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
$\begingroup$ $\endgroup$
Add a comment |
2 Answers
$\begingroup$ $\endgroup$
0 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: $P_7 + P_{11}= \frac{1}{6}+ \frac{1}{18}$ $= \frac{2}{9}$
$\begingroup$ $\endgroup$
1 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 $11$?
- $\begingroup$ 3/36+ 1/36= 4/36...reduced fraction 2/9....Thank you so much. $\endgroup$kimj– kimj2012-09-30 23:34:08 +00:00Commented Sep 30, 2012 at 23:34