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.