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 That's a Load of Craps! (Posted on 2003-10-22)
Craps is a 1-player dice game that is played as follows: Roll two 6-sided dice; their sum becomes your "initial" roll. If this initial roll is 2, 3, or 12, you lose. If the initial roll is 7 or 11, you win. Otherwise, keep rolling the dice until you reroll you initial number (and win) or until you roll a 7 (and lose).

You're betting that your adversary is going to lose his game of craps, which should be a favorable bet for you. But you receive an anonymous tip that he's secretly loaded one of the dice, so that it will always come up 5. This increases his chances of winning to 2/3.

Having learned of his evil deed, you're going to secretly load his other die so as to minimize his chance of winning. With what probability should you load each of the six faces? And how does that change his probability of winning?

 No Solution Yet Submitted by DJ Rating: 4.4167 (12 votes)

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 solution | Comment 9 of 12 |
Key idea is to minimize the chance of rolling 2 or 6 in te initial roll while maximizing the chance of rolling 2 (and avoiding rolling the same number as in initial roll) in next rolls.

Obviously, we can ascribe probability 0 to rolling 6, because we will lose in first roll and it does not help in next rolls, either.

Probability of rolling 1,3,4,5 should be the same as it is indifferent whether we roll 1 initially and try to get 2 in next roll or we roll 3 (or 4, or 5) in first roll and try to get 2 in next roll.

Question is, what is the desired probability of rolling 2? We will find this by maximizing our winning probability. Formula for winning probability is this: (1-p)*(p/((1-p)/4+p)), where p is probability of rolling 2. Explanation: we dont want to roll 2 in first roll, thus (1-p) for first roll. In next rolls, probability of winning is p (if we roll 2), probability of losing is (1-p)/4 if rolling the same number as in initial roll (probabilities of rolling 1,3,4,5 are the same, therefore (1-p)/4 is probability of rolling one of these numbers). Every game must end by winning or losing (even if the game continues to infinity) so probability of winning is p/((1-p)/4+p). Multiplying the winning probability for initial roll and next rolls we get (1-p)*(p/((1-p)/4+p)). We find the 'maximizing' p by first derivation of this formula equalling 0, which gives p=1/3.

The answer is then: we should load probability of 1/3 to face 2, probability of 0 to face 6 and probability of 1/6 to faces 1,3,4,5 of the other die. Our chance of winning will be then 4/9, which is still not worth the effort :)
 Posted by Saso on 2003-10-28 11:05:11

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