THE GAME OF CRAPS
Throughout the game, dice are always rolled in pairs and the sum (which must lie in the range {2,3,...,12}) noted. If on the first throw the sum is 2,3 or 12 the game is lost, while if it is 7 or 11 the game is won. If it is anything else (4,5,6,8,9 or 10) the game goes into a second phase. The second phase after an initial throw of x, say, goes like this: keep throwing the dice until either x reappears (in which case the game is won) or a sum of 7 is obtained (in which case the game is lost). The obvious question concerns the probability that the game is won when both dice are fair (each outcome for each dice has a probability of 1/6). It can be shown that the probability of winning this game with a fair die is less than 0.5 or 50% chance (specifically, approximately 49.3% chance of winning) which seems to be unfair in the long-run to play such a game.
Challenge to this problem:
It is possible to increase the win probability in a game of craps by playing with dice biased in such a way as to increase the chances of 3 or 4 (which appear on opposite faces) and to decrease the chances of 1,2,5 and 6. Hence, imagine that the probabilities of 3,4 are increased to 1/6 + 2ε for some ε > 0 for both dice while that of the other outcomes are reduced to 1/6 - ε .
Question: How large does ε need to be to make the win probability for craps equal to 0.5 ?
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