Puzzle on rolling dice game

Question

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.

Answer 1

That works.

A faster, but essentially the same, route would be to say $$5 \times 1 + 3 \times 15 + 1 \times 75 -1 \times 125 = 0$$.

A slightly more sophisticated approach is to say that you get $\$2$ each time a particular die shows your number and you pay $\$\frac13$ to play (fair) and you have to play in batches of three

Answer 2

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 $-1+2E[N]=-1+2(1/2)=0$.