement that the door opened reveals a goat. -- Antaeus Feldspar 15:33, 20 December 2006 (UTC)

Well it does explain where the 2/3 went.--Lorenzo Traldi 23:51, 20 December 2006 (UTC)

Clarifying that instead of having three results, all of which are possible from our randomizer and all of which are possible according to the evidence, we have six results, all of which are possible from our randomizer but only four of which are possible according to the evidence, is indeed a good thing. But to speculate on what happens "when" those two results which our problem statement makes clear don't happen, happen, is not really clarifying anything. -- Antaeus Feldspar 14:18, 21 December 2006 (UTC)

It doesn't matter if there is a do over, because if there is, then there is no choice made. A choice can only be made if he picks the goat, eliminating the chance that he picks a car. Also, the page says he still picks a goat and it is one half. This is impossible. If you think it is not, then you probably should spend the rest of the doing what my user name implies, to knock some sense into you (no offense, it's probably just you confusing yourself. The average person would end up with a similar result if under the same condition). Feldspar, you are completely correct. I will not go into further depth on this situation because it has been explained above. Squarethecircle 00:13, 6 January 2007 (UTC)

"It doesn't matter if there is a do over" is simply not correct. For instance, suppose we toss a coin and call a do over every time we get heads. Then the only toss that counts is tails! --Lorenzo Traldi 19:28, 14 January 2007 (UTC)

In this case, do over or not is not specified. All we can, and should, do is compute the conditional probabilities in the case where the host accidentally does not reveal the car. As has been explained above several times, given the host picked randomly and given the host did not reveal the car, the resulting situation is a 50/50 case. -- Rick Block (talk) 21:33, 14 January 2007 (UTC)

Many people work their way through problems like this one by doing thought experiments. You are absolutely right, Rick, that "imagine that Monty chooses a door at random, and happens to pick a door with a goat instead of a car" gives exactly the same probabilities as "imagine that Monty chooses a door at random, and if he picks the car, we do the whole problem over and over until he picks a goat." However, if you ask someone who isn't already following to imagine that random chance just "happens to" come out a particular way, they will almost certainly think that you are asking them to twist the rules of probability in order to lead to an incorrect conclusion. If you tell them that they can simulate the random choice themselves, using any randomizer they like and simply treating any result which does not correspond to the conditions specified in the problem as a signal for a "do-over", they are more likely to be able to carry out an appropriate thought-experiment and reach the correct answer. -- Antaeus Feldspar 01:02, 15 January 2007 (UTC)

Personally I think "just happens to show a goat" is clearer, and it is the do over that is introducing a new rule that seems twisted ... of course this is merely an expository distinction, but I like the idea (discussed elsewhere) of first considering the general situation including the possibility that Monty Hall might show the car by accident, and then restricting one's attention to those cases in which he happens not to. Lorenzo Traldi 01:07, 16 January 2007 (UTC)

I may sound stupid, but...

...If the host shows one of the goats the game provability is no more 1/3 or 2/3 or whatever: it becomes 1/2, right? In fact the game was never 1/3 because one of the doors with a goat will always be eliminated, so the goat door 1 and the goat door 2 are, in a kind of way, the same door (aka the same result).It doesn't matter if you choose a door with a goat or not, when in the end you have to choose there are just two doors and one goat. In my point of view the game starts when there are only two doors and 50% of changes for each door.
PS: Feel free to correct me
PS2: I'm sure you will ;)

What you're saying isn't stupid; however, it is incorrect. You are saying that when there are only two doors, there are only two possibilities: either the car is behind the player's door, or it is behind the other unopened door. If the probabilities of these two possibilities are equal, then they must each be 1/2.

However, it is not true that the probability of these two possibilities are equal, because there are twice as many ways to get to "the car is behind the other unopened door" as there are to get to "the car is behind the player's door". If the player picks the first goat (1/3 probability), Monty shows the second goat, and the car is behind the other unopened door. If the player picks the second goat (1/3 probability), Monty shows the first goat, and the car is behind the other unopened door. Only if the player picks the car (1/3 probability) does the other unopened door contain a goat. -- Antaeus Feldspar 20:27, 25 December 2006 (UTC)

Thank you for the reply, now I can see it. I would have read the other disussion threads, but in this article there are too much.Great mike 10:27, 26 December 2006 (UTC)

The explanation is wrong and contains a logical fallacy and arrives at the right answer by coincidence

This problem has probably confused so many people because the typical explanation for it is wrong even though it still gives the right answer. People poke holes in the argument only to find through some other means that the conclusion is still true.

It is fallacious to hold that you originally had a 1/3 chance of choosing the very goat that monty reveals that you did not choose when he opens a door. It is true that the knowledge that you did not choose the goat monty reveals would make it 50/50 you chose what was left, but the additional information that monty revealed that goat (which he is less likely to do because you chose the car than because you chose the other goat) readjusts the odds such that you are more likely to win by switching.

Using an alternate version of the problem where the odds of choosing each are the same but you know monty has a strategy for showing one of the 2 goats (which you can distinguish) shows how this reasoning fails in the general case. Here everything that is said in the explanaition of the original problem remains true but the conclusion is now false - because once again it is fallacious to hold that you had a 1/3 chance of choosing a goat to begin with even after it has been proven to you that you did not choose that very goat.

Here is the math, first a real explanation and then a alternate version of the problem where the wiki explanation fails.

A REAL sample space might look like

1/3 You chose Goat 1 Monty reveals Goat 2 Switching Wins

1/3 You chose Goat 2 Monty reveals Goat 1 Switching Wins

1/6 You chose car Monty reveals Goat 1 Switching Loses

1/6 You chose car Monty reveals Goat 2 Switching Loses

You can see how this is accurate because if Monty reveals Goat 2, then you can cross out thatyou chose goat 2 and that you chose car and monty reveals 1 i e

1/3 You chose Goat 1 Monty reveals Goat 2 Switching Wins

X1/3 You chose Goat 2 Monty reveals Goat 1 Switching Wins

X1/6 You chose car Monty reveals Goat 1 Switching Loses

1/6 You chose car Monty reveals Goat 2 Switching Loses

and then you get

2/3 You chose Goat 1 Monty reveals Goat 2 Switching Wins

0 You chose Goat 2 Monty reveals Goat 1 Switching Wins

0 You chose car Monty reveals Goat 1 Switching Loses

1/3 You chose car Monty reveals Goat 2 Switching Loses

Laying out the sample space in this correct manner also allows you to solve any other similar problem including the one where Monty chooses goat 1 with 2/3 probability when you choose the car:

1/3 You chose Goat 1, Monty reveals 2

1/3 You chose Goat 2, Monty reveals 1

2/9 You chose the car, Monty reveals 1

1/9 You chose the car, Monty reveals 2

Note in this case that although you had originally a 1/3 chance of choosing each goat or the car, collapsing the 1/3 chance of the revealed door into the unrevealed unchosen door to get 2/3 chance of car, or following any of the other rediculous suggestions in the wikipedia article does not give you the right answer. The odds can even be 50/50 for switching if monty always chooses one goat over the other.

-The author of this post abstains from signing his posts such as not to promote appeal to authority and ad hominem fallacy. — Preceding unsigned comment added by 69.252.158.32 (talk • contribs)

I'm not sure I completely understand your objection, but let's try an example case where Monty chooses goat 1 with 100% probability when you choose the car. I think you're saying this means the starting probabilities are:

1/3 You chose Goat 1, Monty reveals 2

1/3 You chose Goat 2, Monty reveals 1

1/3 You chose the car, Monty reveals 1

0 You chose the car, Monty reveals 2

So, if Monty reveals Goat 1 then you have a 50% chance of winning the car by switching and (similarly) if Monty reveals Goat 2 then you have a 0% chance of winning the car by switching. Although I agree these are the correct conditional probabilities, I'm not sure this is a helpful analysis since the goats are indistinguishable. Since you don't know which goat Monty has revealed, I think you need to combine these two probabilities and when you do this you end up needing to look at all of the possibilities (not just those involving the goat that Monty has revealed). -- Rick Block (talk) 19:14, 28 December 2006 (UTC)

It is stated in my examples that the goats are distinguishable. It is irrelevant whether or not you can distinguish the goats as far as the validity of the given solutions to the problem are concerned. Why should the fact that the goats are distinguishable suddenly invalidate a claim like "collapsing the 1/3 chance of the revealed door into the unrevealed unchosen door to get 2/3 chance of car"? It wouldn't - this claim was already invalid to begin with. It is no different than saying if you dance on your head on the night of a fool moon you will have a 2/3 chance of winning by switching. See look - it has been verified by experiment.

All you have to do to disprove the dancing on your head claim is do the problem without dancing on your head and see if you still get 2/3 chance to win by switching. And all you have to do to disprove the collapsing 2/3 between 2 doors claim into one door is to do a version of the problem where everything this explanation is based on is still true but a different answer is arrived at. In my version of the problem each door still has a 1/3 chance of holding the car so therefore if the statement regarding collapsing probabilities was ever true it should still hold. It does not hold so therefore it was never true. -The author abstains from signing posts so as not to endorse ad hominem and appeal to authority fallacies. Let everyone be no more or less than the strength of their arguments. — Preceding unsigned comment added by 69.252.158.32 (talk • contribs)

I hereby propose that this section of the talk page be deleted as any attempts to discuss actual mathematics with 69.252.158.32 will only result in more people getting death threats. -- Antaeus Feldspar 01:15, 29 December 2006 (UTC)

It seems to me that a user violating terms of service wouldn't be a reason to eliminate past work done by someone, even when it was the case. But having followed your link it appears that you have made a false accusation - a straw man. A death threat is when someone threatens to kill you. Not when someone argues that you deserve to be killed. It appears it was this writer's intention to cause you to face the reality that repeated use of passive-aggressive, directly aggressive, or any other tactics designed to overcome reason with force only motivates acts in kind and that it is rediculous to encourage behavior like yours yet prevent other acts of force in retaliation. Although death may seem a bit extreme, it is no less moral than acts of deception and other means of forcing people to act according to your views instead of their own. The only difference is that no amends can be made after it has been carried out, and often the prospect of it is enough to force someone to realize that trying to force their ignorant views on others is not worth the associated costs. -K99

It seems to me that when you made a death threat, Kriminal99 / 69.252.158.32 / 69.180.7.137, you should not have been surprised that people might not choose to treat you as a sane and civil contributor anymore, because you proved instead that you were a very sick and disturbed person. You're still proving your sick and disturbed nature. You're the one who changed the equations in my comments: and then falsely accused me of doing so: , and yet you talk about "acts of deception" as if it was something you stood against instead of something you committed. You talk about "tactics designed to overcome reason with force" as if your death threats were not the most blatant example possible. -- Antaeus Feldspar 06:37, 30 December 2006 (UTC)

I am K99 and noone else, the IP address represent the ideas of my brother and roomate. YOU purposely misrepresented his arguments on several occasions using editing tactics to decieve others. Of course people will use similar tactics as you in self defense. Use of force to overcome reason cannot be defended against by being reasonable. You can only either be reasonable or unreasonable towards someone. If you choose to manipulate the system to overcome superior reasoning you open yourself to all kinds of tactics being used against you in self defense. I agree with my brother that your actions are a disease of Misplaced Pages that should be stopped by any means necessary. -K99

Which is why no one believes that you are a separate person from your supposed "brother", because you hold all the same beliefs as "him" when no one except a psychopath in denial would believe those things in the face of the evidence. No one but you could look at the very clear evidence of you editing my comments and accuse me of "using editing tactics to decieve others". No one but you could look at your vicious "Kill Feldspar" rant -- and your subsequent defense of it -- and not recognize you as the one guilty of "use of force to overcome reason". -- Antaeus Feldspar 16:25, 30 December 2006 (UTC)

Fallacious?

