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Given 1 roll in a game of craps, a win is if, the sum of outcomes on a fair pair of dice is 7 or 11, otherwise you are said to crap-out . Compute the probability of a win on just 1 roll? Im second guessing myself on my answer

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up vote 1 down vote accepted

I believe you will be able to obtain the answer by adding the probability of rolling a 7 with two dice and the probability of rolling an 11 with 2 dice.

Therefore: $P_7 + P_{11}= \frac{1}{6}+ \frac{1}{18}$ $= \frac{2}{9}$

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Thank you so much – kimj Sep 30 '12 at 23:34

Imagine you have a red die and a green die. Each can come up $6$ ways, so there are $6^2=36$ possible rolls. How many sum to $7$? How many sum to $11$?

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3/36+ 1/36= 4/36...reduced fraction 2/9....Thank you so much. – kimj Sep 30 '12 at 23:34

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