Repeated versions of the Monty Hall problem

Question

As per previous questions, there is a discussion about the Monty Hall issue that opens $k$ doors for $n$ doors.link

In this article, the probability is referred to as follows:

$$ P = \frac{n - 1}{n(n - k - 1)}$$

However, wouldn't this probability be the same if the host and player repeated Monty Hall $k$ times?

(Repeating Monty Hall $k$ times here means repeating the transactions in the Monty Hall problem $k$ times. In other words, after the player re-selects a door from the remaining $n-1$ doors, the host opens another door from the other $n-2$ doors that have no cars in it, and the player re-selects a door from the remaining doors, and so on, repeating this process $k$ times.

Answer 1

No, it is not the same.

Since it is not clearly indicated in the question, assume:

After the player selects a door at any point of the game, the host will never open it.

and

The player will never select a door that is selected at any point of the game.

So:

We have $k< \frac{n}{2}$. After the selection with $m$ doors left (not chosen before or opened), suppose the car is still in the unselected doors, the problem reduces to a Monty-Hall problem with $m-2$ doors and $k=1$.

This prob. of selecting the car at the $1,2,3...,k,(k+1)$th selection is

$p(1),p(2),p(3),...p(k),p(k+1)=\frac{1}{n},\frac{n-1}{n(n-2)},\frac{n-3}{(n-2)(n-4)},...,\frac{n-2k+3}{(n-2k+4)(n-2k+2)},\frac{n-2k+1}{(n-2k+2)(n-2k)}$

And then we need to check the cases where the car is excluded in an early selection. Then we have

$p'(m)$=prob. of selecting the car at exactly the $1,2,3...,(k+1)$th selection

given by

$p'(1)=p(1)$

$p'(m)=p(m)[1-\sum_{i=1}^{m-1}p'(i)]$

you will be looking for $p'(k+1)$. It is quite different from the original case.

Answer 2

It depends on the details of the rules and on the player's strategy.

In this answer I make the usual assumptions the player knows in advance exactly how many doors Monty will open (and that Monty will never open the player's most recently chosen door, etc.) and that the player has no other knowledge that would differentiate the doors (such as the show-runners put the car behind door 2 more frequently, or Monty always opens the lowest-numbered available door).

I also assume that Monty is allowed to open a previously-chosen door provided that it is not the door the player has most recently chosen. (The roles of the numbers $n-1$ and $n-2$ in the question are a little unclear.)

One strategy the player can follow under these (extended) standard assumptions is to pick a door at random and then always "re-select" the same door ("stay") until Monty has opened $k$ doors, at which point the player should "switch" to one of the remaining closed doors.

That strategy creates an outcome identical to the one where the player makes one choice and then Monty has to open all $k$ doors before offering the player another choice, and where Monty decides which $k$ doors to open by choosing one door, then another, then another, and so forth until he knows which $k$ doors he plans to open. That is, it's just a way to choose $k$ items randomly from among $n - 2$ items.

Other strategies can give different results, however.

Consider the case $n = 4$, $k = 2$. The player initially chooses a door at random. Let $A$ be the event that the car is behind that door; $P(A)=\frac14$. Now Monty opens a door, and suppose the player "switches" to one of the remaining doors at random; Let $B$ be the event that the car is behind that door, $C$ the event that the car is behind the remaining closed door (not $A$): then $P(B \cup C \mid \text{events so far})=\frac34$, and since the player chose randomly, $P(B \mid \text{events so far})=\frac38$.

Now Monty opens a second door. The probability the car is behind that door is $P(A \cup C \mid \text{events so far})=\frac58$. Therefore the player should switch again, but players who follow this strategy ("switch twice") win only $\frac58$ of games whereas players who followed the "stay then switch" strategy win $\frac34$ of games.

With more doors the computations of probability get even more complicated.