Maybe a Monty Hall kind of problem?
Solution 1:
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,
$$E = \frac{1}{3}\cdot (1 + (20 + E) + (260 + E)) = \frac{1}{3}\cdot (281 + 2E) \implies E = 281.$$