I've copied the section just added to the article below. Before adding this analysis to the article I request a source be provided. Without a source it looks like original research to me (which is not allowed). Furthermore, it looks incorrect (per my comments above). The deleted section follows. -- Rick Block (talk) 04:23, 30 December 2006 (UTC)

I've re-removed the section, per Rick Block's comment. A source needs to be provided for this; if we do include it, it needs some cleanup—I just don't understand the point that it's trying to get across. TenOfAllTrades(talk) 04:57, 30 December 2006 (UTC)

Ricks comments were already clearly addressed above and he failed to provide a response yet acts as though his thoughts were unaddressed. This is clearly passive aggressive behavior. I assume TenOfAllTrades is a friend of his as you reguarly appear wherever he removes other people's work and vice versa. Besides, other people have had no trouble understanding the section. If you honestly have problems understanding the section, perhaps you should be more specific. -K99 —The preceding unsigned comment was added by Kriminal99 (talk • contribs).

My comments were not clearly addressed. This whole thing is not worth arguing about. Please provide a reference substantiating this analysis. If it's not original research then there is a reference you should be able to provide. With 40 or more academic articles about this problem, it shouldn't be difficult to find one supporting your view if it has any validity whatsoever. Thanks. -- Rick Block (talk) 05:34, 30 December 2006 (UTC)

Your comments were clearly addressed, and your refusal to read or respond to the comments which obviously purport to address your question clearly demonstrate that you are attempting to decieve here. "Not worth arguing about" clearly communicates that you do not wish to read the opposing comments to even see if they address your question. The section is supported by the citations ALREADY provided for the rest of the article. Of course you wouldn't be able to see that unless you bothered to read and understand the argument and responses to your questions about it. The only signifigant claim in this section is that it is fallacious to simultaneously hold that there is a 1/3 chance of Monty revealing something and a 0 chance of him revealing the very same thing. It is not a lack of clarity at all that is the issue here, but your belief that you do not have to consider opposing viewpoints as long as you have the capability to manipulate the system to squelch them instead. -K99 —The preceding unsigned comment was added by Kriminal99 (talk • contribs).

I have read your response and fail to see how it addresses my comments. The odds in question are not conditional odds given a known exposed goat. In the problem as stated, the goats are not distinguishable. Any analysis involving distinguishable goats is simply irrelevant and confusing to the point. If you claim this analysis is already supported by the existing citations, please cite a specific one (paragraph or page number would be nice as well). Please sign your comments (or if you refuse to, at least don't delete the "unsigned" tag others add - who makes what comments can be easily determined by looking at the history of the article so you're not accomplishing anything by not signing). -- Rick Block (talk) 05:58, 30 December 2006 (UTC)

What is irrelevant is the fact that an alternate version of the problem is needed to demonstrate the fallaciousness of the current explanation. By your same reasoning, one could provide an explanation of the problem that says "Because the word Monty is used in the problem, there is a 2/3 chance to win by switching". This is obviously quite rediculous, and one could disprove it by changing the host's name in the problem to see that the result is still the same, or keeping the host name the same but changing the truly relevant aspects of the problem. Your response is equivalent to saying that "In this case the name Monty is used, and all other factors are what they are, so therefore the explanation is valid for this problem"-K99

We already know from external experience that it is impossible for there to be a one/third chance of having chosen a goat and that same goat having been revealed by monty, so the current explanation contains a logical contradiction. Additionally everything that the explanation purports to be dependent on can remain the same, and the result different which also shows that the explanation is fallacious. IE there can be a 1/3 chance to choose each goat and the car initially, and only choosing the car result in loss by switching, and yet the probability to win by switching can be different than 2/3. Whether or not the goats are distinguishable is not relevant to the validity of this explanation-K99

I'm an unbiased bystander (not that it should matter on WP) and I have no idea what the deleted section (quoted below) or the relevant discussion is talking about, and yet I have a full understanding of the analyses given in the article. Where, exactly, is the reader required to accept that Monty opened the door that the reader chose? At the outset of the problem, the player can choose either goat 1, goat 2, or the car. Monty can choose, respectively, goat 2, goat 1, or either goat 1 or goat 2, correspondingly. There is no flaw in the analysis that I can see. Perhaps you could explain your issue again in a more grammatically unambiguous manner? Maelin (Talk | Contribs) 13:53, 30 December 2006 (UTC)

It matters because Misplaced Pages has become diseased with small groups of tribal moralist/facist (In the sense that they want their opinions represented regardless of the opinions' logical value simply because it is theirs) editors who have agreements to squelch all opposing beliefs by calling on each other to bypass the 3 revision rule. It is simply human nature that this is unavoidable given the current system, since the ability for people who have come to know each other here to overcome dissenters who know less other editors will result in moral decay such that these people will put less and less effort into understanding others viewpoint's when they "don't really have to" (IE are physically capable of squelching the opinions instead). IMO the only way to avoid this is to prevent people from siding with the same person more than once per a certain time period, though I don't know how such a system might be implemented. -K99

Anyways to get back to the point, here is where the contradiction occurs:

At the point when the player is asked whether to switch, there are three possible situations corresponding to the player's initial choice, each with equal probability (1/3):

The player originally picked the door hiding goat number 1. The game host has shown the other goat.

The player originally picked the door hiding goat number 2. The game host has shown the other goat.

The player originally picked the door hiding the car. The game host has shown either of the two goats.

This proposed sample space is wrong in that it attributes a 1/3 probability of something happening that is already known to have not happened. Note at the top it says "At the point when the player is asked whether to switch" That means Monty has already eliminated one of the above possibilities - yet at the same time it is attributed a 1/3 probability. -K99 —The preceding unsigned comment was added by Kriminal99 (talk • contribs).

The three possible situations are equally likely and mutually exclusive and describe the complete universe of possible states. If we number these possibilities 1-3 (corresponding to the player having chosen goat 1, goat 2 or the car, respectively), if Monty reveals goat 1 we now know that we must be in state 2 (with a 1/3 probability) or state 3 (with a 1/6 probability assuming Monty shows a random goat if the car has been picked). The conditional probability of winnning by switching, given distinguishable goats, in this state, is still 2/3. Similarly, if Monty reveals goat 2 we must be in state 1 (1/3 probability) or state 3 (1/6 probability), and switching wins with a 2/3 probability. Given distinguishable goats, these are the only possible outcomes. This "distinguishable goat" analysis is effectively included (slighly more rigorously) in the section titled "Bayes' theorem". With indistinguishable goats, there is no way to know which of the three states you're in. You've picked a door (in each) and Monty has revealed a goat (in each). You don't know which goat. I believe what you're saying is something like "if I flip a coin but don't show you the result, and ask you to pick whether it's heads or tails, at the point you're picking we can ignore one of the two possible cases because the coin has already been flipped". Since you don't know the outcome, you must consider all possibilities. It might be heads or it might be tails. It's clearly only one. Similarly, when Monty exposes an indistinguishable goat, we must consider all possibilities. If the goats are distinguishable it's a different problem, but still ends up with a 2/3 chance of winning by switching. -- Rick Block (talk) 15:53, 30 December 2006 (UTC)

When you say that "Monty has already eliminated one of the above possibilities", what do you mean? From the player's perspective, all three situations reveal to the player one goat; the player has no information that distinguishes (or rules out) one of those three cases.

There's no assumption that Monty chooses to reveal a specific goat in advance (a choice which may be thwarted by the player's choice, thus ruling out one situation...), nor is it assumed that the goats are distinguishable (rendering a 'numbered goat' solution unnecessarily complicated.)

Incidentally, if you can't compose an answer without the word 'fascist', please don't reply at all. TenOfAllTrades(talk) 15:39, 30 December 2006 (UTC)

Is the above explanation fallacious?

The above explanation may be fallacious in that it requires the reader at a certain point to hold that the contestent had a 1 in 3 chance of choosing the goat that Monty has already revealed that the contestent did not choose. This is in violation of the principles of probability which would have someone change the probabilities of their initial choice with the new information given when the goat is revealed.

In this specific version of the Monty hall problem, the additional information given by the fact that the goat Monty reveals could not have been chosen is exactly offset by the information that Monty reveals that goat (which he is less likely to do when the contestant has chosen the car) Therefore the right answer can be arrived at incorrectly.

When Monty has a known strategy for revealing one of two distinguishable goats it is still true that there is initially a 1/3 chance of choosing each object and only in one of those 3 situations would switching allow you to win. However that has nothing to do with the probability of winning when one switches as this case shows, since the odds are no longer 2 to 3 in favor of switching. And the reason it has nothing to do with it, as mentioned before, is because it is fallacious to simultaneously hold that you had a one third chance of choosing a given goat and that you did not choose said goat because Monty has revealed that you did not.

Fallacious explanations for the outcome in the Monty Hall problem may have been the source of earlier widespread disagreement regarding the problem in which many members of the academic community expressed disagreement. Many people realize there is something wrong with common explanations of it only to find that there is still a 2/3 chance of winning by switching doors using other means of investigation.

Does being able to distinguish the goats make some other strategy optimal?

This is a question that has already been asked and answered but it seems it needs to be addressed again -- if Monty has some particular pattern or method by which he picks a goat to show, and the player knows this and can distinguish between the goats, does it ever make some strategy other than "always switch" an optimal strategy?

The answer is "no". To prove that this is so, we will assume the opposite. We will assume that one of the goats is black and one is white, and that Monty has a bag with B black marbles and W white marbles; when he picks a black marble, he shows the black goat and when he picks a white marble he shows the white goat. B and W may be 0 but they cannot, of course, be negative numbers.

B+W represents the total number of times Monty chooses between the two goats, which he only does when the player initially picks the car. The player's chance of picking the black goat is equal to his chance of picking the white goat is equal to his chance of picking the car; therefore B+W is also the total number of times that the player sees a particular goat because he chose the other goat.

Therefore:

• B+W+B times, the player sees the black goat (B+W times because he chose the white goat, and B times because Monty had a choice and chose to show the black goat)
• B+W+W times, the player sees the white goat (B+W times because he chose the black goat, and W times because Monty had a choice and chose to show the white goat)

Now, let us assume that we can find values for B and W such that seeing a particular goat means the player should stay, and not switch. If there is such a goat, we can devise a strategy in the form "if you see this goat, stay, and otherwise switch" that beats the strategy of "always switch". If however the optimal strategy for each individual goat under all values of B and W is "switch" it means that the optimal strategy for the game is "always switch". So, we assume that for some goat (say the black goat) the player is actually more likely to have picked the car and therefore maximizes his chances by staying, rather than switching.

But here we run into a problem. Out of the B+W+B times the player sees the black goat, B is the number of times the player picked the car and B+W is the number of times the player picked the white goat. For staying to be an optimal strategy, B must be larger than B+W. W would have to be negative to fulfill this condition but we noted at the beginning, W represents a number of marbles and clearly cannot be negative. Even choosing a value of 0 for W only makes it possible to encounter a situation where switching provides no particular advantage, but: 1) no strategy in that situation can make the player's chances greater or less than 50%; 2) for every two times the player faces that situation, he gets a situation where switching pays off 100% of the time.

Can we even pick any values for B and W such that the overall success of a strategy of always switching even becomes something other than 2/3? No, we cannot. What is the proof? The proof is that staying always pays off when the player originally picked the car; switching always pays off when the player originally picked a goat. The ratio of times the player picks the car to times the player picks one of the two goats is constant, and is not affected by how Monty chooses which goat Monty chooses to show in those situations where he has a choice of goats. -- Antaeus Feldspar 02:27, 31 December 2006 (UTC)

Not to confuse matters (I agree with the analysis above), but there are "host choice" protocols for which there are better strategies even with indistinguishable goats. For example, if the host is known to always open the highest numbered door containing a goat the strategy of "initially pick door 1, stay if the host opens door 3, switch if the host opens door 2" always wins. The point that there are many variants of the problem depending on the exact host behavior is well covered in the paper by Mueser and Granberg. -- Rick Block (talk) 03:24, 31 December 2006 (UTC)

Perhaps an easier way to show that (in the standard host choice protocol) being able to distinguish the goats does not alter the result is given in the new "Bayes Theorem" section. Notice, in particular, that no goats appear in the expression for the conditional probability P(Hij | Ck, I). The proposition "Ck = the car is behind door k" and its negation sufice.The Glopk 16:52, 2 February 2007 (UTC)

Graphic Model - Monty Hall Problem

[In an earlier version of this section, I submitted a graphic model, with the idea that there had been an error in the original article. Another user, Rick Block, responded with a comment showing me how I was failing to model the actual question, "Does it improve your odds of getting a car if you switch your initial choice after Monty reveals a goat?" using my proposed graphic style. With his suggestions, the graphic did solve the problem for me. Possibly, his correction shows something about how best to illustrate the Monty Hall question.

