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Revision as of 19:44, 20 September 2017 editDa Vinci Nanjing (talk | contribs)Extended confirmed users2,045 edits Oknazevad reverted my edit instead of showing inconsistencies in my explanation.← Previous edit Revision as of 22:34, 20 September 2017 edit undoOknazevad (talk | contribs)Extended confirmed users106,317 edits →Average dice throwNext edit →
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] (]) 19:44, 20 September 2017 (UTC) ] (]) 19:44, 20 September 2017 (UTC)

:To be honest, I am having difficulty following your argument because of your grammar. Regardless, as I said in my edit summary, the sentence is about the name of the variant, which derives not from looking at the outcome of rolls, regardless of how many, but merely the average of the values of the faces, merely that (1+2+3+4+5+6)/6 is the same as (2+3+3+4+4+5)/6. There's no statistical probability involved, so any mention of expected outcomes is irrelevant. ] (]) 22:34, 20 September 2017 (UTC)

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There aren't mentions about De arte aleae Dawid2009 (talk) 13:28, 26 March 2017 (UTC)

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Average dice throw

Section 5.1.2 of the article states:

>>> A variation on the standard die is known as the "average" die. These are six-sided dice with sides numbered 2, 3, 3, 4, 4, 5, which results in the same average result as a standard die (3.5 for a single die, 7 for a pair of dice), but have a narrower range of possible values (2 through 5 for one, 4 through 10 for a pair). They are used in some table-top wargames, where a narrower range of numbers is required. <<<

To roll a dice is a random event/process and in statistics, these events are not called average result, expected value would be the suitable/proper team. In short, to call it average result would be wrong. — Preceding unsigned comment added by Da Vinci Nanjing (talk • contribs) 14:26, 17 September 2017 (UTC)

That's not what the use means. It's not talking about likeliness of outcomes, which, but the actual average (arithmetic mean) of all the possible sums. The use of the technical term needs better explanation. oknazevad (talk) 14:59, 17 September 2017 (UTC)

Hello Oknazevad,

I don't agree.

The paragraph it not referring to the likelihood of outcomes. The likelihood for each number for throwing a dice is 1/n. For a dice n=6. So the probability is 1/6.

To roll a dice is in probability theory an event described by a uniform distribution. To find an average of a UD you have to sum up all possible outcomes and divide this by the count of outcomes:

n=6

1+2+3+4+5+6=21

21/6= 3.5

This average what you get is called expected value.

To get the expected value of throwing a dice in real you would have to roll the dice an infinite amount of throws than calculate the average of all throws.

It doesn't matter if a 6 sided dice has 1-6 on its sides or 2,3,3,4,4,5. The expected value for both variants of dices is 3.5.

Further, you state, the use of this technical term needs better explanation:

This means you agree it is the proper term, but it's too difficult to understand? If the term expected value needs a better explanation what about to change the term into a hyperlink referring to an online encyclopedia (e.g. Misplaced Pages) explaining it? Da Vinci Nanjing (talk) 14:06, 18 September 2017 (UTC)

to Oknazevad:

I got to know this stuff in Quantitative Methods - Statistics at Koblenz University of Applied Science (Germany) the Faculty of Business and Management. Thanks to Prof. Dr. G.S.

Da Vinci Nanjing (talk) 18:36, 19 September 2017 (UTC)

Hello Oknazevad,

you did not make any comment to my explanation: You did not write, I am wrong or I am right or somewhere in between. Your edit summary says:

No, the sum of the possible outcome values divided by the number of possible outcomes is an arithmetic mean, not an expected value. That's the point of the passage.

That's wrong, why? Look at my next-to-last post in the Talk section related to the article. Referring to your explanation, it all comes down to a random event. Additionally, you reverted my edit instead of doing a new one will deteriorate some of my edit statistics. It seems like, you try to tease me on purpose.

If you got the courage to challenge me, defuse my arguments.

Let's assume you throw a 6 sided dice: You could calculate the expected value before you throw the dice as every outcome got the same probability. You could calculate the average e.g. after you throw the dice 9 times:

2 4 3 2 3 3 6 2 3

Generated with the randbetween function from MS Excel 2013 The sum is 28 / 9 = 3.111111... The 3.111111... is the average. The 3.5 is the expected value, even if your average differentiates, you knew this figure even before you threw the dice. The paragraph doesn't state how often you would throw and the outcome so it is referring to the expected value.

Hereby I provide a source to calculate the expected value from a dice throw:
http://www.mathwords.com/e/expected_value.htm

Da Vinci Nanjing (talk) 19:44, 20 September 2017 (UTC)

To be honest, I am having difficulty following your argument because of your grammar. Regardless, as I said in my edit summary, the sentence is about the name of the variant, which derives not from looking at the outcome of rolls, regardless of how many, but merely the average of the values of the faces, merely that (1+2+3+4+5+6)/6 is the same as (2+3+3+4+4+5)/6. There's no statistical probability involved, so any mention of expected outcomes is irrelevant. oknazevad (talk) 22:34, 20 September 2017 (UTC)
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