Fine the value of k such that this game is fair (E(X) = 0)

Question

From an ordinary deck of 52 cards, cards are drawn at random and with replacement until three aces are drawn. If it takes more than ten draws to obtain the three aces, the player loses k dollars; otherwise, he wins 100 dollars. Let X denote the amount of money that the player wins per game.

(a) Find the value of k such that this game is fair (that is, such that E(X) = 0).

(b) For the fair value of k, find the variance of X.

Hello people. The first question is throwing me off and preventing me from being able to start the problem. Can I get hints towards what the first question is asking me to do.

Answer 1

Instead of actually stopping the game once the third ace is drawn, let us instead simply draw ten times. Notice the following:

The player wins if and only if there are at least three aces drawn within 10 attempts.

Instead of calculating the probability that he wins, let us instead calculate the probability that he loses. There are three cases to consider:

• In ten attempts, he draws exactly two aces
• In ten attempts, he draws exactly one ace
• In ten attempts, he draws no aces

What is the probability that he draws exactly two aces?

Since we are drawing with replacement, pick which two spots out of the ten are occupied by the aces. $\binom{10}{2}$ choices. In order, pick which card it is that appears in each non-ace spot. $48$ choices each time. In order, pick which ace appears in each ace-spot. $4$ choices each time. There are then $\binom{10}{2}4^248^8$ different possibilities. There are $52^{10}$ possible ways to draw ten cards with repitition. Thus, the probability of drawing exactly two aces is $\binom{10}{2}\frac{4^248^8}{52^{10}}$

Continue by finding the probability of exactly one ace and no aces, and then adding these together to find the probability of a loss. The probability of a win is then $1-P(\text{loss})$

Note: instead of the brute force counting above, we could have simply used a binomial distribution with $\frac{1}{13}$ as the chance of success (drawing an ace) and $\frac{12}{13}$ as the chance of failure (not drawing an ace). The chance of exactly $k$ successes in 10 attempts is then $\binom{10}{k}(\frac{1}{13})^k(\frac{12}{13})^{10-k}$

Now, letting $X$ denote the random variable for how much money he wins, $E[X] = Pr(\text{Win})\cdot 100 - Pr(\text{Lose})\cdot k$. If it were fair, then we want $E[X]=0$. Use the information found above to solve for $k$.