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?)
Best 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!