# How much should this player expect to win over the long run?

The problem goes like this: In a game, a player can bet one dollar on any one of the numbers 1,2,3,4,5, and 6. Three dice are rolled. If the player's number appears $k$ times where $k\geq 1$, the player gets $k$ dollars back, plus the original stake of one dollar. Otherwise, the player loses the one dollar stake. So over the long run, how much should a player, who plays many times, expect to win or lose??

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Let the random variable $W$ be our net gain in one throw of the three dice.

With probability $\dfrac{1}{6^3}$, our number comes up $3$ times, and $W=3$.

With probability $\dbinom{3}{2}\dfrac{1}{6^2}\dfrac{5}{6}$ our number comes up twice, and $W=2$.

With probability $\dbinom{3}{1}\dfrac{1}{6}\dfrac{5^2}{6^2}$ our number comes up once, and $W=1$.

Finally, with probability $\dfrac{5^3}{6^3}$ our number comes up $0$ times, and $W=-1$.
Use the above numbers to find $E(W)$. We have $$E(W)=3\cdot \frac{1}{216}+2\cdot\frac{15}{216}+1\cdot \frac{75}{216}+(-1)\cdot\frac{125}{216}.$$ That gives us the expected amount of money we "win" per game. (This is a realistic problem. The number $E(W)$ is negative. On average, the more we play the more we lose.)

Another way: We could instead let $Y$ be the amount of money the casino hands over. Then in the various cases $Y$ is $4$, $3$, $2$, or $0$. Find $E(Y)$ using the same basic method, and at the end subtract the $1$ dollar we had to pay to play the game. The result will be the same.

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Wouldn't it be with probability $\frac{1}{6^3}$, our number comes up 3 times, and $W = 4$? Because don't we want to also take into account that we win back our dollar? And then similarily with $W = 2$ and $W = 1$ –  TheHopefulActuary Oct 18 '12 at 3:10
I had let $W$ be our net win. We could also look at the gross wins $4$, $3$, $2$, $0$, find the expectation of that, and subtract at the end the $1$ dollar it cost to play the game. Same end result. –  André Nicolas Oct 18 '12 at 3:16
Oh I see what you did, thank you! –  TheHopefulActuary Oct 18 '12 at 3:17

The probability to win when $k$ = 3: $$p = 6/6^3 = 1/36$$

The probability to win when $k$ = 2:

There are 3 ways to select 2 dice from a set of 3 dice($_3C_2$). For each of these, there is a 1/6 chance that the 2 dice will show the same number: $$p = 3*{1 \over\ 6} = 1/2$$

To calculate our average winnings($w$), we multiply the probability of the event by the payoff if the even occurs: $$w = 3*{1 \over 36} + 4*{1 \over 2} = {25 \over 12} \approx 2.08$$

The average winnings per round will be about 2.08.

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