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.

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    $\begingroup$ no. The reason its not 50/50 in the original problem is because he knows where the goats are. Here hes just opening them at random. The fact the car hasn't shown up doesnt make it any more likely that its behind your door, and not the other one. $\endgroup$ – XRBtoTheMOON Nov 30 '17 at 3:35
  • $\begingroup$ @btcgrl No, he's not just opening all the doors at random. There's still the condition that he can't open our chosen door. He can only open 98 of the remaining 99 doors at random. And the contestant knows this. $\endgroup$ – Ryder Rude Nov 30 '17 at 3:45
  • $\begingroup$ @RyderRude That makes no difference. He is opening up the remaining 99 doors at random---we gain no additional information from his choices. I think that you falling into a trap related to the gambler's fallacy (or, perhaps, some kind of hot hand fallacy?). $\endgroup$ – Xander Henderson Nov 30 '17 at 3:47
  • $\begingroup$ @btcgrl I'm just saying that if we assume, for a minute, that the car is in the remaining 99 doors, the chances that it survived 98 random door openings is very low. Hence our assumption is unlikely. $\endgroup$ – Ryder Rude Nov 30 '17 at 3:49
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    $\begingroup$ Yes. Originally having selected a door your probability of having chosen the car was $\frac{1}{100}$, and each time that he opened a door with a goat behind it, the probability that the car is behind a specific untouched door or behind the specific door that you chose at the beginning all go up by the same amount until you finally are left with the door you picked at the start and the final door that he didn't select both having probability $\frac{1}{2}$. $\endgroup$ – JMoravitz Nov 30 '17 at 3:54

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.

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 $1/2$.

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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.

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

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