During a train trip I cooked up a semi-autobiographical variation of The Monty Hall problem for people who already know the original. Perhaps it is more of a meta-puzzle than a puzzle, but I hope it is suitable for this site. (At least it got some people disagreeing who knew the correct answer to the original quite well.) So here goes.

John M. is a television mogul who at Friday night watches a new game show, produced by one of his production companies and aired on one of his networks. In it the host, Monty Hall, shows the candidate three doors to choose from. Behind one is a car and behind the others are goats.

'Good', murmurs John. 'Two doors with goats, so 2/3 chance for the candidate to make the wrong choice. This will only cost me one car per three episodes on average. Very good.' But then something annoying happens: after the candidate has chosen a door, the host announces that he will always open one of the doors the candidate did not choose, and even more annoyingly, always a door with a goat behind it, after which the candidate can choose again.

'Aaaaargh!' Shouts John at the monitor. 'Does Monty think cars grow on the tree in my garden?! This will cost me at least one car every two shows! But this will not stand! I will tell Monty to stop this nonsense tomorrow first thing in the morning!'

So the next morning John takes the train to Delft (where Monty lives) to give him a stern talking to. After said talking, on his way back home, John arrives again at Delft station (see picture) and finds himself in the situation of needing to choose between two blue stairways: the rightmost stairway leading to platform 1 or the leftmost stairway leading to platform 2 and 3. Too busy to check his phone he takes the left stairway so that he has 2/3 chance for his train to arrive at a railway that is adjacent to where he is standing.

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But then something annoying happens. John suddenly remembers that he arrived, earlier that morning, on platform 2 and hence his train will not return form that platform (as apparently trains on rail 2 travel away from his house instead of towards it) leaving platforms 1 and 3 as the only viable options for his train to arrive.

Question: is (considering losing a car 'as bad as' missing a train) this situation on Saturday equivalent to the situation on Friday, i.e. to the Monty Hall problem? If yes, should he change platforms? If no what is the difference?

  • $\begingroup$ #LateralThinking we could just go online ourselves and see which tracks go opposite track 2. I don't think that's what you want, though. $\endgroup$ – Brandon_J Apr 9 '19 at 17:42
  • $\begingroup$ "Too busy to check his phone he takes the left stairway so that he has 2/3 chance for his train to arrive at a railway that is adjacent to where he is standing." Where does 2/3 come from? Does the train randomly choose a platform with each being equally likely? $\endgroup$ – noedne Apr 9 '19 at 22:17
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    $\begingroup$ Yes. Strictly speaking the train has already chosen its platform and there is no more randomness there, but from John's perspective all three platforms are equally likely to be the right one so he can 'safely' model it as a probability question. Maybe a more correct analogy is that taking the left stairs yields a 66.7% Confidence Interval around the location of the right platform. $\endgroup$ – Vincent Apr 10 '19 at 7:21
  • $\begingroup$ Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :) $\endgroup$ – Rubio Apr 12 '19 at 8:23

A difference between the Monty Hall scenario and the train platform scenario is that

Monty will always select a door the contestant didn't pick, whereas (if I'm understanding the train-platform scenario correctly) the platform you suddenly remember can't be the right one might be the one you're already on.

After you remember that the train will definitely not be leaving from platform 2,

it is equally likely (as far as your state of knowledge goes) to leave from platform 1 or from platform 3. So in the situation described -- where you are choosing between {1} and {2,3} -- you shouldn't care whether you switch or not. (So you might as well stay put and save yourself some walking.) This is not the same as in the Monty Hall situation, where the new information you gain from the host's revelation of a goat is enough to give you a 2/3 chance of winning if you switch, versus a 1/3 chance if you stay.

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  • $\begingroup$ It is common expand this problem to 100 doors for people to understand the Monty Hall problem. This also helps in this case: if you had 100 platforms and you chose just one, and then remembered that 98 of them were not correct, the platform you chose and the other one left have equal probability of being correct, because the other platforms were not excluded based on your choice (in oposition to the Monty Hall problem). $\endgroup$ – cinico Apr 10 '19 at 10:36
  • $\begingroup$ This gets to the core of it. The biggest difficulty people have with understanding the Monty Hall problem is that the rule you enunciate in the first spoiler block often isn't explicitly stated by the person posing the riddle. $\endgroup$ – Monty Harder Apr 10 '19 at 14:51

Here's how the original Monty Hall worked:

You pick a door. (2/3 chance of it being a goat, 1/3 chance of it being a car)

Monty Hall opens a door that he knows has a goat in it, and then asks you if you'd like to switch to the other door.

2/3 of the time, the door Monty Hall opened is the only other door with a goat - because 2/3 of the time, you picked a goat with your first guess.

Thus, if you switch, the probability of getting a car is exactly 2/3. If you don't switch, the probability is the expected 1/3.

Here's how your situation works:

Let each track be a "door". Let the wrong tracks be "goats" and the right track be "the car".

You are allowed to pick either doors three and two or simply door one. You logically pick doors three and two, knowing that at least one of them is not correct. Knowing that door two is a "goat" does not give you an incentive to switch doors. You already knew that at least one of them was going to be incorrect; however you didn't know which door it would be. Now it's down to doors 1 and 3 - a 50-50 chance.


As Gareth pointed out - there's no guarantee as to which platform you remembered, which is different from the original M, where it was always on a platform you didn't pick.

In conclusion,

The situations are completely different. Switching gives you a 50-50 in your new scenario; it gives you 2/3 in the original.

I also ran a simulation to back up my final answer.

