# Monty Hall problem extended with expectations i.e. prior probabilities

I am fascinated by the Monty Hall problem and its variants such as N-doors version here.

Now suppose expectations. How does the Monty Hall problem changes with expectations?

Simple example

Contestant believes that prize is behind the door A with 0.01% probability, B with 10% probability and C with 89.99% probability.

Now the smart contestants pick up door A first because they know that host will show them B or C to be empty and they are planning to switch to C when B shown empty. Suppose host shows that the door C is empty -- all of a sudden, the contestants are skeptic about the prior probabilities

by prior probabilities, A is 100 times less probable than B in the original setting but how does the fact that C is empty changes the situation?

Tinkering questions

1. What kind of game strategy should be taken with prior probalities/expectations?

2. Suppose you are shown N-2 times to be wrong in all of your expectations with N doors, you make new expectation after each door opening. What is the optimal strategy?

3. Suppose you are shown N-2 times to be right in all of your expectations with N doors, you make new expectation after each door opening. What is the optimal strategy?

• You need to specify how Monty Hall chooses the doors in cases where there are more than 3. No clear answer unless you address that. – user76844 Sep 29 '14 at 12:09
• @Eupraxis1981 Good question. I have developed following. Strategy A is: "Monty chooses the door with the smallest expected probability until the last two doors. When only two doors left, Monty chooses the door with the highest probability." Strategy B is: "Monty chooses any door but not door with the highest probability and not the door with the lowest probability. When two doors left, Monty chooses the door with the highest probability." I don't know the optimal strategy, this may have been researched in the context of subgame perfect equilibrium in game-theory. Inspiring. – hhh Sep 29 '14 at 18:53

I will answer the following question for the $3$ door case.

What kind of game strategy should be taken with prior probalities/expectations?

It is reasonable to assume that the strategy we are looking for is the optimal one, i.e. the one that maximises the chance of the contestant getting the prize of a sports car, rather than being stuck with a goat. In this case, the optimal strategy is indeed that of the "smart" contestant mentioned in the question. That is, simply pick the door with the lowest probability, then switch. To be a bit more precise, suppose the probabilities of the prize being behind each door are:

$$p_1, p_2, p_3$$

Without loss of generality we can assume $p_1 \leq p_2 \leq p3$. Now, assuming the probabilities of the player are accurate, then the probability that the prize is in either door $2$ or $3$ is $1 - p_1$. Since the host, Monty, must eliminate one of these doors after door $1$ is chosen, then the probability of the prize being behind the remaining unopened and unchosen door is still $1 - p_1$, conditional of course on the accuracy assumption already stated. Now, we want to maximise this probability, which is $1 - p_1$, which is clearly maximal when $p_1$ is the smallest. Notice that for $p_1 = p_2 = p_3 = \frac13$ this yields exactly the "standard" Monty Hall problem solution, as expected.

As correctly pointed out by user76844, the $N > 3$ case is more complicated. We need the strategy Monty uses to eliminate doors, and then we need to consider all possible combinations of initial door chosen with prize door to calculate the optimal solution. My gut tells me that for some "common sense" strategy by Monty, we would still be best picking the door with the lowest probability to begin with. In performing the calculations for this case, it might be useful to note that if door $p_j$ is chosen initially, and then door $p_k$ is eliminated, then all the remaining probabilities scale by a factor of:

$$\dfrac{1 - p_j}{1 - p_j - p_k}$$