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Timeline for Which game is more advantageous?

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Nov 21, 2019 at 16:29 comment added Sextus Empiricus @whuber I agree, it does not make a lot of sense. However the post does contain a format like "While the expected Value of B is higher than A, the probability ....". So, by using the word 'while' the OP explains which are the two different points of view. The expected value on the one hand and the probability on the other hand. But it is indeed not very clear. It is not made clear why the expected value - of the number of dice with a "6" - matters (so the post sort of indicates that the problem lies in that particulas aspect, although not very clear how exactly).
Nov 21, 2019 at 15:10 comment added whuber @Sextus The post did not indicate how one might use the moments. It's difficult to see how they are even relevant to the question.
Nov 21, 2019 at 4:02 comment added fectin @BruceET well, it fits perfectly into a Bernoulli distribution, so that would be my guess…
Nov 21, 2019 at 1:28 history edited Alexis CC BY-SA 4.0
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Nov 20, 2019 at 23:54 history edited Alexis CC BY-SA 4.0
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Nov 20, 2019 at 23:41 vote accept Anil K.
Nov 20, 2019 at 23:19 comment added Sextus Empiricus @whuber the other one is "I calculated the first moments which are 4/6 for game A and 8/6 for game B." So the contrast (between two solutions) is in game B getting a higher mean number of dice rolls with a six, but on the other hand the probability to win is lower. (the trick in this difference that for game B we both raise the bar to win and the mean number of dics rolls by a factor two, but... at the same time the distribution get's more narrow.)
Nov 20, 2019 at 23:11 comment added BruceET Winning seems clearly defined for each variant of the game. Those rules for winning have to do with probabilities of various outcomes, not directly related to the binomial means. // I seem to recall this is a variant of one of the early betting games solved by probability pioneers (Bernoulli, Laplace, etc.) but I can't immediately find an exact reference, does anyone recall that?
Nov 20, 2019 at 22:54 answer added BruceET timeline score: 3
Nov 20, 2019 at 20:51 comment added MSIS Ultimately, without clear criteria the question cannot be reasonably answered that I can tell. Or you can give conditioned answers: Under conditions X,Y,Z , game A is advantageous, under other conditions, B is more advantageous.
S Nov 20, 2019 at 20:50 history suggested Yohanes Alfredo CC BY-SA 4.0
Changed formating a bit
Nov 20, 2019 at 20:36 comment added quester as for second question for factor $c=2$ $P(game\, with\, c=2) = 2*(P(X_{Bin(4,1/6)}=2)*P(X_{Bin(4,1/6)}=0)) + P(game\, with\, c=1)^2$ or something quite close to this equality
Nov 20, 2019 at 20:32 comment added quester probably we would like to have answer to $P(X_{Bin(4, 1/6)}=1) ??? P(X_{Bin(8, 1/6)}=2)$
Nov 20, 2019 at 20:28 review Suggested edits
S Nov 20, 2019 at 20:50
Nov 20, 2019 at 20:21 comment added whuber I see only one solution where you make a conclusion about which game is better. What is the other one you are asking about?
Nov 20, 2019 at 20:18 comment added Anil K. But why cant we say that P(X = x) for X ~ Bin(n,p) = P(X = 2x) for X ~ Bin(2n,p) ?
Nov 20, 2019 at 20:17 comment added Anil K. Yes, the game with the higher probability of winning, which seems to be A
Nov 20, 2019 at 20:13 comment added MSIS But what notion of advantage do you then use? Just to see which one you're more likely to win?
Nov 20, 2019 at 20:11 comment added Anil K. There is no payoff it just asks which game is more advantageous
Nov 20, 2019 at 20:10 comment added MSIS What is the payoff for each outcome?
Nov 20, 2019 at 20:10 review First posts
Nov 20, 2019 at 21:11
Nov 20, 2019 at 20:08 history asked Anil K. CC BY-SA 4.0