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?

Respuesta :

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:

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

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