Xrhrft

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.

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?

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 tracks. You already knew that at least one of them was going to be incorrect; however you didn't know which platform it would be. Now it's down to platforms 1 and 3 - a 50-50 chance.

Also,

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;

            }while(remembered==correct);

            if(remembered==2||remembered==3) {

                ourScenario++;

                if(correct==2||correct==3) {

                    rightTrack++;

                }

            }

        }

        System.out.println("In 1,000 trials, our scenario occured " + ourScenario +

                " times. Out of that, we were already on the correct platform " + rightTrack + " timesn");

        ourScenario=0;

        rightTrack=0;

    }

}

Here is a sample output of the code:

In 1,000 trials, our scenario occured 676 times. Out of that, we were already on the correct platform 312 times

In 1,000 trials, our scenario occured 677 times. Out of that, we were already on the correct platform 346 times

In 1,000 trials, our scenario occured 666 times. Out of that, we were already on the correct platform 330 times

In 1,000 trials, our scenario occured 666 times. Out of that, we were already on the correct platform 372 times

In 1,000 trials, our scenario occured 657 times. Out of that, we were already on the correct platform 333 times

In 1,000 trials, our scenario occured 669 times. Out of that, we were already on the correct platform 325 times

In 1,000 trials, our scenario occured 661 times. Out of that, we were already on the correct platform 350 times

In 1,000 trials, our scenario occured 676 times. Out of that, we were already on the correct platform 353 times

In 1,000 trials, our scenario occured 667 times. Out of that, we were already on the correct platform 346 times

In 1,000 trials, our scenario occured 676 times. Out of that, we were already on the correct platform 345 times

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 2. 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.

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.