In the game of craps two dice are rolled If the first roll i

In the game of craps, two dice are rolled. If the first roll is a 7 or an 11, the player wins. If the first roll is a 2, 3, or 12, the player loses. If any other outcome is observed on the first roll, the player wins if that outcome is rolled again before a 7 is rolled; otherwise, he loses. What is the probability of winning this game?

Solution

We start by finding the probability of winning at the first roll, that is obtaining 7 or 11: Pr(7) 1 6 6/36 = 1/6 probability of getting 7 2 5 3 4 (use fractions rather than decimals) 4 3 5 2 6 1 Pr(11) 5 6 2/36 = 1/18 probability of getting 11 6 5 So the probability of winning on first roll = 1/6 + 1/18 = 2/9 Now we find the probability of losing at the first roll (i.e. getting 2, 3, 12): Pr(2) Pr(3) Pr(12) 1 1 = 1/36 2 1 = 2/36 6 6 = 1/36 1 2 The probability of losing on first roll = 1/36 + 2/36 + 1/36 = 1/9 Then the probability of making a point will be 1 - 2/9 - 1/9 = 2/3 The individual probabilities for the numbers 4, 5, 6, 8, 9, 10 are shown below. Pr(4)(3/36 = 1/12) Pr(5)(4/36 = 1/9) Pr(6)( = 5/36) 3 1 4 1 5 1 2 2 3 2 4 2 1 3 2 3 3 3 1 4 2 4 1 5 Pr(8)( = 5/36) Pr(9)(4/36 = 1/9) Pr(10) (3/36 = 1/12) 6 2 6 3 6 4 5 3 5 4 5 5 4 4 4 5 4 6 3 5 3 6 2 6 Now ADD these probabilities. P(4u5u6u8u9u10) = 3/36 + 4/36 + 5/36 + 5/36 + 4/36 + 3/36 = 24/36 = 2/3 (as we have already shown above) This is the probability of making a point. We now consider the individual probabilities depending on which \'point\' has been obtained. (E.G., probability of getting 4 as the \'point\' = 1/12) I shall work through this problem in detail and the result can then be applied to the various \'point\' probabilities. The probability of gaining 4 as the \'point\' = 1/12 \" getting 7 = 1/6 \" game continuing = 1 - 1/12 - 1/6 = 3/4 So the probability of winning in this case is (1/12) + (3/4)(1/12) + (3/4)^2(1/12) + ..... to infinity = (1/12)[1 + (3/4) + (3/4)^2 + ..... to infinity] = (1/12)[1/(1 - 3/4)] = (1/12)(4) = 1/3 So the total probability of gaining the \'point\' 4 and then winning = (1/12) (1/3) This will be the same probability if the initial \'point\' was 10. If the \'point\' was 5 or 9 the initial probability of 5 or 9 is 1/9 and the probability of continuing thereafter is 1 - 1/9 - 1/6 = 13/18 and by the same argument as above the probability of winning is (1/9)[1/(1 - 13/18)] = (1/9)(18/5) = 2/5 and the total probability is (1/9)(2/5) If the \'point\' is 6 or 8 the probability of continuing is 1 - 5/36 - 1/6 = 25/36 and the probability of winning is (5/36)[1/(1 - 25/36)] = (5/36)(36/11) = 5/11 and the total probability is (5/36)(5/11) We can now find the overall total probability of winning = 2/9 + 2[(1/12)(1/3) + (1/9)(2/5) + (5/36)(5/11)] = 244/495 = 0.493 which as you would expect will be less than 1/2 so that over the long run you will lose to the bank.