MATH SOLVE

3 months ago

Q:
# 1. A game is set up as follows: All the diamonds are removed from a deck of cards, and these 13 cards are placed in a bag. The cards are mixed up, and then one card is chosen at random (and then replaced). The player wins according to the following rules. If the ace is drawn, the player loses $20. If a face card is drawn, the player wins $10. If any other card (2β10) is drawn, the player wins $2. How much should be charged to play this game in order for it to be fair?

Accepted Solution

A:

Answer: the game must be played in groups of 13 draws, with a cost of $28 for the 13 draws. Step-by-step explanation:Probability of drawing any of the 13 balls is the same. So the expected value of play is the average of the values of each card. The only losing card is the ace; the player loses $20 if that card is drawn.
Each of the 3 face cards wins $10; each of the remaining 9 cards wins $2.
So the average value of each draw is[tex]\frac{-20+3\times10+9\times2}{13}[/tex]= 28/13Therefore, on average the player will win $ 28/13. Β to be a fair game the player must pay Β $28/13 for a game. that is not possible, unless the game must be played in groups of 13 draws, with a cost of $28 for the 13 draws.