In a certain dice game, three fair dice are rolled and the total on all three dice is counted. Let the random variable x be the total count on all three dice.
You pay one dollar to roll the dice. If you roll a 16 or greater, you will win $10. If you roll a 6 or lower, you win $5. If you roll anything else, you lose your money.
What is the expected value of this game?

Question

In a certain dice game, three fair dice are rolled and the total on all three dice is counted. Let the random variable x be the total count on all three dice. 
You pay one dollar to roll the dice. If you roll a 16 or greater, you will win $10. If you roll a 6 or lower, you win $5. If you roll anything else, you lose your money.
What is the expected value of this game?

anonymous · Answer

Well we know $X$ always gives some value on the interval $[3\ ..\,36]$, with $34$ different sums that are all equally likely.
We know that for $X\ge16$, our payout $P$ will be $10$ (this happens with probability $21/34$ since there are $21$ possible numbers that work here).
For $X\le6$, we have $P=5$ (this happens with probability $4/34=2/17$ since there are $4$ possible numbers that work here).
We also have the event that $6\lt X\lt 16$, we have $P=0$ (probability $9/34$).

anonymous · Answer

Our *expected value* of $P$ is just an average of its possible values, each weighed by their probability, so:$$E(P)=\frac{21}{34}\times10+\frac2{17}\times5+\frac9{34}\times0=\frac{115}{17}$$

anonymous · Answer

weighTed

anonymous · Answer

Thank you!!!!