Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?)

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

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!

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.