[Math] the expected value? Three dice are rolled. For a 1 dollar bet you win 1 dollar for each 6 that appears (plus dollar back). No 6, lose dollar.

probability

Three dice are rolled. For a 1 dollar bet you win 1 dollar for each 6 that appears (plus your dollar back). If no 6 appears you lose your dollar. What is your expected value?

Practice midterm question.

So I know that the formula for expected value is:
E(X) = x1p1 + x2p2 + x3p3 + . . . + xnpn.

and the probability of rolling a six is 1/6.

So would I calculate the expected value by inputting:
E(X) = 0(1/6)+1(1/6)+2(1/6)+3(1/6)
where each x1, x2,…xn is the amount of 6's rolled?

But how do I account for the dollar lost when no 6's are rolled?

edit: Figured some more parts out:

So I calculated that the total possible outcomes for rolling three dice is 216 and the chances of rolling no 6's (so losing a dollar) is 125/216 because 5^3/6^3=125/216. So that must mean there's a 1-(125/216) = 91/216 chance of rolling at least 1 six and gaining one dollar. I'm still not sure how to calculate the expected value though.

Best Answer

Three dice are rolled. For a 1 dollar bet you win 1 dollar for each 6 that appears (plus your dollar back). If no 6 appears you lose your dollar. What is your expected value?

So your winnings are $w(X) = \begin{cases} -1 & : X=0 \\ X+1 & : X\in\{1,2,3\}\end{cases}$

And $\mathsf P(X=x) = \dbinom{3}{x}\dfrac{5^{3-x}}{6^3} \qquad$ because we have a binomial distribution.

Calculate: $\mathsf E(w(X)) = \sum\limits_{x=0}^3 w(x)\;\mathsf P(X=x) \\[1ex] = (-1)\mathsf P(X=0)+2\,\mathsf P(X=1)+3\,\mathsf P(X=2)+4\,\mathsf P(X=3)$

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