# Carnaval Probability Question - Minimum Pricing

A carnival operator is constructing a new game.

In this game, the player will place a bet. Then a ball is selected from an urn containing balls of three colours: $2$ green balls, $3$ red balls and 5 white balls.

If a red ball is drawn, the player wins $\$3$, if a green ball is drawn, the player wins$\$4$, and if a white ball is drawn the player wins nothing.

What should the carnival operator set the bet at such that on average he makes $\$0.50$profit a game? (That is, so the player loses$\$0.50$)

Hint. One way to approach this is to imagine that ten games are played, with everything coming out exactly in proportion. That is, in the ten games, two players draw green balls (each winning $\$4$), three players draw red balls (each winning$\$3$), and five players draw white balls (each winning nothing). What is the cost to the carnie? How much does he need to make to earn $\$5$total in the ten games (that would be$\$0.50$ of profit per game)? How much of a charge per game does that work out to?
You are to calculate the player's expected winnings. What is the chance the player wins $\$3?$How about$\$4, \$0$? Add up the expected winnings of each case and you have the player's expected winnings. Now add he$\$0.50$ profit.