I have the following problem:
Consider the following game with three doors. Behind one of the doors there is \$20, behind a second door there is \$260, and behind the third one is a plastic mouse.
In each round you can choose a door at random. If behind that door there is money, you can collect it, then the door will be closed again, the money behind it is reinstated, the contents behind the three doors is permuted at random, and you play another round, and so on so forth.
The game finishes when you choose the door with the plastic mouse. How much should you be willing to pay for this game so that it is a fair game (and the value of the plastic mouse is \$1)?
I suspect that I should a discrete time process $(X_n)_{n \in \mathbb{N}}$ with state space $\{\$20, \$260, \$1\}$ and a martingale related to it, but I do not know how to proceed. Any clues? Thank you!
To elaborate on @lulu's answer:
Let $E$ be the expected value.
From the information we've got, each outcome is equally likely ($\to$ probability $\frac{1}{3}$).
In one of these outcomes, where behind the door we find the mouse, the game terminates with an extra 1\$. In the other two outcomes, we get to continue play. Observe that the expected value of money we get from the next rounds is exactly $E$. So,
$$E = \frac{1}{3}\cdot (1 + (20 + E) + (260 + E)) = \frac{1}{3}\cdot (281 + 2E) \implies E = 281.$$