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 lust roll, (the player wins if that outcome is rolled again before al is rolled: otherwise, he loses. What is the probability of winning this game?

Solution

Let Z denote the sum of scores on a given roll.

There are 2 cases in which winning is possible.

First case : They roll a 7 or 11 in first roll.

P₁ = 6/36 + 2/36 = 2/9

Second case: They roll a 4 in first roll.

They will win if on second roll they roll a 4 or rolls anything else than 4 or 7 on second roll and rolls 4 on third roll or they roll anything else than 4 or 7 on second and third roll and roll a 4 on fourth roll and this continues on.

P₂ =(3/36)*(3/36) + (3/36)(27/36)(3/36)+(3/36)(27/36)(27/36)(3/36)+......

Third Case: They roll a 5 in first roll.

They will win if on second roll they roll a 5 or rolls anything else than 5 or 7 on second roll and rolls 5 on third roll or they roll anything else than 5 or 7 on second and third roll and roll a 5 on fourth roll and this continues on.

P₃ = (4/36)*(4/36) + (4/36)(26/36)(4/36) + (4/36)(26/36)(26/36)(4/36) + ......

Fourth Case : They roll a 6 on the first roll.

They will win if on second roll they roll a 6 or rolls anything else than 6 or 7 on second roll and rolls 6 on third roll or they roll anything else than 6 or 7 on second and third roll and roll a 6 on fourth roll and this continues on.

P₄ = (5/36)*(5/36) + (5/36)(25/36)(5/36) + (5/36)(25/36)(25/36)(5/36) + ......

Similar cases for first roll of 8,9, and 10.

Thus the winning probability will be,

P=P₁+2P₂+2P₃+2P₄ = (2/9)+2(1/36)+2(4/45)+2(25/396) = 244/495 = 0.49

z	2	3	4	5	6	7	8	9	10	11	12
P(Z=z)	1/36	2/36	3/36	4/36	5/36	6/36	5/36	4/36	3/36	2/36	1/36