# Probability of getting a “double” in at least two throws of two dies

We play a game where we throw two distinct dies twice. A player wins $\$3$if he gets at least one time a double between the two throws, and loses$\$1$ if he doesn't get a double in any throw of the two. What is the expected value of the player winnings in a game?

I am having a bit of a problem knowing what is the probability of getting a double in at least one of the two throws. I thought about it like this: Our $\Omega$ is all the sequences on $\{1,2,3,4,5,6\}$ with length 4 (for the two throws). $| \Omega | = 6^4$. $Pr(\text{to get a double in at least one throw}) = 1 - Pr(\text{to not get a double at any of the throws}) = 1 - \frac{36}{6^4} = \frac{35}{36}$ which is obviously wrong. Any help will be appreciated.

P.S.: (Can we solve it using an Indicator Random Variable?)

## 1 Answer

The probability of rolling a double on 2 six sided dice is $\frac{1}{6}$. So the probability of not rolling one is $\frac{5}{6}$.

The probability of this happening twice is $\left(\frac{5}{6}\right)^2=\frac{25}{36}$. The expectedayoff is $0.22 - deal me in! • Just to add: To calculate the expected value:$E[f] = \sum_{a \in \mathbb R} { a \cdot Pr(f=a)} = (-1) \cdot \frac{25}{36} + 3 \cdot \frac{11}{36}\$. – TheNotMe Jun 30 '13 at 9:25