Find the random variable, value function, and value you would pay to break even... In a game you receive three cards, $\omega$ , from a well-shuffled deck. You then receive $10 if the hand contains at least two face cards. In order to determine how much you would be willing to pay, per game, to play a large number of hands, you need to create a suitable random variable X and a value function f. What is the random variable, the value function, and the value you would be willing to pay to break even, namely E[f]?

Let $X=1$ if you get $2$ or more face cards, and let $X=0$ otherwise.

There are $\binom{52}{5}$ equally likely poker hands.

There are $\binom{12}{2}\binom{40}{1}$ hands that have $2$ face cards, and $\binom{12}{3}$ hands that have $3$. Thus $$\Pr(X=1)=\frac{\binom{12}{2}\binom{40}{1}+\binom{12}{3}}{\binom{52}{5}},$$ and therefore $$E(X)=\frac{\binom{12}{2}\binom{40}{1}+\binom{12}{3}}{\binom{52}{5}}.$$

Let $f(x)=10x$. Then our income from the game is $f(X)$. And $$E(f(X))=10\frac{\binom{12}{2}\binom{40}{1}+\binom{12}{3}}{\binom{52}{5}}.$$

The number $E(f(X))$ is the the amount one should pay per game to make the game a "fair" game.

**Remark:** Alternately, we could let $Y=10$ if we get $2$ or more face cards, and $0$ otherwise. We then want $E(Y)$. Of course we get the same number.