Recently I was doing a problem that I thought might be similar to the Monty Hall problem. In it were five intervals $\left[0,2\right], \ \left[2,10\right], \ \left[10,30\right], \ \left[30,70\right], \ \left[70,90\right] $ in which a person could die. The person could choose an interval to die in uniformly, so the probability of any interval being chosen is $0.2$. Suppose a person lived till $30$, then the probability of death is now $1$ over the remaining two intervals, so he can choose either of them with probability $0.5$.
All correct till this stage (I hope). Now replace this with the Monty Hall doors. The first three doors have been eliminated, so the probability of "winning" is $1$ over the remaining two doors.
Do I switch? It seems no, but the answer is yes, as we know. I've been thinking of why. It's because the two problems aren't really the same at all. The event "dying" isn't pre-determined to be in any interval -- it occurs with probability $1$ in any of the intervals that are still remaining. Whereas, the car's position is pre-determined -- no matter what, the car is always behind a door (which will never be opened by the host before he asks his question). In other words, the car isn't randomly spread over two doors -- it's located for sure behind a single door.
Said another way, the conditional probablity of death in the remaining intervals is $1$, while the conditional probability of the car being in the door we didn't choose is $2/3$, because that's the nature of the variables we're dealing with.
Is this a correct intuition for these problems? Can someone give a more formal/less rambling explanation?