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??
[Math] How much should this player expect to win over the long run
probabilityprobability theory
Best Answer
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.