There is a similar graphic in the original article (http://upload.wikimedia.org/wikipedia/commons/thumb/9/9e/Monty_tree.svg/350px-Monty_tree.svg.png) that has been through a couple of iterations. Based on Rick Block's corrections, I'd like to suggest another improvement to that graphic, one that might have kept me on the right path. I'll do this if I ever figure out how to insert a copyright information tag. Until then, you could reasonably skip the rest of this section (Graphic Model - Monty Hall Problem). I'm impressed with the wiki process. Etarking 20:00, 2 January 2007 (UTC)

This discussion goes around and around without a definitive end, like an argument over whether you should make clam chowder with milk or tomatoes. I think a graphic of the possible sequences of events puts an end to it, provided that people can agree that the graphic is an accurate model of the system. Here's my proposal:

File:ModelOfEvents-Probabilities.jpg

If this model is right, then its behavior is what you should program if you want to simulate the system. You can also infer the probabilities of each possible event in the Monty Hall sequence, like this:

File:ModelOfEvents.jpg

The probability of each option (joint event) shown in this model is calculated by multiplying the probability of the single event times the probability of the prior event. The probability of a hidden prize is 1.

The process is determined by three events:

1: I choose among the three doors.
2: Monty reveals a goat.

3: I choose one of the remaining two doors (by switching or not switching).

If this model is valid (modelling what it's supposed to model), then it disproves the original, peer-reviewed article. This is a mathematical proof, which is a pretty strong form of proof for this type of problem. Someone has already pointed out that disproof would be an interesting outcome, due to what it reveals about the limits of peer review. It's rare that a peer-reviewed decision can be tested in this way.

On this model, I have a .5 probability of coming out of the game with a car. This probability is determined by summing the probabilities of the sequences in which I get the car (/₆+/₆+/₁₂+/₁₂).

If this model is not valid, then someone should propose a different decomposition of the possible events in Monty's system.

I take it as axiomatic that:

1: The system can be analyzed as a series of three events.

2: The probabilities of the possible options for each event must sum to 1.

If someone does propose another decomposition of the possible sequences, the second requirement is a good logical test of the model.

I think that the model is correct and exhaustive, in which case we should be done with the question of what is right. That would leave us with a charming article that is fundamentally wrong. I can't suggest an equally charming article based on my analysis. I do have some suggestions, but they are inappropriate until commentators reach a consensus. I have been known to make mistakes from time to time.Etarking 03:38, 2 January 2007 (UTC)

The problem is your "I switch" line. The correct probabilities for this line (left to right) are 1/3, 0, 1/3, 0, 0, 1/6, 0, 1/6. With this correction, switching shows a 2/3 probability of winning the car. Isn't this the same analysis already included in the "Decision tree" section of the article? -- Rick Block (talk) 04:28, 2 January 2007 (UTC)

Thanks very much, that clarifies things. I was asking the wrong question. There is actually other discussion archived elsewhere that I missed, that would have straightened me out. Didn't find it -- I'm a newb, sorry. Actually, the first decision tree that I saw in an earlier version of the article (still available at http://en.wikipedia.org/Image:Monty.GIF) was a little different from what shows now. The newer one is definitely better. Adding up the probabilities on the bottom line of that older diagram and getting > 1 is what led me to believe that the author got it wrong. I wrote myself some notes and put the whole thing aside to come back to after about 8 months, I guess. I get no glory, anyway.

My contribution doesn't seem to contribute anything, I'm open to advice on how to remove it. Etarking 05:27, 2 January 2007 (UTC)

You could just remove this whole section if you'd like, or prune it to the comment you added at the top. -- Rick Block (talk) 01:13, 3 January 2007 (UTC)

Forgotten goat section

I've restored the section about the variant recently discussed in Vos Savant's column. There's discussion about this in two separate sections (above). If the host doesn't know or forgets where the car is and opens a door anyway, it makes no difference whether you switch or not and you have a 50/50 chance either way. Please see the sections above for more discussion about this. -- Rick Block (talk) 01:39, 3 January 2007 (UTC)

Nope, the original solution remains. To win, a switcher must still pick a goat on the first guess (2/3). To win, a non-switcher must still pick the car on the first guess (1/3). The host's knowledge changes absolutely nothing and no amount of bad dice analogies or flawed diagrams can change that. The section should be removed before it confuses more people. Tailpig 15:34, 3 January 2007 (UTC)

This is discussed in more detail above, but in the 'forgetful host' variant 50/50 is correct. This assumes that if the forgetful Monty inadvertently opens the door with the car – which will happen 1/3 of the time – the contestant gets a 'do-over'.

So, the outcomes are as follows.

• 1/3 probability. Contestant chooses car. Monty opens one goat door. Switching loses.
• 1/3 probability. Contestant chooses goat (probability 2/3) and Monty opens door with other goat (probability 1/2). Switching wins.
• 1/3 probability. Contestant chooses goat (probability 2/3) and forgetful Monty opens door with car (probability 1/2). We go back and try again until we get one of the first two outcomes.

There you have it. Switching wins and loses with equal likelihood in this 'forgetful Monty' variant. TenOfAllTrades(talk) 15:52, 3 January 2007 (UTC)

Unfortunately that is not what the variant is. You should read the external article at ]. I don't know who first introduced the idea of a 'do-over' but it is never mentioned in the article. In the variant, the only thing that happens is the host forgetting where the car is and correctly picking at random a door with a goat. In this case, switching is still the better choice. The section should be removed before it confuses more people. Tailpig 15:59, 3 January 2007 (UTC)

The "do-over" is not a part of this problem. What matters is that we in this case get some additional information: We are not in one of those cases where the host accidentally reveals the car (as he will in half the cases where the guest chose a goat to begin with). This "prunes" the probability tree, leaving a fifty-fifty chance whether the guest switches or not.--Niels Ø (noe) 16:10, 3 January 2007 (UTC)

Oops, you're right. Even in the more general case – without a do-over – the 50/50 odds work. If one assumes that Monty doesn't hit a door with the car, there's equal odds in favour and against switching. TenOfAllTrades(talk) 16:28, 3 January 2007 (UTC)

Then the variant should be defined as such. From the article, I was under the impression that the host would never accidentally reveal the car. I no longer think that the section should be removed. Tailpig 17:03, 3 January 2007 (UTC)

I apologize if I confused things by introducing the concept of the "do-over" to discussion of this variant. It seemed to be the easiest way to explain a difficult concept (or at least a concept that someone was indeed expressing difficulty with.)

The concept in question is that even though there are certain possibilities which could clearly come about by random chance, these possibilities are shown not to have happened by the evidence given as part of the problem statement. In this, the "forgotten goat" variant Monty Hall problem, there are six combinations of doors that can be chosen between the player and Monty; two of these are combinations where Monty does mistakenly show the car. However, the problem statement says that Monty does not show the car; thus, we are dealing only with the four possibilities that are not contradicted by the problem statement.

This seems to puzzle many people, who are not sure whether it is "legal" to simply 'prune' the probability tree and say that a thing doesn't happen when clearly it could. Many of us learned to work through probability puzzles through thought experiments employing an appropriate randomizer ("OK, if I assign two faces of the die to each of the three doors, and if I get every face of the die once...") However, the forgotten goat variant poses difficulty for those choosing this approach: how do you simulate a randomizer not coming up with results that, quite obviously, it could generate at any time?

The answer is to incorporate the "do-over" into the simulation: if you get a result that is contradicted by the evidence of the problem statement, simply ignore it and keep going until you get a result that isn't contradicted. However, it was only ever intended as an aid to help people run the simulation for themselves and see in a convincing way why the answer is what it is; it was never meant to be implied that the "do-over" was part of the actual problem. -- Antaeus Feldspar 22:15, 3 January 2007 (UTC)

One of the things that I had a hard time swallowing about this is that it's not sensible that the host's knowledge somehow effects the probability distribution. I think it's because the real defining factor is A. that the host always chooses a door with a goat and B. the contestant knows this. In other words, it's whether the contestant knows that the host always chooses a goat that matters, and not what the host knows. This is easy to demonstrate. If you rephrase the problem such that the host's knowledge is not mentioned or explicitly stated as being an unknown, the host's knowledge at the time of choosing is irrelevant because the contestant doesn't know whether the host knew or just randomly chose a door or if he only shows a goat when the correct choice is made. I think if this distinction were made, it would make it easier for people to understand. Dubwai 03:13, 24 January 2007 (UTC)

Except this is wrong. If the contestant doesn't know whether the host knows, the probability of winning by switching is not known (it's either 1/2 or 2/3). It doesn't matter whether the contestant knows, but it does matter whether the host knows. The difference is whether the host is introducing a random selection or not. If the host introduces a random selection, the probability is 1/2. If the host does not introduce a random selection, the probability is 2/3. -- Rick Block (talk) 03:50, 24 January 2007 (UTC)

The way you're expressing this, Dubwai, makes me wonder if you're really understanding the logic of the problem. To make sure we're clear, let's specify that what we are asking is whether there is an optimal strategy under a given set of conditions -- not whether any given contestant could be expected to or would have the knowledge necessary to figure it out. Whether the contestant knows how the host chooses a goat isn't relevant -- but whether the host knows how to always pick a goat or has just happened to pick a goat is essential.

You say "it's not sensible that the host's knowledge somehow effects the probability distribution". You are partially correct. The host's knowledge does not affect the probability distribution. However, the host's state of knowledge does affect the portion of the probability distribution which could possibly have happened.

Suppose that instead of being told "the player picks a door at random; the host picks one of the remaining doors at random; the host's door is opened to reveal a goat", we are told "the player picks a door at random; the host picks one of the remaining doors at random; the host's door is opened to reveal the car". Now, before the host's door is opened, we would have said that the probability distribution gives a 1/3 chance that the car is behind the player's door. When the host's door is opened to show the car, however, the chance that the car is behind the player's door is clearly zero! The probability distribution has not changed, but the portion of the probability distribution that matches the evidence has.

Now let's get back to our original problem: the player picks a door at random, the host picks one of the remaining doors with knowledge of what's behind each door, and opens the picked door to reveal a goat. Does this eliminate anything from our probability distribution? No, it does not: The player has a 1/3 chance of picking the car, and a 2/3 chance of picking a goat; in each of these cases the host can and will choose a goat from one of the remaining two doors. But then let's look at what happens if the host is picking at random like the player: like the player, the host has a 1/3 chance of picking the car. However, when the host's door is opened and shows a goat, we know that this 1/3 chance didn't happen (similar to our previous example, where seeing that the host did choose the car eliminated the 1/3 chance that the player had.) -- Antaeus Feldspar 05:09, 24 January 2007 (UTC)

Problem

But once the host has opened one of the doors, the choice has become out of TWO doors. The opened door can be forgotten. It's as if a new choice has been given: "A door has a car and a door has a goat" CHOOSE! :/ it 's a 50% choice!!! please help —The preceding unsigned comment was added by 194.204.127.214 (talk • contribs).

Have you read the entire article? Have you read some of the other explanations on this talk page? The point is the opened door is not opened randomly. If you do this over and over and stay with your first pick every time you'll win the car about 1/3 of the time (the host might as well not even bother to open a door). If you always switch, the opposite must happen, i.e. you'll win about 2/3 of the time. -- Rick Block (talk) 21:23, 6 January 2007 (UTC)

It would only be "as if a new choice has been given" if the first choice was no longer having any effect on the problem. However, that is not true; the first choice has in fact already determined whether staying or switching is the correct choice for this round.

Consider it this way:

1. The player's choice separates the doors into two sets: the player's set, which has one door and thus a 1/3 chance of containing the car; and Monty's set, which has two doors and thus a 2/3 chance of containing the car.
2. Monty eliminates a door that does not contain a car from Monty's set. This step cannot change which set contains the car; thus the player's set still has a 1/3 chance of containing the car, and Monty's set still has a 2/3 chance of containing the car.

