Question

(9 pts) In a casino game, you pay $5 to play. You draw a card from a standard deck. If you draw any card ace (worth 1) through 10, you receive twice the number of dollars as the card’s number (so if you draw a 7, you get $14). You lose if you draw a face card. Make the probability distribution for this game, find the expected value of a bet, and decide if you should play.

Answer #1

below is probability distribution of X (net gain from this game):

P(X=2*1-5=-3)=P(draw ace) =4/52

P(X=2*2-5=-1)=P(draw a 2) =4/52

P(X=2*3-5=1)=P(draw a 3) =4/52

P(X=2*4-5=3)=P(draw a 4) =4/52

P(X=2*5-5=5)=P(draw a 5) =4/52

P(X=2*6-5=7)=P(draw a 6) =4/52

P(X=2*7-5=9)=P(draw a 7) =4/52

P(X=2*8-5=11)=P(draw a 8) =4/52

P(X=2*9-5=13)=P(draw a 9) =4/52

P(X=2*10-5=15)=P(draw a 10) =4/52

P(X=-5) =P(draw a face cards) =12/52

expected value of a bet

E(X)=ΣxP(x)=(-3)*(4/52)+(-1)*(4/52)+(1)*(4/52)+(3)*(4/52)+(5)*(4/52)+(7)*(4/52)+(9)*(4/52)+(11)*(4/52)+(13)*(4/52)+(15)*(4/52)+(-5)*(12/52)=**3.46**

since expected value is positive for the game, one should play this game.

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