Paying money to choose the box with \$100 There are 4 boxes. One box has $100. The other 3 has nothing. You pay \$X to choose any box. If it doesn't have the money, you can choose another box. How many should \$X be to ensure a fair game if the person plays optimally? The solution is \$40.
The solution is as follows:
You can define $E_1, E_2, E_3, E_4$ to be the events that the first, second, third, fourth boxes you choose has the money, respectively. So we want the expected payment for playing the game to be equal to $100. So we have
$$
100 = XP(E_1) + 2XP(E_2) + 3XP(E_3) + 4XP(E_4) \\
= 0.25X + 0.50X + 0.75X + 1.0X \implies X = 40
$$
Now suppose if you choose 2 boxes and haven't gotten the money, the price changes to $Z$. What would $X$ and $Z$ be to ensure a fair game? I am stuck on this problem. According to the above equation, the equation becomes
$$
100 = XP(E_1) + 2XP(E_2) + 3XP(E_3) + 4XP(E_4) \\
= 0.25X + 0.50X + 0.75(2X + Z) + 1.0(2X + 2Z)
$$
So we have 1 equation and 2 unknowns and if we bound $X,Z \in [0, 100]$, then we have an infinite number of solutions. Is this the answer, or am I missing something here?
 A: Suppose you have two boxes.  One has nothing and one has $100.  Otherwise, the rules are the same as in your first game.
The fair price $Z$ for each play is now given by $$100 = \frac{1}{2}Z + \frac{1}{2}(2Z) = \frac{3}{2}Z \implies Z = \frac{200}{3}.$$

Now, with your equation, solve for $X$ in $$100 = 0.25X + 0.25(2X) + 0.25(2X+Z) + 0.25(2X+2Z)$$ to find $$X = \frac{400-3Z}{7}$$
and then use our value of $Z$ to find $$X = \frac{200}{7}$$
A: The key is the condition "a fair game if the person plays optimally".
You have analyzed only one strategy: playing the game until you win the money. In the first version of the problem, that is necessarily the optimal strategy: if it was worth playing the game with $4$ boxes, it becomes worth playing the game with $3$ boxes, as well.
In general, we can consider $4$ strategies; the $k^{\text{th}}$ strategy is to stop after $k$ boxes have been opened (if you haven't yet won). In the second version of the problem:

*

*The $k=1$ strategy can be disregarded: if it was worth opening the first box, it is worth opening the second box, which has the same price but better chances.

*The $k=3$ strategy can be disregarded: if it was worth opening the third box, paying $Z$ for a chance at $\$100$, then it's worth opening the fourth box and paying $Z$ for a certain $\$100$.

But the $k=2$ strategy and $k=4$ strategy are both valid. So the constraints on fairness are that:

*

*The $k=2$ strategy does not earn free money: $$\frac14(100-X) + \frac14(100-2X) + \frac12(-2X) \le 0.$$ Equivalently, $7X \ge 200$.

*The $k=4$ strategy does not earn free money: $$\frac14(100-X) + \frac14(100-2X) + \frac14(100-2X-Z) + \frac14(100-2X-2Z) \le 0.$$ Equivalently, $7X + 3Z \ge 400$.

*There is a way to break even: equality holds in at least one of the cases above.

I would consider any strategy that satisfies all three conditions "fair, if the person plays optimally". This means that either $X = \frac{200}{7}$ and $Z \ge \frac{200}{3}$, or $X \ge \frac{200}{7}$ and $Z = \frac{400-7X}{3}$. This includes:

*

*The case $X=40$ and $Z=40$, which is unsurprisingly fair, because it's the same as the first version of the game.

*The case $X=\frac{200}{7}$ and $Z = 100000$, which is fair if the person plays optimally and stops after two boxes. (In this game, it is a bad idea to open the third box, since you pay more money to open it that you could possibly win.)

Possibly the intended solution, though, is $X = \frac{200}{7}$ and $Z = \frac{200}{3}$, which is fair if the person plays either the $k=2$ or the $k=4$ strategy.
Alternatively, you could argue that even if equality does not hold in either case, we have "a fair game if the person plays optimally". The game with $X=100000$ and $Z=100000$ is a fair game if the person plays optimally: if the person walks away and does not play at all, they break even.