Hopefully it's clearer now why it is in fact a 2/3 choice and not a 50%/50% choice. -- Antaeus Feldspar 01:25, 7 January 2007 (UTC)

One thing that is important to remember is that just because there are two options doesn't mean it's a 50/50 chance. Suppose I hold out my hands, and tell you that in one fist I'm holding a $1 coin but in the other I'm holding nothing, and that you choose one of my hands and have whatever is in it. If I tell you nothing else, it's a 50/50 chance (in your mind). But suppose I then tell you (in honesty) that 90% of the time, I hold the coin in my left hand. That changes the situation for you. It's still a choice between two options, but it's no longer a 50/50 situation, because you have new information. Now you know, the left hand is a better choice.

In the same way, when Monty opens a door, he is giving you information. It's cleverly concealed information (which is what makes the puzzle tricky), but it's information nonetheless. He is telling you, "Of the two doors that you didn't choose, this one definitely had a goat." Now you know something you didn't know before. This changes the situation and it tells you that the remaining closed door that you didn't choose is a better choice, that is, that you should switch. Maelin (Talk | Contribs) 01:44, 7 January 2007 (UTC)

Oooh thanks a lot! :D - markbri

FAR nomination

Monty Hall problem has been nominated for a featured article review. Articles are typically reviewed for two weeks. Please leave your comments and help us to return the article to featured quality. If concerns are not addressed during the review period, articles are moved onto the Featured Article Removal Candidates list for a further period, where editors may declare "Keep" or "Remove" the article from featured status. The instructions for the review process are here. Reviewers' concerns are here. Gzkn 10:57, 7 January 2007 (UTC)

Flash simulation

I wrote a flash simulation of the Monty Hall problem. Is there a way for me to upload it to wikipedia? —The preceding unsigned comment was added by Lax4mike (talk • contribs) 03:22, 9 January 2007 (UTC).

Venn diagrams

These aren't Venn diagrams in the strict modern sense, but I'm not sure what else to call them. Venn might have disagreed. Septentrionalis PMAnderson 04:18, 9 January 2007 (UTC)

Bertrand's Box Paradox

The following text was deleted:

A similar problem appeared in Joseph Bertrand's Calcul des probabilités (1889), known as Bertrand's Box Paradox, which is the . In this you have three boxes, each with two drawers on opposite sides. Each drawer contains a coin; one has a gold coin on both sides, one a silver coin on both sides, and the third gold on one side and silver on the other. You open a drawer and find a gold coin; what is the chance of the other side being silver? <:ref>Michael Clark, Paradoxes from A to Z, p.16; for a comment on the relationship, see Howard Margolis Wason, Monty Hall, and Adverse Defaults <:/ref>

Like Monty, one box is eliminated; you can't have picked the silver/silver box. Likewise, the probability of your having picked the gold/silver box is one-third to begin with, so the chance of a gold coin on the other side is two-thirds

on the grounds that it is not related, except in being a cognitive illusion. I disagree; I must not have made it clear, but I think (and so do several sources, including the paper cited, that it's the same cognitive illusion. The fallacious argument in the Paradox is

• if you find a gold coin, the box must be GG or GS,
• These are equally likely,
• So the chance of a silver coin on the other side must be 50%.

Discuss? Septentrionalis PMAnderson 02:56, 10 January 2007 (UTC)

I removed the mention of Bertrand's Box Paradox from Monty Hall problem#History of the problem. Like the Monty Hall problem, the BBP involves cognitive illusions, but the illusions involved are very different. The cognitive illusion in the BBP is the player not realizing that, while there are two boxes that have not been eliminated, they may be looking at one of three coins, two of which happen to come from the same box. Despite the superficial similarity in the MHP of the player eventually facing two choices, one of which is twice as likely as the other, the reasons why are not the same.

By the way, the header should probably be reduced simply to "History", since "of the problem" is redundant with the subject of the article. -- Antaeus Feldspar 03:05, 10 January 2007 (UTC)

More importantly, I think, there is no indication that the Bertrand thing is really part of the history of the Monty Hall problem, even if it is an older and somewhat similar problem. Thus, I don't think Bertrand belongs in a "History" section (but is of course relevant enough to be in the "See also" section).--Niels Ø (noe) 09:23, 10 January 2007 (UTC)

Repetition

Abscissa brings up a point worth discussing. Much of this article consists of different explanations of the same result, and they really are themselves quite similar.

• Do we need all of them?
  • We may; one reader will understand one, and another reader the next.
• If not, which do we keep?

Septentrionalis PMAnderson 03:36, 10 January 2007 (UTC)

I would advise moving the "Simulation" section upwards, since we have a citation available (Howard Margolis Wason, Monty Hall, and Adverse Defaults, page 17) specifically identifying it as a successful method for inducing understanding -- as specifically contrasted with verbal explanations, actually. "If subjects actually play

the MH game a few times (handling the cards themselves: watching someone else is not very effective), that physical sense of the situation easily switches the intuition. After a few plays where you handle the cards yourself, it will become obvious that it pays to switch. But mere verbal instructions or even merely watching someone else handle the cards will easily fail. So a person in doubt about whether it really pays to switch ought to take a couple of minutes and deal the game a few times." -- Antaeus Feldspar 05:55, 14 January 2007 (UTC)

MontyHallProblemMadeEasy.jpg

Is this picture helpful to anyone? When I first started to understand the problem this hindered more than helped. I understand that some people might have different ways of understanding the solution, but I believe this image is peddling a bad explaination on how to arrive at the solution. This problem has absolutly nothing to do with what is suggested in the picture. I move for its deletion, or at least a recreation which explains the solution better. This is not a case of just having 2 seperate guesses at the correct door (and thus 2/3 probability) 202.10.86.59 20:50, 13 January 2007 (UTC)

• On the contrary: the concept cartoon was fair because '2 seperate guesses' is exactly what Monty gives the contestant. The fact that it's Monty who opens one of the other two doors and not the player is irrelevent, since it's always a losing door. The cartoon does not negate the need for the rigorous analysis that follows; it sits well with the short, superficial explanation near the top of the article. Misplaced Pages pitches itself at the casual reader as well as the academic. Those wedded to the idea of equal probability, but who do not wish to read in depth, will find the cartoon helpful. I believe it should be re-instated; but then again I would say that, as I drew it. StuFifeScotland 21:36, 13 January 2007 (UTC)

• I have no problem with the way of considering the problem presented in the cartoon but I think the presentation is unencyclopedic. Illustrations should be illustrations, not textual analyses presented in a JPEG. If you can think of a good way to present this argument in a properly pictorial form, I will support it. This image, however, does not belong here. Maelin (Talk | Contribs) 23:47, 13 January 2007 (UTC)

• I agree. Presenting this analysis in the form of a JPEG adds nothing to presenting it as text. At the current size, the text is not even readable. -- Rick Block (talk) 23:57, 13 January 2007 (UTC)

• It is right that warnings are issued before the solution to a puzzle is revealed. Hence the text in the concept cartoon should be kept deliberately small so that it will not 'catch the eye' of those who prefer to work out the problem for themselves. It's analogous to printing the answers upside down at the foot of the page. Those who wish to read it will click on the picture, just as they will enlarge thumbnail photos to see more detail. As regards the picture itself: there is a narrow distinction between the opening cartoon, which shows a goat and three doors, and the one with the think bubbles; I believe they sit well together. In one the text is predominantly in the caption, while in the other it is purposefully shown as what a contestant might be thinking during that crucial moment following Monty's offer. Encyclopaedias do not have to be wordy and dry. School and college text books increasingly use this kind of concept cartoon to get over lines of reasoning, because they work. A good Misplaced Pages article, like this one, should cater for all readers, from casual browsers to university academics, and all ages. Blending the approaches together should be the challenge, not aiming for some kind of literary purity at all costs. StuFifeScotland 14:01, 14 January 2007 (UTC)

• • I'm very disappointed that Booyabazooka saw fit to delete the image without first discussing it here. A whole different way of looking at the MHP has been lost. StuFifeScotland 13:25, 12 February 2007 (UTC)

Well, then, since the people who have already spoken above saying "Presenting this analysis in the form of a JPEG adds nothing to presenting it as text" are apparently just not enough for the cartoon's provider, let me add yet another voice to that contingent. Perhaps then the cartoon's provider will be able to grant that yes, other editors have considered the merits of the cartoon and still don't think it's appropriate and -- here's the kicker -- should therefore not be re-added. The cartoon does not in any way shape or form illustrate the problem and therefore should not be in the article in the form of an illustration. This leaves the question of whether the text should be in the article in the form of text. Frankly, I do not think so; it is not nearly as clear as its author seems to think. It is hardly intuitive that "to open both the other doors instead ... exactly the choice Monty is offering"; in fact, I am rather skeptical that it could even be described as true since "is exactly" and "is equivalent to" are two very different things. Now of course we could explain this equivalence, but I think the section "Combining doors" is already doing exactly that, very well. For these reasons, my recommendation is leave the cartoon out.

If I sound frustrated it's because I am. "I can put it back against consensus" does not mean "I should put it back against consensus". -- Antaeus Feldspar 01:20, 21 February 2007 (UTC)

I think the cartoon is excellent. It detracts nothing from the article and instead adds a little bit of humour as well as clever explanation in pictorial form. I can't imagine why anyone would be dead set against it. The arguments offered above to that effect seem to me to be peculiar. To start arguing about identity versus equality when talking about a cartoon is bizarre. And I'm not even from the Kingdom of Fife.Davkal 01:47, 21 February 2007 (UTC)

About the Venn diagrams

Infact, I should have been more general. I do not believe the Venn diagram is an accurate way to represent the problem. In the Venn diagram, doors 2 and 3 are grouped, seemingly because they are physically next to each other. What if door one were next to door three? If there is indeed a sound reason why, I think it should be presented in the article, in the Venn diagram section. There is a 2/3 chance it could be behind doors 2&3, then it must follow that there is a 2/3 chance of it being behind doors 1&3. So then, doesn't the Venn diagram present a choice that is 50/50? Clearly, this does not agree with the actual solution. In conclusion, the Venn diagram seems like an over-simplification, if it is even correct. WIth regards to the picture, any tool that seems to aid understanding is fine with me, however, I don't think the Venn solution is accurate, and therefore the same is said for the picture, in my opinion. 202.10.86.59 05:58, 14 January 2007 (UTC)

The Venn diagram images divide the doors into two sets that have (as sets) the same probability of containing the car both before and after the host opens a door, i.e. the door the player has chosen and those the player hasn't chosen. The analysis and the diagram go together. Certainly other diagrams are possible, for example doors 1 & 3 can be grouped. In this case, before door 3 is opened they have a combined probability of ⅔ (as do any two doors before any doors are opened). However, to say this set of doors still has a ⅔ probability after door 3 is opened would be saying the player's initial choice was both these doors. The player's choice was only door 1. Given a 1&3 grouping, this group now includes a door with a ⅓ probability (door 1) and a door with a 0 probability (door 3), for a combined probability of ⅓. Since the probabilities must sum to 1, this means door 2's probability changes to ⅔. The issue is that this simply isn't a very useful grouping (similarly, grouping doors 1&2 is possible, which together start with a ⅔ probability and end up with a probability of 1). -- Rick Block (talk) 17:09, 14 January 2007 (UTC)

Ok, I don't think I am getting my point clearly across.

"Given a 1&3 grouping, this group now includes a door with a ⅓ probability (door 1) and a door with a 0 probability (door 3), for a combined probability of ⅓."

How is this different from:

"Given a 2&3 grouping, this group now includes a door with a ⅓ probability (door 2) and a door with a 0 probability (door 3), for a combined probability of ⅓."

If the combined probability of 2&3 is 2/3 and the probability of door 3 is 0, it holds that door 2 has a probability of 2/3. Now, this is true, but when you are trying to prove that door 2 has a 2/3 probability, it does little to help.

