SOLUTION: 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 ra

Question 1161335: 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 (2 - 10) is drawn, the player wins $2.
How much should be charged to play this game in order
for it to be fair?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

1 of 13 diamonds is an Ace, 3 of 13 diamonds are face cards, the other 9 cards are 2 through 10.

Card    X      P(X)   
Ace   -20      1/13
Face   10      3/13
2-10    2      9/13

The expected value is the sum of X times P(x) for each of the three possible outcomes, Ace, Face, or 2-10. A perfectly fair game has an expected value of zero. A positive expected value favors the player and a negative expected value favors the owner of the game. In this case, it is not possible given that you need to round the charge to play the game to the nearest cent to charge an amount that makes the game perfectly fair. You will find, when you do the arithmetic, that charging $2.15 to play very slightly favors the player and $2.16 very slightly favors the owner of the game.

John

My calculator said it, I believe it, that settles it

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