The following question is taken from glassdoor.
Three doors, it takes you $1$ minute to go through the first door, it also takes you $1$ minute to go through the second door, it make your speed slow down $\frac{1}{2}$ after going through third door. Only the first door is the right way to go out. You choose each door with the equal probability. How much time will you go out in expectation?
Let $E(X)$ be the required expectation. By law of total expectation, $$E(X) = P(\text{first door})E(X\,\vert\,\text{first door}) + P(\text{second door})E(X\,\vert\,\text{second door}) + P(\text{third door})E(X\,\vert\,\text{third door}).$$ Each probability above is $\frac{1}{3}.$ We also know that $$E(X\,\vert\,\text{first door}) = 1, \quad E(X\,\vert\,\text{second door}) = 1 + E(X).$$ However, I do not know how to calculate $$E(X\,\vert\,\text{third door}).$$ Any hint is appreciated.
Assuming each pass through the third door cuts your speed in half but you go through that door instantly you have $$E(X|\text{third door})=2E(X)$$ When you plug this into your equation you will get a contradiction, which means that the sum $E(X)=\sum_{t=0}^\infty tP(t)$ where $P(t)$ is the probability of escape after time $t$ does not converge.