# Craps game probability

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

-

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