A casino game works as follows: A player pays exist1 to play. Then, the player d

Question

A casino game works as follows: A player pays exist1 to play. Then, the player draws a card randomly from a standard playing deck. It the card is an ace, the player gets back exist10. Otherwise, the player gets back nothing. a) Calculate, to the nearest cent, the expected net result to the player from playing this game. b) Calculate, to the nearest cent, the payoff (instead of exist10) that would make this a fair game. c) Another way to make this game more attractive to players, other than changing the payoff amount (as in Part (b)), is to remove non-ace cards from the deck, to increase P (drawing an ace). Using the original payoff amount of exist10 for an ace, calculate the minimum number of non-ace cards that would have to be removed from the deck, to make this a game that favours the player.

Explanation / Answer

Solution:-) Player pays $1 to play the game.

In a deck of cards there are 52 cards and 4 cards are ace. n(x)=4 , n(s) =52.

Probability of getting an ace card:-

P(x)=n(x)/n(s)=1/13. If the drawn card is an ace the player get back $10. Let it be x.

The expected value of getting back something:-

E(X)= 10*(1/13)+0*(12/13)=0.77

1.Thus the expected value for the game is (-1+0.77)= -0.23 cents

2. The game will be fair if the expected value of the game will be $1. Let E(X)=e

e-1=1 ; e=2. So,E(x)=2.

Let the amount he wins for getting an ace is b.

b/13 =2 ; b=26,

So, the nearest payoff should be $32.

3.As there are 4 ace cards, let s be the non-ace cards. Total cards are s+4

E(x)= 10*(4/(s+4))

2 =40/(s+4)

2s+8= 40 ; s=16;

Minimun number of non-ace cards that have to be removed from deck is (48-16)= 32.

Cheers!!

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