The following question is taken from Mark Joshi et al. Quant Job Interview.
Let’s play a game. There are four sealed boxes. There is $100$ pounds in one box and the other are empty. A player can pay $X$ to open a box and take the contents as many times as they like. Assuming this is a fair game, what is the value of $X$?
Fair game means that expected gain is zero (expected gain $=$ expected cost). So, we must have $$100 = 2.5X$$ I manage to solve this part. I got $X = 40.$
Instead of maintaining a constant price, if you choose to continue playing after two incorrect guesses the price changes to $Z$. What is the fair value of $X$ and $Z$?
This is the part which i have difficulty. Same reasoning as above gives us that \begin{align*} 100 & = X/4+2X×1/4+(2X+Z)×1/4+(2X+2Z)×1/4 \\ & = \frac{7X}{4} + \frac{3Z}{4}. \end{align*} However, we have only one equation with two variables $X$ and $Z.$
Any hint is appreciated.
Indeed, the conditions don’t determine a single solution for the parameters. However, there is still some information that you haven’t used. It must be impossible to gain by stopping early.
If the first guess is worth paying $X$, the second guess is certainly also worth paying $X$. So we have to check the expected payoff after two guesses. This is
$$ \frac14\cdot(100-X)+\frac14(100-2X)+\frac12(-2X)=50-\frac74X\le0\;. $$
Thus we must have $X\ge\frac{200}7\approx28.57$. Then, if the third guess was worth paying $Z$, the fourth guess is certainly also worth paying $Z$, so the only remaining condition is the one that you derived for stopping at the fourth guess. Thus any pair of parameters with $X\ge\frac{200}7$ and $7X+3Z=400$ (and thus $Z\le\frac{200}3$) makes the game fair.