There are $n$ doors, only one has a prize behind it. You can pick one at the beginning. At every round, host will open a door (which is empty) and ask you if you want to switch. The game will last $n-2$ rounds.
I think the optimal strategy is to stay put till the last turn and then switch but it is hard to prove.
At first I think there is a simple recurrence relation:
if the current door contains prize with probability $p$, and there are $m$ closed doors . After switching, the probability becomes $\frac{1-p}{m-1}$.
However, I realized that player can always switch back to a previously chosen door if it is not being opened yet. So the probability will depend on your decision (chose a new door or switch back to previously chosen doors) and also the strategy of the host (for example, a mixed strategy where host has probability $q$ of opening the door player just left and $1-q$ of opening a random empty door).
Can anyone give me an idea of this problem?
Thanks.
Since Monty is required to open all unpicked bad doors, when he opens a bad door to open, you know that he chose it because it is not your pick and it is not the prize. That is the information you gain. For each of the doors that you have not picked, this tells you that the pool they were in has gotten smaller, so their probability has gotten larger. But your door is not in that pool. Monty didn't pick it because it was your current choice. you gain no new information about it, so its probability does not change. No door ever goes down in probability. It will either go up, or if it is your choice, it will stay the same.
As long as you keep it, your inital door has probability $\frac 1n$. The probability of all the other doors rose from that to $\frac 1{n-1}$ at the first reveal. If you have any other door at the final swap, its probability will be at least $\frac 1{n-1}$, since the door can never go down in probability. If at some time you leave your initial door, then while you are holding another door, your initial door also goes up in probability. If you switch back, its probability will still be greater than $\frac 1n$.
You could maximize the probability of your door at the final swap by picking a new never-chosen door each time. But it will never have probability $>\frac 12$, so you are always wisest to swap at the final opportunity. Thus your best strategy is to minimize the probability of your door at this time, thereby maximizing the probability of the other door.
Therefore, your best strategy is to stick with your original door until the last opportunity, maintaining its original $\frac 1n$ probability as the prize door, then switch.