A gambler goes to bet. The dealer has 3 dice, which are fair, meaning that the chance that each face shows up is exactly 1/6.
The dealer says: "You can choose your bet on a number, any number from 1 to 6. Then I'll roll the 3 dice. If none show the number you bet, you'll lose \$1. If one shows the number you bet, you'll win \$1. If two or three dice show the number you bet, you'll win \$3 or \$5, respectively."
Is it a fair game?
PROPOSED APPROACH:
Let A be event when all three dice show the given number, B be the event that only two dice show the same given number, C be the event that only one dice shows the given number, D be the event otherwise. Now
P(A) = $1/216$
P(B) = ${3 \choose 2} * [1*1*5/216] = 15/216$
P(C) = ${3 \choose 1} * [1*5*5/216] = 75/216$
From the law of complementary event the P(D) = 1 – P(A) – P(B) – P(C) = $1 – 1/216 – 15/216 – 75/216 = 125/216$
Now expected value for winning is $\$1*P(C) + \$3*P(B) + \$5*P(A) = 125/216$
expected value for losing is $\$1*P(D) = 125/216$
Given the two values are equal, it is a fair game.
I am thinking if there is a quicker or smarter way to do it.
Best Answer
Your payoff is $-1+2N$, where $N$ is the number of dice showing your chosen value. But $E[N]=3(1/6)=1/2$, by linearity of expectation; so your expected payoff is $-1+2E[N]=-1+2(1/2)=0$.