Here's the code for it:

public static void main(String[] args) {

    int correct = 0;
    int remembered=0;
    int ourScenario=0;
    int rightTrack=0;

    for(int j=0; j<10; j++) {
        for(int i=0; i<1000; i++) {
            correct = (int)(Math.random()*3)+1;
            do {
                remembered = (int)(Math.random()*3)+1;
            if(remembered==2||remembered==3) {
                if(correct==2||correct==3) {
        System.out.println("In 1,000 trials, our scenario occurred " + ourScenario +
                " times. Out of that, we were already on the correct platform " + rightTrack + " times\n");

Here is a sample output of the code:


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  • $\begingroup$ The actual behavior of real-life game show host Monty Hall was to sometimes allow contestants to switch and sometimes not. Whether switching would be beneficial or not would depending upon the odds that Monty Hall would allow a contestant to switch after picking the car, and the odds that he would allow a contestant to switch after picking the goat. Switching could be a guaranteed win, a guaranteed loss, or anywhere in-between. $\endgroup$ – supercat Apr 9 '19 at 21:09

When doing Bayesian updating on new information, we should change our probability assignment for a hypothesis only if the probability of seeing the new information depends on whether the hypothesis is true.

Suppose we choose Door A, and Monty Hall reveals a goat behind Door B. If Monty Hall always randomly chooses between the unchosen doors when we initially choose a car, and reveals the other goat when we initially choose a goat, then the probability of seeing a goat behind Door B is 50% regardless of whether Door A has a car or not. If Door A has a car, then Monty Hall has a 50% chance of choosing to reveal Door B, and 50% of choosing to reveal Door C. If Door A has a goat, then Door B has a 50% of being a goat as well, forcing Monty to reveal it. So seeing that Door B has a goat doesn't change the probability that our initial choice was correct.

However, it does give information about Door C. If Door C has a car, then we will definitely see that Door B has a goat. But if Door C has a goat, then Monty Hall might reveal that goat, rather than showing a goat behind Door B. Since the probability of seeing a goat behind Door B, given that Door C has a car, is greater than the probability of seeing a goat behind Door B, given that Door C has a goat, seeing a goat behind Door B increases the probability of C having a car.

In the train case, the probability of the previous train coming on Platform 2, given that Platform 1 is the current correct platform, is 50%, which is larger than the probability given that Platform 1 is not the correct one (33%). Both Platform 1 and Platform 3 are made more likely by Platform 2 being incorrect, and they are symmetric, so both are 50%.

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This problem is

equivalent to the original Monty Hall problem.

Suppose a contestant is playing on Monty Hall's game show. Furthermore, their game is linked with your world in the following way. Each of the game's three doors corresponds to one of the three train platforms, where the door hiding the car corresponds with the platform that will have your desired train. When the contestant first picks a door, say No. 1, then you are led to the other corresponding train platforms, No. 2 and No. 3. Then, Monty reveals that one of the doors the contestant didn't pick hides a goat, so one of your train platforms is the wrong one.

Just like the original Monty Hall problem, the probabilistic answer depends on what assumptions we make about this underspecified game. The first assumption we will make is that when the host makes a random choice from a set of doors, he is equally likely to choose any one of them.

  • If the host can reveal the door the contestant chose or the door hiding the car, then in the current situation doors No. 1 and No. 3 are equally likely to hide the car.
  • However, if the host always reveals a door that the contestant did not choose and hides a goat, then in the current situation door No. 3 is twice as likely to hide the car as door No. 1 is.

(To see this second situation more intuitively, imagine there were originally 100 doors, and the host reveals that all but one of the doors the contestant didn't pick hide a goat. Now the contestant should definitely switch!)

Thus, the answer of whether he should change platforms is that

it depends on how the situation is modeled: like the original Monty Hall problem, the question as stated is missing crucial details.

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    $\begingroup$ Your two spoilered boxes don't appear to agree (to me). $\endgroup$ – Brandon_J Apr 9 '19 at 20:18
  • $\begingroup$ @Brandon_J One says "equivalent to the original Monty Hall problem" and the other "like the original Monty Hall problem." What is the discrepancy? $\endgroup$ – noedne Apr 9 '19 at 20:19
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    $\begingroup$ The original MH problem is not missing details; there was no "depends" in your first box. $\endgroup$ – Brandon_J Apr 9 '19 at 20:19
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    $\begingroup$ Ah, I'm on mobile r n. Pretty sure the wikipedia article said that the host knew to open a goat door. Will look when I get off mobile. See my answer for my version also $\endgroup$ – Brandon_J Apr 9 '19 at 20:24
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    $\begingroup$ The problem doesn't seem equivalent to me. If you chose door/platform 2 then platform 2 would still be revealed as a wrong option, despite the fact that you chose it. Also you can't actually choose just platform two or just platform three. You have two choices which are "platform 1" or "platforms 2 and 3". Perhaps you could expand your answer to be clearer on how these two choices map to the monty hall problem's doors... $\endgroup$ – Chris Apr 10 '19 at 10:10

This situation:

Differs because initially he had a 2/3 chance of a good outcome and 1/3 chance of bad, which was then reduced to the 1/2 choice either way.
In Monty Hall it's a 2/3 chance of bad outcome and then 1/2 choice. So this situation would be similar to there being 2 cars and one of the car options being shown and removed.
If I remember correctly, in Monty Hall, the chance isn't actually 1/2, but a combination of 2/3 and 1/2 for the bad outcome that is slightly better than 2/3 but not as good as 1/2. >! Therefore in this slightly different situation, he shouldn't switch as it's the "opposite" of the first problem.

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    $\begingroup$ In the original Monty Hall, the probability of getting a car after switching was exactly two-thirds. $\endgroup$ – Brandon_J Apr 9 '19 at 16:05

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