# What is 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.

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.

• You are probably expected to find the probability of $0$ six, which you did, the probability of $1$ six, which is $\binom{3}{1}(1/6)(5/6)^2$, the probability of $2$ six, which is $\binom{3}{2}(1/6)^2(5/6)$, and the probability of $3$ six, which is $(1/6)^3$. Then you can use the formula you quoted. There are other ways. Oct 18 '15 at 23:36
• On average, you win $1/6$ from each die hence total of $1/2$. Now you need to compute how much you lose on average and subtract it from $1/2$
– A.S.
Oct 18 '15 at 23:36
• But perhaps you are already supposed to know that the mean number of $6$'s is (binomial distribution) $np$, where $n=3$ and $p=1/6$. Oct 18 '15 at 23:42

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

In the formula for expected value, the probabilities p1, p2, etc. represent the probability distribution of all the possible outcomes. Together they should add up to 1. (You might see calculations that add up to less than one if some outcomes have no payoff, but technically there is another probability that is being multiplied by a payoff of 0.)

In your first calculation E(X) = 0(1/6)+1(1/6)+2(1/6)+3(1/6), you appear to be using the probability of each die rolling a six. This is incorrect, as indicated by the fact that the sum of your probabilities is on 4/6.

Begin this problem by calculating the probably of rolling 0, 1, 2, or 3 sixes on three dice. Then you can use your formula.

You account for the lost dollar by using a negative value for the payoff (x=-1).

• Ah, the dollar is lost making it negative. I will adjust the formulae I presented accordingly. Oct 18 '15 at 23:41

So, this get's broken down into cases, let's take 2 dice for the time being $$P(X=(2)) = 1/36\\ P(X=(1)) = 10/36\\ P(X=(0)) = 25/36$$ the expected value of this is $-1*25/36 + 0 * 10/36 + 1 * 1/36$, which is -2/3.

To calculate this probability in general you use the binomial distribution https://en.wikipedia.org/wiki/Binomial_distribution.

In this case the probability distribution function is ${n \choose y}\frac{1}{6}^y(1-y)^{n-y}$ where y is the number of 6's rolled on n dice.

The overall expected value is $$E(X) = \sum_{i=0}^y{{n \choose i}\frac{1}{6}^y(1-i)^{n-i}*(i-1)}$$

As this is the probability of y 6's times the payoff of y 6's across all possible number of 6's on n dice