If I can understand this, I understand how this solution is determined in this fashon. If you do manage to say something that helps me understand, it should probably appear in the article (I will put it there). I am no slouch when it comes to maths and logic, and if it isn't clear to me, I am not too sure it will be clear to the average reader. 202.10.86.59 18:21, 14 January 2007 (UTC)

The text around the diagrams is intended (but seemingly fails) to make this clear. The reasoning is that the probability of the player's initially chosen door doesn't change when door 3 is opened, so the combined probability of doors 2&3 can't change either. This might be more clear with more doors. Let's say there are N doors. The user selects one, which we'll call door 1 and has a 1/N chance of this door being correct. The remaining doors, as a group, have a (N-1)/N chance. If we open one of these, the group still has a (N-1)/N chance now spread over N-2 doors rather than just N-1. As we open more and more of these doors the (N-1)/N chance of the group remains constant, but the number of doors involved grows less and less until there's a single door in this group with a (N-1)/N chance. The key is that the initial choice divides the doors into two groups, the one chosen door and the rest. This is the only division that results in two groups with winning probabilities that don't change as losing doors are opened. -- Rick Block (talk) 21:04, 14 January 2007 (UTC)

You can say that, but that is what I don't understand. How does the probability of a goat being behind the first door NOT change, and the the probabily of it being behind the second does? Stating that the player chose door 1 doesn't help me at all. Let's just say the host reveals the car instead (he stuffed up (it doesn't matter)). There obviously cannot be a car behind door 1, yet, according to you, there is STILL a 1/3 chance. You might think this is different because there are multiple goats, but it really isn't. The 2/3 probability cannot be explained by the Venn diagram in this case. 202.10.86.59 05:46, 15 January 2007 (UTC)

Well, I can concede that the fact a car will never be revealed changes this. Not sure how though... 202.10.86.59 07:47, 15 January 2007 (UTC)

Let's back up. If I know my strategy going in is to not switch, I have a 1/3 chance of picking the car with my initial choice (before any door is opened). The car is not moved when the door is opened, so whether this door (guaranteed not to be the car) is opened or not if I don't switch I'll win 1/3 of the time (right?). This means opening the door does not change the probability of the initially chosen door. The reason is there are 2 unchosen doors, so everybody already knows at least one of them is not the car. The 2 doors have a 2/3 chance between them of hiding the car. At least one is not the car, so if we know which one is not the car the other one must have a 2/3 chance. This is what the diagrams are intended to show. -- Rick Block (talk) 14:51, 15 January 2007 (UTC)

I'm still not convinced. But I will not take up any more of your time. Let the records show that I do not agree :) 202.10.86.59 16:15, 15 January 2007 (UTC)

Do you at least agree if I decide beforehand that I'm not going to switch, I'll win the car 1/3 of the time based on the initial pick being right 1/3 of the time? If so, I think you're agreeing opening the door doesn't change the probability. -- Rick Block (talk) 00:58, 16 January 2007 (UTC)

About the Introduction, and online simulators

Because of the very nature of the paradox, the usual 50:50 response and the amount of debate that it generates, I would like to see more text in the intro on the background leading into the analysis. The analysis is so long and uses so many methods, without fully explicating the controversy that has been generated in the public sphere, that many readers will not bother with the analysis, and will write if off -- some people were prompting to take the page down as incorrect and a near-hoax, for instance. I would suggest using text based on the following:

"When Marilyn vos Savant quoted this puzzle in the US a few years ago , she received over 10,000 letters mostly telling her she was wrong.

One was from Robert Sachs, a professor of mathematics at George Mason University in Fairfax, Va. who said "As a professional mathematician, I'm very concerned with the general public's lack of mathematical skills. Please help by confessing your error and, in the future, being more careful."

However a week later and Dr. Sachs wrote her another letter telling her that "after removing my foot from my mouth I'm now eating humble pie. I vowed as penance to answer all the people who wrote to castigate me. It's been an intense professional embarrassment."

Well, we said it was counter-intuitive. Even professional mathematicians get it wrong." (http://www.grand-illusions.com/monty.htm) That is just a framework suggestion, not suggesting direct plagiarism of the text. Or similar words around what has driven the controversy most recently, being the 'Marilyn vos Savant' article. The so-called Marilyn vos Savant character is presumably a pen name and may represent a number of contributors .

Also, it would be good to include links to online simulators in a separate section, or make reference in the current heading to them. (examples also found at the same site: http://www.grand-illusions.com/monty2.htm) --Sean01 07:37, 15 January 2007 (UTC)

Awesome article

I think this is an excellent article, I like the formal discussion and and also the artwork, even the goofy one at the top. You might want to consider moving the history section up a bit and add to it the above posting (on this talk page) about why this has caused confusion. Maybe a good place would be just before the "understanding aids", especially if the above information is added then the reader will be even more interested to read the excellent analysis. But I mean, even as it currently stands, it's awesome, simply awesome! --Merzul 02:28, 17 January 2007 (UTC)

Bayesian analysis

Is the extended Bayesian analysis recently added from some particular source? If so, can someone please add a reference? The article is currently undergong a featured article review, which includes making sure that everything that should be, is referenced. I fear some folks might view this extended analysis as approaching original research, which is prohibited (see WP:NOR). -- Rick Block (talk) 14:24, 17 January 2007 (UTC)

No particular source, and definitely not original research. It is a very straightforward application of basic probability theorems and concepts (Bayes theorem, marginals and conditionals, normalization). This is the kind of problem I may have given to my students as an exercise after the second lecture of an undergraduate class on probability back when I used to teach. I don't think any references are necessary other than the Misplaced Pages articles already linked to - it's really plain algebra. The Glopk 20:17, 17 January 2007 (UTC)

BTW, I hope that extended does not mean too long. I could have compacted the whole proof in about half the lenght, but I assume that the average reader is somewhat familiar with concepts of probability, but not too well versed in the formalism (especially the Bayesian kind). So I took some care in spelling out every step as clearly and simply as I thought proper. The previous version of this section, which mine replaced entirely, was defective in this respect, I think. But I haven't contributed much to Misplaced Pages yet, so I'd welcome some feedback on this regard.The Glopk 20:32, 17 January 2007 (UTC)

"Straightforward application of basic probability theorems and concepts" sounds arguably like original research (I wouldn't argue this, but given the current level of scrutiny on this article I'm not sure no one else would). My understanding is that this problem is actually used as an example in a variety of probability texts - if you could find it (even as an exercise, with the answer in the back of the book) it would be clearly not original (I know it's not in the probability text I used). I don't mean "too long" by "extended", just that it now is much more formal (and longer) than the previous section. It has a math textbook sort of tone ("we denote ...", "we can state ..." ), which I'm not sure is appropriate. I haven't had a chance to look at any other math-related featured articles to compare the tone. If you're interested in doing this yourself, the articles that are featured-class are listed at Misplaced Pages:Featured articles. These are meant to be the very best Misplaced Pages has to offer, satisfying the criteria listed at Misplaced Pages:Featured article criteria. The Monty Hall problem article was nominated as a featured article in July of 2005 (it looked like this at the time). Since then, it's suffered a bit. IMO, the current FA review is meant to goad the folks who care about it into spending some time fixing up some of the problems that have accumulated in the intervening year and a half (FA review comes with a threat that if the concerns are not addressed the article will be removed from featured status). -- Rick Block (talk) 04:01, 18 January 2007 (UTC)

Done some googling around - easy now that we can look at books. Here is one with an analysis similar to the one I wrote up: . Can we point to Google Books previews from Misplaced Pages references? Here is another web page with a similar one: . And yes, it is given as an exercise to undergraduate students of probability theory - see here: . Notice that it is at page 31, right at the end of the first chapter, because it really is an easy application of first principles and definitions. There isn't much in there, at least from a mathematician's point of view, but it's a cute enough problem, maybe because so many non-mathematicians find it so hard (there's a reason why so many social scientists have tested how people respond to it). There is only one way to do a correct Bayesian analysis of this problem, the difference is only in the symbols, and I tried to write it up in a way that would be at the same time correct, complete and easy to follow for a curious non-mathematician. I checked the July 2005 nominated version, and its Bayes Theorem section is almost identical to the one I replaced. The reason I replaced it is that it wasn't easy to read, even for a specialist like me. The reasoning was quite abbreviated and hard to follow, and it didn't really do justice to the insight that a Bayesian treatment of the problem provides. Besides, I don't see references for it there either. I believe the author had the correct intuition, but just couldn't formulate it well, IMHO. As for the tone, well, I should think that a section with a mathematical title should have some mathematical flavor too - what's the point of having one if the content is mathematically unappealing? ;-) The Glopk 05:49, 18 January 2007 (UTC)

As I feared, there's now an objection to this section (others as well) as being "original research". The Google Books references shouldn't be directly used since it's a "signup" site. Assuming you're on a campus someplace, if you could find one of these references in the library and add a direct citation (e.g. ) that'd be perfect. -- Rick Block (talk) 04:08, 26 January 2007 (UTC)

Done. Also removed two dead references to probability theory textbooks.The Glopk 17:30, 26 January 2007 (UTC)

How to render simple fractions?

I have noted that there has been some unproductive edit skirmishing over how to render simple fractions. I'd like to see us discuss the relevant advantages and disadvantages of our options, rather than just reverting each other without such discussion. There are three main approaches, and I'm just going to list the advantages and disadvantages that I see to each. -- Antaeus Feldspar

• Plain text with only algebraic markup. Example: 1/3 , 2/3
  • Advantages: Renders on everything.
  • Disadvantages: Can get hard to read in complex equations.
• LaTeX markup. Example: <math> \frac{1}{3} </math> (which becomes ${\displaystyle {\frac {1}{3}}}$); <math> \tfrac{2}{3} </math> (which becomes ${\displaystyle {\tfrac {2}{3}}}$)
  • Advantages: Renders more precisely than plain text.
  • Disadvantages: Harder to edit; can cause odd line spacing.
• Special characters. Example: ⅓ , ⅔
  • Advantages: Available to all editors via the box under the save and preview buttons.
  • Disadvantages: Renders poorly on some systems (such as the modern desktop machine I'm using right now, where in the editbox they render so small I couldn't tell what these fractions were if I didn't place them there myself) and not at all on some systems (such as my PDA's web browser.)

As you might be able to guess, the last option is my least favorite, but I think my reasoning (readers should be able to read the math we are presenting) is rather unassailable. Anyone else? -- Antaeus Feldspar 01:42, 20 January 2007 (UTC)

Although I recently made a pass through the article converting to the Unicode special characters (unnumbered option #3), I actually prefer the plain text approach (unnumbered option #1). I'm not fanatical about this, but I think anything involving even remotely unusual markup (or anything that disturbs line spacing) should be avoided. -- Rick Block (talk) 05:00, 20 January 2007 (UTC)

I think the option that renders best for online fractions is using HTML super and subscript codes, like this ⁄₃₇. It is already used in the article itself, see the Venn diagram section. It's a bit hard on the fingers though, but I expect that any longish expression would rather be done not inline, but as a separate equation with LaTeX markup. I don't know how it works on a PDA screen though. The Glopk 06:54, 20 January 2007 (UTC)

Urk. I, uh... I can't get behind that at all. Each of our other three choices directly designates, in some way, what we're showing the reader. Algebraic markup uses the human-readable markup standard in mathematics; the LaTeX markup identifies the intended piece and their intended inter-relations. Even the Unicode solution, if you can identify which Unicode character you're looking at you can look up which fraction it represents. What you're suggesting is that because a superscript and a subscript and a virgule between them happen to resemble a fraction (at least in the current rendering engine!) that we use them to show our fractions? I don't think this is a good idea. -- Antaeus Feldspar 08:11, 20 January 2007 (UTC)

Not sure about other browsers, but In the browser I normally use (Safari) this version disturbs the line spacing as well - which is why I converted from this format to Unicode characters. -- Rick Block (talk) 15:35, 20 January 2007 (UTC)

In a nutshell, our choices seem to be

1. Accept that the fractions will be almost unreadably tiny (if they render at all); or
2. Accept that the line spacing will be affected.

