Question

6.32 In the game of Chuck-a-luck, three dice are rolled. The player selects a number between 1 and 6. If the player's number comes up on exactly one die, the player wins $1. If the player's number comes up on exactly two dice, the player wins $2. If the player's number comes up on all three dice, the player wins $3. If the player's number does not come up, the player loses $1. Let X denote the player's win or loss. a. What is E[x]? b. In some versions, the player wins $10 if all three dice show the player's number. Now what is E[x]?

Answer 1

here probability of a number shown on one dice =1/6

hence probability that your number is on none of die =P(X=0)= $\binom{3}{0}(1/6)^{0}(5/6)^{3}$ =0.5787

P(X=1) = $\binom{3}{1}(1/6)^{1}(5/6)^{2}$ =0.3472

P(X=2) = $\binom{3}{2}(1/6)^{2}(5/6)^{1}$ =0.0694

P(X=3) = $\binom{3}{3}(1/6)^{3}(5/6)^{0}$ =0.0046

hence a) E(X) = $\sum$ xP(x) =(-1)*0.5787+1*0.3472+2*0.0694+3*0.0046 =$-0.08

B)

E(X) =(-1)*0.5787+1*0.3472+2*0.0694+10*0.0046 =$-0.05