Reverse Monty Hall problem

Question

First, we pick one door out of 100 doors. Then the host Monty Hall opens 98 doors out of the remaining 99 doors one by one. But this time he doesn't intentionally open the doors with goats. He just happens to. He didn't know which door had the car. He was just opening 98 out of the remaining 99 doors at random.

Also, if the host opens the car door, we lose. But luckily, Monty happens to reveal 98 goats. Now, is it rational to believe that if the car was in the remaining 99 doors, chances are that it'd have been revealed by now? So, sticking to the originally chosen door is an advantage?

EDIT Suppose the car is in one of the remaining 99 doors. Then the probability that it's not revealed in 98 random door openings is $\frac{1}{99}$. So, 98 out of 99 times the car would have been revealed. The fact that it wasn't revealed makes it very unlikely for the car to be in the remaining one door.

Answer 1

Let $A$ be the event of picking the door hiding the car.

Let $B$ be the event of Monty revealing 98 goats.

Let $C$ be the condition that Monty cheats.

We seek $\mathsf P(A\mid B, C)$ and $\mathsf P(A\mid B,C^\complement)$, the probability that the contestant picked the door hiding the car when Monty reveals 98 goats under either condition (cheat, or not).   By Bayes Rule, and the Law of Total Probability, these are:

$$\begin{split}\mathsf P(A\mid B, C) &~=~& \dfrac{\mathsf P(B\mid A, C)\mathsf P(A\mid C)}{\mathsf P(B\mid A, C)\mathsf P(A\mid C)+\mathsf P(B\mid A^\complement,C)\mathsf P(A^\complement\mid C)}\\\mathsf P(A\mid B, C^\complement) &~=~& \dfrac{\mathsf P(B\mid A, C^\complement)\mathsf P(A\mid C^\complement)}{\mathsf P(B\mid A, C^\complement)\mathsf P(A\mid C^\complement)+\mathsf P(B\mid A^\complement,C^\complement)\mathsf P(A^\complement\mid C^\complement)}\end{split}$$

Now, in either case, the probability for picking the door hiding the car is $1/100$.   Further, if the contestant does so, only goats can be revealed.

$$\begin{split}\mathsf P(A\mid B, C) &~=~& \dfrac{1}{1+99\mathsf P(B\mid A^\complement,C)}\\\mathsf P(A\mid B, C^\complement) &~=~& \dfrac{1}{1+99\mathsf P(B\mid A^\complement,C^\complement)}\end{split}$$

Now, when Monty cheats he can certainly avoid revealing the car, so $\mathsf P(B\mid A^\complement, C)=1$.  However, when Monty does not cheat, it is quite unlikely that he would miss the car when the contestant has not picked its door. $\mathsf P(B\mid A^\complement, C^\complement)=1/99$.

$$\begin{split}\mathsf P(A\mid B,C)&~=~&\dfrac{1}{100}\\\mathsf P(A\mid B, C^\complement) &~=~& \dfrac{1}{2}\end{split}$$

So if Monty is 'cheating' then you should switch doors, and if Monty is 'honest' then, eh, you may as well switch.

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Now, is it rational to believe that if the car was in the remaining 99 doors, chances are that it'd have been revealed by now?

Yes.   However, it is more likely that the car was among those 99 doors than that it was behind the 1 door the contestant picked.   It balances out when Monty does not cheat.

There is a probability of $1/100$ that the contestant picked the car and Monty reveals $98$ of the $99$ goats behind the other doors.   There is a probability of $(99/100)(1/99)$ that the contestant missed the car and that Monty did so too. So it is rational to believe it unlikely that either of these events would happen ($1/50$). However, when given that one from the two did happen, the (conditional)probability that the former happened is $1/2$.

Answer 2

In this situation, the chances of the car being behind either of the remaining doors is 50%.

Here's a simple way to see why this is true: we can think of an equivalent problem in which two players get to select doors. Suppose you select door 1 and your adversary selects door 2. Monty then opens the other 98 doors, revealing only goats. Thus, either you or your adversary has won. Clearly, there is no bias between you and your adversary; you have an equal chance of winning.

To clearly articulate why this is different than the classical Monty Hall problem; here, your adversary has randomly chosen a door to open. This is equivalent to Monty randomly selecting 1 of the 99 doors to not open, then opening all the rest. In the classical problem, Monty's choice isn't random, so this equivalence does not hold and thus you have the famous advantage for switching in that scenario.

Answer 3

Since the OP apparently doesn't want an intuitive explanation as to why the answer should be $50\%$, approach instead by definition.

Let $A$ be the event that the door you selected happened to be the door with the car. Let $B$ be the event that among the $98$ doors that Monty opened, the car was not among them. We are tasked with calculating $Pr(A\mid B)$.

By Bayes' theorem: $Pr(A\mid B)=\frac{Pr(B\mid A)Pr(A)}{Pr(B)}$

Calculating each individual term:

• $Pr(B\mid A)$, the probability that the 98 doors he opens all have goats given that the car was in fact behind the door that we selected, will trivially equal $1$.

• $Pr(A)=\frac{1}{100}$, by direct observation.

• $Pr(B\mid A^c)$, the probability that the car is not found within the doors Monty opens given that there is a goat behind the door you chose, can be seen to equal $\frac{1}{99}$, either by relating it to the probability that the the door Monty decided not to open containing the car, or by expanding it out for each door in sequence that was chosen not having the car which would expand as $\frac{98}{99}\cdot\frac{97}{98}\cdots\frac{1}{2}=\frac{1}{99}$

• By total probability, $Pr(B)=Pr(B\cap A)+Pr(B\cap A^c)=Pr(A)Pr(B\mid A)+Pr(A^c)Pr(B\mid A^c) = \frac{1}{100}\cdot 1 + \frac{99}{100}\cdot\frac{1}{99}=\frac{2}{100}$

So, $Pr(A\mid B)=\frac{1\cdot\frac{1}{100}}{\frac{2}{100}}=\frac{1}{2}$