Well, which flavour of ugly do we want? TenOfAllTrades(talk) 16:08, 20 January 2007 (UTC)

Is there some reason you're rejecting

3. Use plain text with algebraic notation, e.g. 1/3

-- Rick Block (talk) 16:26, 20 January 2007 (UTC)

Er...careless inattentiveness? I suppose that there are cosmetic concerns there, too. Still, that might well be the best option. TenOfAllTrades(talk) 17:27, 20 January 2007 (UTC)

The special characters are there on the Edit page for a reason and I think they should be used. In the case of ó, ò, ô, ö, õ, ō, ŏ, ő, and the like, there is no alternative. If there is a problem with viewing ⅓ or ⅔ then the same problem exists with superscripts, such as x³. Should these be shown as x^3? I think not! Perhaps it's time for wikipedia as a whole to look at the font and 'accessibility' issues. The browser on my Palm T|x PDA won't display the site at all, so simple fractions aren't the major issue. StuFifeScotland 14:24, 21 January 2007 (UTC)

I'm sorry, but I do not find "The special characters are there on the Edit page for a reason" to be a convincing argument. That's equivalent to saying "What someone came up with as an option several years ago, without knowledge of our situation, is automatically better for our situation than anything that we can come up with with knowledge of our situation." Unicode characters are a tool in our toolbox; we are completely free to use different tools if we find them more appropriate. Your superscripts analogy isn't very convincing, either, especially as it's factually inaccurate (the browser I mentioned which doesn't render the Unicode fraction characters does render superscripts with no complaint at all.) -- Antaeus Feldspar 00:51, 22 January 2007 (UTC)

Since this article primarily includes only very simple fractions (mostly 1/3 and 2/3), I think the advantages of using plain text which renders everywhere outweigh any disadvantages with this approach. Does anyone violently object to this solution? -- Rick Block (talk) 02:40, 23 January 2007 (UTC)

Given the above discussion, I'm going to BOLDly convert all simple fractions to plain text with algebraic notation. -- Antaeus Feldspar 15:45, 1 February 2007 (UTC)

Remaining FAR concerns

I believe all current specific suggestions from Misplaced Pages:Featured article review/Monty Hall problem have been addressed, except this comment from user:SandyGeorgia:

I'd like to see some adjustment to the lead if possible; the spoiler in the midst of it is distracting, and citations aren't usually provided in the lead (with some exceptions). Would it be possible to summarize the article, per WP:LEAD, without using the direct quotes, the citations, and the spoiler? The lead should be a stand-alone summary of the entire article, without getting into too much detail in any one area. There should be no text in the lead which isn't expanded in the article. Rather than summarizing the article, the lead seems to be introducing the problem (with text that is not included elsewhere in the article).

There's a thread above, #About the Introduction, and online simulators, with a suggestion for the lead as well. I'm not quite sure what to with these comments. The problem statement in the lead is deliberately a quote to deter wordsmithing (which has been a problem). The spoiler is there since the solution is specified. I've recently added a sentence about the response to the 1990 Parade column. I'm considering shortening the lead by deleting everything after the spoiler warning (and the warning), which would include deleting the cartoon about the solution and revising the text following the quoted problem statement, sort of like:

The problem is also called the Monty Hall paradox; it is a veridical paradox in the sense that the solution is counterintuitive. For example, the correct answer offered by vos Savant resulted in approximately 10,000 readers, including several hundred mathematics professors, writing to tell her she was wrong. Despite explanations, simulations, and formal mathematical proofs, the correct answer is met with disbelief by most people.

Thoughts on this? -- Rick Block (talk) 17:21, 20 January 2007 (UTC)

One of the great things about this article is the way it satisfies both the general reader and the academic. The general reader has everything they need to browse at the top of the screen. If they wish to learn about the many ways the problem and its solution can be analysed mathematically/logically, they can read on. Contrast that with the many other wikipedia articles that appear to be cribbed straight from university text books and make no concession to the non-academic reader at all. They pose the question: who is wikipedia's target audience? I would hope the answer is 'everyone!'

Any change that removes the quick-and-easy solution from the intro will put off the general reader and I think that would be unfortunate. A good article should have the 'inverted pyramid' structure of a newspaper story, and that's what this page current has. StuFifeScotland 14:46, 21 January 2007 (UTC)

I've invited user:SandyGeorgia to comment here. Her comments about the lead are quoted above. She clearly wants the lead to not have a "spoiler" warning. Others have expressed a strong preference for including a spoiler warning if the solution is described in the lead. It seems we could:

1. Delete the solution from the lead (as I proposed above).
2. Ignore user:SandyGeorgia's comment.
3. Present the solution in the lead without a spoiler warning.
4. Include the solution in a show/hide box.

Any other possibilities I'm missing? Of these, I actually prefer #3 but I'm interested in other comments. -- Rick Block (talk) 18:38, 21 January 2007 (UTC)

I frankly would rather see the solution deleted from the lead. There are of course many different standards for what should go in the lead and what should be saved for after the lead, but one that I favor is that the lead should be what a programmer calls a "black box" -- one sees its inputs and outputs but not its inner functioning. Describing how it has confused people because of its counter-intuitive solution -- that's input/output. Describing what its counter-intuitive solution is and why -- that's the inner functioning.

That being said, the solution should be at most the second thing after the lead, since if someone is interested in anything else besides the "inputs and outputs" of the problem, it will be the solution. The only reason for making it the second thing after the lead, rather than the first thing, is that the solution will often appear nonsensical or wrong to people if they've misunderstood the exact problem constraints, so there is a strong argument to be made for presenting the unambiguous problem statement first and then spelling out the solution within those constraints. -- Antaeus Feldspar 02:46, 22 January 2007 (UTC)

Sorry for the delay - just returning from travel, trying to catch up - I've printed the article for my bedtime reading <fun, huh?> Any consensus here yet? I'm looking at the printout, and the lead appears visually more attractive and engaging than what was there before, all cut up with the spoiler and the extensive ref info about the quote. I'll read tonight. SandyGeorgia (Talk) 22:10, 23 January 2007 (UTC)

hmmmmm ... most strange. While my printout doesn't contain a spoiler, I see the onscreen article does - don't know if that's a bug or by design. SandyGeorgia (Talk) 22:59, 23 January 2007 (UTC)

Printing is inhibited for the spoiler warning, but only in the monobook skin (it's of class "metadata" defined in MediaWiki:Monobook.css). -- Rick Block (talk) 00:49, 24 January 2007 (UTC)

Suggestion for the lead

In response to further comments from user:SandyGeorgia at Misplaced Pages:Featured article review/Monty Hall problem, I think she's suggesting condensing the lead down to a paragraph or two, perhaps something like:

The Monty Hall problem is a puzzle involving probability, loosely based on the American game show Let's Make a Deal. The name comes from that of the show's host, Monty Hall. The problem is also called the Monty Hall paradox; it is a veridical paradox in the sense that the solution is counterintuitive. For example, when the problem and correct solution offered by Marilyn vos Savant appeared in her Ask Marilyn column in Parade Magazine approximately 10,000 readers, including several hundred mathematics professors, wrote to tell her she was wrong. Some of the controversy was because the Parade magazine statement of the problem was technically ambiguous since it failed to fully specify the host's behavior. However, despite completely unambiguous problem statements, explanations, simulations, and formal mathematical proofs, the correct answer is met with disbelief by many people.

Immediately followed by the section describing the problem (moving the quoted version from Parade and the associated image there), then followed by the section describing the solution (with the cartoon graphic). Anyone have any objections to this approach or alternate suggestions for how to address SandyGeorgia's comments? -- Rick Block (talk) 14:37, 25 January 2007 (UTC)

That section arrangement sounds excellent to me. At first, not describing the problem statement itself in the lead seemed a strange idea. The more I thought about it, however, the more I realized the advantages of describing the problem unambiguously, once, outweigh any strangeness of not having that description in the lead.

I have some suggestions regarding the phrasing in the proposed lead, but those can probably wait. (As well, I'd like to discuss at some point the carefully considered changes I made to the phrasing of the explanation, which were all without exception removed without any discussion. But that too will have to wait for more time and energy.) -- Antaeus Feldspar 15:23, 25 January 2007 (UTC)

Yes, that's the general direction I was heading; the idea was to make the lead a "cleaner" general summary, while introducing the problem more fully, once, in the body. Is there a chance we can get this wrapped up soon so Marskell/Joelr31 won't have to move it down? Maybe someone could post a note to the FAR that a delay is requested, work in progress? SandyGeorgia (Talk) 15:37, 25 January 2007 (UTC)

Aids to understanding

I removed this addition by an IP from aids to understanding - not sure if it's needed or a repeat:

Another way of thinking of this is: If I initially choose the car, he will show either one of the goats, in this case it is best to stay. If I initially choose the first goat, he will show the second goat, it is best to switch. If I initially choose the second goat, he will show the first goat, it is best to switch. Notice that even after he exposes a goat you don't know if your door has a goat or car. Since ⅔ of the time it is best to switch, switch! Thus, simply put, your first choice was likely a goat (a ⅔ chance), so when he shows you where the other goat is, it is in your best interest to choose the other door.

SandyGeorgia (Talk) 04:03, 26 January 2007 (UTC)

I removed this addition showing a connection between two different aids to understanding sections:

Note that the both the prior probabilities ${\displaystyle P(C_{i}|I)\,}$, all equal to 1/3, as well as those conditional probabilities ${\displaystyle P(H_{ij}|C_{k},\,I)}$ which are nonzero are displayed in the above decision tree. The Bayesian analysis which follows at this point is essentially equivalent to the decision tree analysis given previously.

This, and similarly motivated paragraphs are redundant. All proper proofs of the winning strategy are equivalent, but this does not imply that the article should spell out at length the N*(N-1)/2 ways in which N "aids to understanding" relate to each other: it would only add unneeded clutter. Rather, the "aids to understanding" should, IMHO, be left as alternatives to each other - different ways for different minds to arrive to a common conclusion.The Glopk 16:50, 26 January 2007 (UTC)

But... some proofs are more difficult for the average reader to understand than others, and my point was to draw the close parallel between the Bayesian proof and the decision tree proof - since the average reader will find the Bayesian argument more complex and confusing than the decision tree argument. And, I add with a grin, is it necessary to use the word "proper" in "proper proofs"? Finally, you haven't defined what "equivalent" is for proofs, but I can assure you that there are theorems in mathematics that have widely different proofs that many mathematicians would hardly call "equivalent". In the situation at hand, I was trying to point out that, except for wording, the two arguments (proofs) proceed almost identically. I've used both proofs in my own probability classes for years, and students often find the observation about parallel reasoning helpful. Chuck 17:40, 26 January 2007 (UTC)

But I could write a similar argument pointing out the equivalence of the Bayesian argument to the Venn diagram one, or of the latter one to the door combination one, and so on for all the pairs, directly and inversely. Where does it stop? The fact that different people may find the proofs more or less difficult to understand is addressed in this article by having many proofs, in the hope that the reader will "get the clue" from at least one of them, without an expectation that she will understand all of them. Say, the Venn diagram argument, which is often cited ans the most "intuitive", would likely be rather meaningless for a blind person, who may instead find the purely verbal/symbolic Bayesian argument quite easy to follow. That some person would not gain much by knowing that the argument she understands is "equivalent" (in whatever sense of the word you care to choose) to one she can't even conceptualize.The Glopk 18:13, 26 January 2007 (UTC)

Aquarium analogy

I've moved the following text here from the article:

The aquarium analogy

An excellent analogy helping to understand is suggested by Andrea Gennari of Rome, Italy. This is the so called "Aquarium analogy". An aquarium contains 100 mugs of water (the example can be made using different measurement units without affecting the result, in the example a "mug" of water can be a 1 liter mug or a 1000 liter mug). Now assume that the player cannot see what is inside the aquarium i.e. the aquarium is hidden to sight. The aquarium contains a fish. For the sake of simplicity, assume the fish can be contained in a large enough mug of water. As for the 100 doors example, the player is given the possibility of finding the fish by filling a mug with water so that he has to pick up a mug and put it in the aquarium. What is the chance of the fish being in the mug the player has just filled in with water from the aquarium? Clearly the probability is 1 out of 100 or 1%. Now we place the hidden mug aside without showing the content and we progressively fill 98 mugs with water from the same aquarium, this time showing that in each mug there is no fish. Progressively the aquarium is emptied but we take care of never picking the fish when we fill in the 98 mugs. The fish remains in the aquarium. So at the end of the process we have the original first mug filled with water, the content of which is hidden to the player, the 98 mugs filled with water and no fish, the content of which is shown to the player, and the aquarium filled with the equivalent of only one mug of water and always hidden to the player. The player is then given the possibility of switching between the original mug chosen and the aquarium (containing a volume of water equivalent to a mug). What is the most logical action to maximise the probability of finding the fish? Or, in other terms, is it more likely that the fish is in the original one mug chosen at the beginning or in the one mug volume of water contained in the aquarium? From the example it is clear that the fish can only be in one of the two, but it is now intuitive that, while the probability of the fish being in the first original mug picked by the player is 1%, the probability of the fish being in the last remaining hidden volume of one mug (contained in the aquarium) should be 99%, because when the first mug was filled at the beginning, the probability of the fish being in the other 99% of the volume of the aquarium (equivalent to 99 mugs) should have been clearly 99% and none of the 98 mugs "extracted" contained the fish. The example mirrors exactly the 100 doors example. This example helps understanding in that it makes use of volumes and "probability density". The probability density of the fish being in any given volume is the same therefore if we split the volume in two smaller volumes (a "1 mug" volume and a "99" mugs volume) the probability of the fish being in the "99" mugs volume is clearly 99%. When we show the content of 98 out of the 99 mugs volume, what we do is to concentrate the probability of 99% in the remaining one mug volume of the aquarium. The same example can be done with a "three mug" aquarium analogy thus reflecting the original Monty Hall problem.

