Question

Remember the formula for expected value is: EV = Outcome1 * Probability1 + Outcome 2 * Probability2… and so on. 23.Suppose you are playing a game with one die at a casino. You win $1 if you get a 2 or a 6. You lose $2 if you get a 3, win $5 for rolling a 4, and lose $3 for rolling a 1. What is the expected value of this game? If you played the game repeatedly, would you expect to win or lose money? Why?

Answer 1

In a standard die with 6 sides, each side is numbered with a value from 1 to 6.

All sides have the same probability of being rolled, therefore each number have a probability of 1/6.

Using the formula for the expected value, we have:

`\begin{gathered} E=\sum_{i\mathop{=}1}^nx_ip_i\\ \\ E=(-3)(1)/(6)+(1)(1)/(6)+(-2)(1)/(6)+(5)(1)/(6)+0((1)/(6))+(1)((1)/(6))\\ \\ E=(-3+1-2+5+0+1)(1)/(6)\\ \\ E=2((1)/(6))=(2)/(6)=(1)/(3) \end{gathered}`

Since the expected value is positive, we can expect to win money.