Despite the fact that the analogy is attributed to "Andrea Gennari of Rome, Italy", it is not cited. We have no idea where this appeared or who Andrea Gennari is. And frankly, even though I understand the Monty Hall problem pretty well, I can't follow this analogy which is supposedly clearer than the existing explanations. -- Antaeus Feldspar 01:13, 1 February 2007 (UTC)

It's actually not a bad analogy if you can work out what it means, but it's a bit unwieldy. Basically, you have a big tank, full with 100 litres of water, and one small goldfish. You blindfold the contestant and give them an opaque 1 litre jug, and tell them to fill it from the tank. You put an opaque lid on it so they can't see whether the fish is inside, and you give it to them to hold. A curtain is then put in front of the tank so the contestant can't see it and the blindfold is removed, and the game host takes a clear 1 litre jug and stands behind the curtain. He fills the jug from the tank, being careful to not get the fish, and each time he holds it up and shows the contestant he has not got the fish, before tipping the jug of water down the drain. Finally, after he has done this 98 times, all that is left in the tank is 1 litre of water (and, with 99% likelihood, the fish), and the contestant is given the choice of taking the lid off his opaque 1 litre jug and having what's in it, or instead taking whatever is still in the tank. Maelin (Talk | Contribs) 23:07, 1 February 2007 (UTC)

Hmmm. With that explanation, I at least understand the analogy. I have to confess I don't see how it's any clearer, though. -- Antaeus Feldspar 01:06, 2 February 2007 (UTC)

It's clearer because it is easier to see the probability transfer onto the remaining options. When the host opens the unchosen doors, it's not obvious that the probability on them is transferred only to the other unchosen doors and not to the chosen door. When the host is tipping out water that was only in the aquarium, however, this becomes more clear - it's obvious that the fish isn't becoming more likely to be in your jug just because the host is emptying the tank. Note that I'm not necessarily arguing for putting this explanation in the article, I'm just trying to explain how it works. Maelin (Talk | Contribs) 03:05, 2 February 2007 (UTC)

In fact, this analogy is perfectly equivalent to the "Increasing the number of doors" argument: just replace "fishtank" with "set of all doors", "fish" with "car" and "The game host then opens 98 of the other doors revealing 98 goats" with "The host fills 98 jugs of water, showing that none contains a fish". Apart for the lack of attribution, the best argument for not adding this section to the article is then redundancy.The Glopk 06:23, 2 February 2007 (UTC)

I found this analogy to be perfect in term of intuitiveness. While the "Increasing the number of doors" argument is already very good but still may appear unconvincing for some, the Aquarium is more intuitive in that it makes the "transfer" to probability clearer: this is due to this analogy being a "continuous" rather than "discrete" example. By all means you can think about removing contiunuous quantities of water thus "shrinking" the "volume" of same probability of 99/100 to smaller volumes , i.e. increasing the probability density associated to the volume. —The preceding unsigned comment was added by Mdimauro (talk • contribs) 09:23, 2 February 2007 (UTC).

There are two things to consider, possibly conflicting:

1. How are thing presented most clearly? (I think the aquarium analogy is more helpful than the 100 door version! We might replace "100 doors" by "aquarium"; having both is overkill.)
2. What can we do without getting into original research? (Are there any references using the aquarium analogy?) —The preceding unsigned comment was added by Noe (talk • contribs) 10:25, 2 February 2007 (UTC).

Continuos, shmontinuous :-) Are we syphoning the water off with a straw now? C'mon, let's not make the explanation more complicate than the problem. The Monty Hall problem is treatable entirely in discrete terms, i.e. using only a finite (and small) number of propositions and their probabilities. I don't see any pedagogical advantage in doing this naive extension to the continuous, particularly since the mathematical concepts underlying it are actually way more complex - discontinuous CDF's, generalized functions for pdf's, marginalization via Lebesgue integration, etc.The Glopk 15:51, 2 February 2007 (UTC)

I don't know what Lebesgue integration is but I do think using a continuous quantity obscures rather than illuminates the problem. Using discrete units-that-might-hold-the-desirable is simple: one such unit can either hold, or not hold, one such desirable. Using a continuous quantity introduces all sorts of unneeded complications: if we remove half a liter from the tank, is there a chance that we've removed half a fish? What if we remove a volume of water from the tank which is less than the volume taken up by the fish? Those of us who already understand the principle behind the problem might find these answers easy to figure out, but I don't think those who do have trouble when we're dealing with integral numbers of prizes and decoys are more likely to grasp the principle from examples involving continuous quantities. -- Antaeus Feldspar 16:30, 2 February 2007 (UTC)

Small Fractions Again

I undid change 104755257 by user David Eppstein, who had changed some of the "inline" fractions in the equations of the Bayes Theorem sections into LaTeX \frac{}{} forms. The result of this change was a marked inconsistency in the equations that looked visually unpleasant. In particular, the multi-line formula evaluating the normalizing constant mixed fractions of the two types, probably because having the final result in a \frac form made for a confusing read next to the previous line.. Also, the line spacing on my monitor became quite "jumpy". Something needs to be done to improve the rendering of math in Wikipoedia, but until it is I think we should edit conservatively....The Glopk 15:24, 1 February 2007 (UTC)

I can't believe that you think "1/6/1/2" is an appropriate way to render math. If you want the vertical fractions bigger, throw \displaystyle into the formula. But I'm not repeating my edit, someone else will have to do it. —David Eppstein 16:36, 1 February 2007 (UTC)

I can't believe your eyes could tolerate the inconsistency in line spacing your edit caused. If you change the style of the formulae, be at least consistent and change them throughout. Or rather, don't, I already did the work for you. For the next time, please consider cooling your kindergarten-style antics before crash-landing on your keyboard. You know, everlasting records, embarassments and all that.The Glopk 17:12, 1 February 2007 (UTC)

Deliberately avoiding getting into an edit war is kindergarten-style antics? *rolls eyes* Anyway, thanks for the fixup. —David Eppstein 17:40, 1 February 2007 (UTC)

Lead paragraph

The lead paragraph should state the problem in a simple and low-drama way. In particular, I think that the mention to any elements of "Let's Make a Deal" should be minimized because this is not about a game show, it is about a logic problem. We should stick to the correct problem statement, the answer, and then mention that some find it non-intuitive. Unless somebody is going to have a heart attack over this, I will attempt to do so.--199.33.32.40 00:17, 2 February 2007 (UTC)

We just completed a major revision of the article in which one of the changes we decided on was moving the problem statement and the answer out of the lead. I would highly advise against deciding, on your own, to reverse that move. Similarly, I cannot see what would be gained by minimizing any mention of "Let's Make A Deal"; the mention is already as minimal as we can get without removing it entirely and it seems to me that if we remove that because "this is not about a game show, it is about a logic problem" then we might as well remove from the lead of Albert Einstein the two words indicating that he was born in Germany because the article isn't "about" how he was born in Germany. -- Antaeus Feldspar 01:42, 2 February 2007 (UTC)

Well, I see that you waited only 45 minutes to see if "somebody is going to have a heart attack over this" before going right ahead. I also see that you didn't care much that the only comment you did get in the short amount of time you waited was against your proposed changes, not for them. Quite frankly, this isn't the kind of cavalier attitude you should take to editing Misplaced Pages, and also quite frankly, edit summaries such as "The problem is not that to state in the lead Section. State it. Correctly. Misplaced Pages is not a game show. It is an encyclopedia." are condescending and will not contribute to productive discussion. -- Antaeus Feldspar 01:42, 2 February 2007 (UTC)

Sorry Anteaus. It is not my intent to present the "game show version" of the problem in the lead section except at a historical curiousity. It is my intent to present the adult "more than three doors" version of the problem in the lead paragraph, and without the doors, the klieg lights, pretty girls, prizes etc. This FA is under the category of Math. Cold, unforgiving, brutal Math. I know it is less entertaining that way, but that is the subject of the article. Misplaced Pages is not a game show. I present the clear, bone-crushingly obvious and boring version of the problem first. The subject is not Monty Hall. The subject is the Math (more specifically Logic) Problem. --199.33.32.40 01:15, 2 February 2007 (UTC)

Anteaus, much of the fighting and confusion of this article is due to the treacle the project puts out via the mentioning in the lead section of females like Marilyn vos Savant and her ludicrous errors in attempting to even address this problem. Here you guys are coming to the consensus that some pretty face is more important than the subject of the article. The project is supposed to be educational, not just a bunch of pot stirring. Consensus is all fine and lovely but Math is unforgiving enough as it is without that female muddying the waters deliberately, mistakenly or simply for the publicity. She barely rates making it into the Trivia section of this article.--199.33.32.40 01:25, 2 February 2007 (UTC)

Your condescension is really doing nothing for your case. Accusing your fellow editors of placing "some pretty face" above everything else is quite insulting. Meanwhile, the irony of you claiming to present "the adult ... version of the problem ... without the doors, the klieg lights, pretty girls, prizes etc." when in fact your proposed version of the lead incorporates more detail about the game show than the previous lead does not give us any reason to think that your complaints about the article are the results of an unbiased, acute appraisal. -- Antaeus Feldspar 01:42, 2 February 2007 (UTC)

This article is a disappointment. Marilyn got it wrong and within this article she should be relegated to some obscure footnote. Based on the dramatic header you have at the top of this talk page, it is no wonder that you fail to educate some readers about the true nature of the subject. --199.33.32.40 01:49, 2 February 2007 (UTC)

If you want to propose a change, please feel free. If you'd like, we could analyze the lead you added, here on the talk page (IMO, it has quite a few problems). As Anteaus mentions, removing the problem (and solution) from the lead was a recent change in response to a Misplaced Pages:Featured article review. All articles are subject to improvement, but making significant changes without prior discussion (and particularly despite unaddressed objections) is not how things are done here. -- Rick Block (talk) 02:46, 2 February 2007 (UTC)

I disagree with 199.33.32.40's apparent conviction that the article should be only about the problem and its solution and not about its history. This is an encyclopedia article, not a maths textbook. We are here to present all notable information pertaining to the problem, and its cultural significance is an important part of that. Maelin (Talk | Contribs) 03:13, 2 February 2007 (UTC)

Problem statement

That some female who did little but cause more confusion gets into the lead section while the problem statement itself does not might be entertaining, but it is not educational. Does this really belong in the "Media" group rather than the "Math" group of FA articles? Well, if it does not belong in the "Math" group, then here is what was tried as the problem statement in the lead section:

The problem presents a guessing game with a finite number, N, of possible discrete responses and only one right answer, such as guessing a number between one and ten where the problem poser has secretly selected one such number as the correct answer. The guesser is forced to make the guess at random. The poser then must then inform the gueser of N-2 wrong responses, which the poser does, and then the poser offers to the guesser the option to switch to the remaining response or stick with the orignal guess. The guesser's best approach is to switch because the guesser, in having had N-2 wrong answer(s) eliminated, has learned more information than what was presented at the beginning of the game.

At least this is about Math rather than about some entertaining "savant". BTW: Once a decent, subject-oriented version of this article is created, I strongly suspect that it will become more obvious that it deserves the "Low" importance rating that it gets from the Math WikiProject and that it is an unworthy subject of a "Math" FA. I actually want this to remain in Math and get permanantly demoted. --64.9.233.132 03:15, 2 February 2007 (UTC)

This is not a statement of the "Monty Hall problem", but a generalized version of it (which is discussed in the section "Increasing the number of doors"). As the article is primarily about the Monty Hall problem, posing this variant in the lead is not appropriate. FA has nothing to do with "mathematical significance". -- Rick Block (talk) 03:28, 2 February 2007 (UTC)

Look, it is a lot of fun to watch some contestants get it wrong we when insist that N=3, but it is ludicrous to suggest that we avoid a clear statement of the problem right up front. Again, if this is about Media, then let's just have whoever maintains Misplaced Pages:Featured articles move the aricle from "Math" to "Media". --64.9.233.132 04:13, 2 February 2007 (UTC)

Anyone else feel it's become obvious that 199.33.32.40/64.9.233.132/Farever is just trolling? -- Antaeus Feldspar 04:18, 2 February 2007 (UTC)

I am trying to improve the quality of the FA "Math" bin. Evicting this article one way or the other for the mid-to-long term would be an improvement to that quality, at least as far as anyone with a college education would be concerned.--64.9.233.132 04:34, 2 February 2007 (UTC)

I don't understand your view that simply being in the Math category means it must be only about math. Can you explain what, exactly, you want changed about the article and why you think this will improve the article for readers? Maelin (Talk | Contribs) 05:03, 2 February 2007 (UTC)

Re the first sentence of the proposed replacement problem statement: "The problem presents a guessing game with a finite number, N, of possible discrete responses..." It is not usually the case that starting a math article with the most formal, precise, and mathematically general statement of the problem is a good way to make the article readable. You'll lose readers on "finite number", "N", and "discrete responses", which make the statement unnecessarily technical compared to "three doors". The generalized version is presented later, to be sure, but it's the wrong way to start. And, like Maelin above, I fail to see how eliminating the story of how the problem came to be famous is an improvement; there isn't and shouldn't be a rule that mathematical articles must be purified of all non-mathematical content that might make them more interesting. —David Eppstein 05:13, 2 February 2007 (UTC)

This is still showmanship. Before you know it, we will be putting {{spoiler}} on the section with the explanation.--68.127.162.111 04:31, 4 February 2007 (UTC)

68:You are too late. They already use {{solution}}. Before you know it, we will have articles with puzzles like "What does mc2 equal?" Then there will be a big, lecturing, knowing display and, ta da! "It equals E!" Much applause.--64.9.237.152 04:29, 5 February 2007 (UTC)

Please stop having your sockpuppets talk to each other. -- Antaeus Feldspar 13:59, 6 February 2007 (UTC)

Uncited References

The References section contains entries that are not cited anywhere in the article body. One was added today, presumably by its author. What is the policy or custom about such references? Leave them alone? In academic papers (at least in my field), having such "dangling" references is highly frowned upon.The Glopk 17:19, 3 February 2007 (UTC)

I removed it. IMO, adding it was a pretty clear violation of WP:NOT#SOAPBOX. -- Rick Block (talk) 18:25, 3 February 2007 (UTC)

What A Pity

I've been away for a couple of weeks. In the intervening period, this article seems to have morphed into a dusty old text book. That's very sad, given that Misplaced Pages has the opportunity to be so accessible and so different. I say: delete things that are wrong, not things that are simply not to your liking. De gustibus non est disputandum, after all! StuFifeScotland 13:18, 12 February 2007 (UTC)

Ehr, could you make some specific examples of "dusty text book"-style text in the article?The Glopk 17:40, 12 February 2007 (UTC)

Cartoon image

Can we reach a consensus for whether the cartoon image should be in the article or not? The reasons I prefer it not to be in the article are listed below. Please, let's not vote about this, but discuss (and respond to) reasons for or against. -- Rick Block (talk) 02:07, 21 February 2007 (UTC)

And, please, less heat. I think we're all reasonable people here. There's clearly divided opinion on this. I'm not sure there's a compromise available, but let's try to stay polite (I'm not accusing anyone of being impolite - just a reminder, argue to the point, not to the person). -- Rick Block (talk) 05:04, 21 February 2007 (UTC)

Reasons for

1. It's a nice pictorial explanation of a sound point re MHP.

   It is not pictorial. It's a textual analysis in a JPEG. Maelin (Talk | Contribs) 03:30, 21 February 2007 (UTC)

   Short response: Balls! Long response: It is a pictorial representation of a) a puzzled man (that's the little thing in the corner with the eyes); and b) his thought process (this is signified by the positioning of the various thought bubbles - i.e, first he thinks this, then he thinks that, then he thinks, hang on, what about this). A textual analysis in a JPEG would simply be text into a jpeg and nothing else - this is not the same thing at all.

   It is pictorial, and it is an explanation. But it is not a pictorial explanation. -- Antaeus Feldspar 04:30, 21 February 2007 (UTC)

2. It adds a little bit of variety to the article.
3. You (I) can read what it says just fine.

   This is not a reason for inclusion. Legibility is a requirement, not a benefit. Maelin (Talk | Contribs) 03:30, 21 February 2007 (UTC)

   This was a response to the argument that you can't read it in the against list.Davkal 04:05, 21 February 2007 (UTC)

   So, as Maelin said, it is not a reason for inclusion. -- Antaeus Feldspar 04:30, 21 February 2007 (UTC)

   Can we agree that this is not a reason for, and should be considered in the context of the alleged reason against (below)? If so, can whoever added this "reason for" indicate it's a concern that is moot (or had been addressed) by striking out the comment (surrounding the original reason with <s> </s>)? -- Rick Block (talk) 05:27, 21 February 2007 (UTC)

4. It does no harm at all.

   This is entirely debatable. It is also not a reason for inclusion. In fact, it is basically taking the whole set of Reasons Against and saying, without explanation, "no it doesn't". Maelin (Talk | Contribs) 03:30, 21 February 2007 (UTC)

   If you read the arguments against you will see that none of them in the list say it does any harm so it is simply not a case of taking the arguments against and saying "no it doesn't". And, it is a perfectly reasonable reason against deletion (i.e. for keeping), as in, why remove it, it's not doing any harm..

5. Cartoons/illustrations are not necessarily meant (probably not meant at all) to make new points, but to illutstrate points already made. This it does in a way that augments rather than detracts.Davkal 03:21, 21 February 2007 (UTC)

   It is debatable whether it augments the article. Either way, it is not a reason for inclusion. Maelin (Talk | Contribs) 03:30, 21 February 2007 (UTC)

   Augment, as far as I understand the term, means, amongst other things, to add to or to strengthen. I can think of no better reason for including something than it adds to and strengthens the article! So I don't really understand what you mean when you say, either way it's not a reason for inclusion.Davkal 04:05, 21 February 2007 (UTC)

   Because you are not presenting any argument for why we should believe your conclusion ("the cartoon augments rather than detracts"). What you are doing instead is presenting a statement that is not really related ("Cartoons to illustrate points already made", but this cartoon does not illustrate any point) and then just repeating your conclusion again as if you'd said something that supported it. -- Antaeus Feldspar 04:30, 21 February 2007 (UTC)

The above are arguments as to why it augments the article. It adds variety, it is a nice pictorial represenation of a sound point along with a humourous way of showing the puzzlement that can be caused by the MHP.Davkal 04:37, 21 February 2007 (UTC)

Reasons against

1. It is inherently text, not graphical, which means it can't be used directly by other language versions.

   Nonsense, it is clearly graphical. See the little face, see the thought bubbles, see the order of the thought bubbles...Davkal 04:41, 21 February 2007 (UTC)

   Try reading the entire sentence. It can't be used in other language versions, and the meaning of the image is conveyed almost entirely through the text. Putting a little face in one corner of a body of text and then wrapping the text in a speech bubble does not turn a textual explanation into an illustration. Maelin (Talk | Contribs) 04:53, 21 February 2007 (UTC)

   Ah, yes. So someone who doesn't read English would see the cartoon, see the puzzled man, see that he has one thought bubble, see the next thought bubble conjoined directly under it, and see the third thought bubble under that, and then jump up from his chair yelling "Gruss Gott! Ich understanden den Monty Hall Problem now!" -- Antaeus Feldspar 04:55, 21 February 2007 (UTC)

2. It doesn't really say anything that can't be (indeed, isn't) said as text in the article itself.

   It's not necessarily supposed to, but it illustrates one of the points in a novel way, as such it adds to the article..Davkal 04:40, 21 February 2007 (UTC)

   No it doesn't. The point is already demonstrated in section 2.4 Combining doors. If you think it could be made clearer, work on that section. Maelin (Talk | Contribs) 04:53, 21 February 2007 (UTC)

   It illustrates nothing. There is not a single point that is illustrated by the cartoon. -- Antaeus Feldspar 04:55, 21 February 2007 (UTC)

3. The text itself can't be edited without editing the graphic (maybe some might view this as an advantage, but editing it requires recovering the exact font that was used to create it in the first place).

   Easy peasy. And why would you want to - what would you like the little man to be thinking.Davkal 04:40, 21 February 2007 (UTC)

   Text is kept on the encyclopedia in text form for the fact that it is easy to edit. This is "the encyclopedia that anyone can edit", not "the encyclopedia that anybody with a graphics manipulation program, access to a wide selection of fonts, and good photoshopping skills can edit". If the image exists only for the text within it, then it should just be stored as text. This will also cut down on server HDD space required. Maelin (Talk | Contribs) 04:53, 21 February 2007 (UTC)

4. Without clicking on the image, the text in the image is too small to be readable which means in a printed copy of the article the text will be too small to read.

   No its not, it can easily be read as is.Davkal 04:40, 21 February 2007 (UTC)

   No. You see, with text, users can use their browser text size functionality to scale the text up until it's appropriately sized for their monitor resolution and vision. Since the image is primarily text, somebody who is unable to read it for whatever reason can't get any meaning from it, which they COULD if the same textual analysis was presented just in plaintext. Maelin (Talk | Contribs) 04:53, 21 February 2007 (UTC)

You're obviously just having a lugh. When you're ready to discuss it sensibly I'll discuss it with you. Davkal 05:10, 21 February 2007 (UTC)

I'm not sure who this comment is addressing, but I don't see anyone here who is commenting in bad faith. -- Rick Block (talk) 05:36, 21 February 2007 (UTC)

Both of you. Unless you really don't know what a picture is, then the only assumption left is that you're having a laugh. What else have we had, oooh the poor Germans, oooh the poor server, oooh what about the poor people with poor eyesight who don't know how to click on an image. What next? If these are supposed to be serious good faith edits then God help you.Davkal 05:47, 21 February 2007 (UTC)

Actually, it's your edits that are getting impossible to believe are serious good-faith edits. "See the little face, see the thought bubbles, see the order of the thought bubbles..."? That is supposed to be a good-faith response? -- Antaeus Feldspar 06:14, 21 February 2007 (UTC)

Suggested replacement

I'm all for having a quick little synopsis of the solution offset from the main text. However, many good points are made above against the image presentation. Personally, I don't like the image's text itself. The explanation poor, the logic unsound, and in its current format it can't be iteratively revised wiki-style. How about using a simple colored box? See demo to the right. (My colors are hideous, but you get the point...) ~ Booya 05:18, 21 February 2007 (UTC)

With a bit of clever layout, we can even keep clipart-guy in there. Set him above or below the box, with his thought bubble alpha-transparency-fading towards the box text. Although, I have an inquiry - how are PowerPoint clipart images licensed? Does the uploader have a right to release such a derivative work into public domain? ~ Booya 05:32, 21 February 2007 (UTC)

• Misplaced Pages featured articles
• Featured articles that have appeared on the main page
• Featured articles that have appeared on the main page once
• Old requests for peer review
• NA-Class mathematics pages
• NA-priority mathematics pages