Yet Another Monty Hall Question - Please advise if alternative scenario proves the same principle

Question

Okay, I'm very embarrassed that there are already 71 questions (based on search of "monty hall") and I'm going to post another one. I read the first 5 before succumbing to choice-overload. I'll try to keep this short and sweet.

• A host and contestant stand before 3 doors. The host advises the contestant that behind 1 of the 3 is a car while the other doors each have a goat.

• The host advises the contestant to choose 2 of the 3 doors to reveal if either has the car.

• The contestant chooses door 1 and door 3.

• The host advises that behind one of the doors chosen is a goat and asks if the contestant wants to keep doors 1 and 3 or switch to only revealing door 2.

Big Question: Is the probability of revealing the car higher if the contestant switches or stays with the original choice?

As I understand the original problem, the above has the same result, so the contestant should switch, but I can't wrap my head around the math and don't want to hurt my brain trying if I'm incorrect about the above fundamentally being the same scenario.

Also, if the above is the same, how is it any different from the contestant saying "3, no wait 2", since no matter which 2 doors are chosen (either by the contestant alone or with the help of the host, as in the original problem), we know that at least 1 door has a goat?

Last bit: If this is the same mathematical scenario, is it even less intuitive than the original or does it help clarify (to someone other than me) why the original works?

Addendum

Original MH problem, simplified:

There are 3 marbles in a bag; 2 are boring and grey, 1 is green. The host asks you to reach in and pull 1 out but not look at it. After doing this, the host, who can look into the bag, pulls out 1 grey marble. He then asks if you want to keep the 1 in your clutched hand or take the 1 still in the bag.

My version, simplified:

There are 3 marbles in a bag; 2 are boring and grey, 1 is green. The host asks you to reach in and pull 2 out but not look at them. After doing this, the host asks if you want to keep the 2 in your clutched hand or take the 1 still in the bag.

In both scenarios, 2 marbles are removed from the bag and 1 of those 2 is definitely grey. If we accept (and we all should at this point!) that in the first scenario the probability of the remaining door or marble being the winning choice is 2/3, shouldn't that hold true in the second scenario? If not, please explain at what point it diverges? If we know 1 of the 2 "out of the bag" is grey or a goat in either scenario, it shouldn't matter if we see which of the 2 it is, right?

Addendum 2

Thanks to Eric T for helping me get my head around this. With either of my modified scenarios, where my logic diverged was I allow the contestant to choose 2 doors and then keep both choices or switch, whereas in the original MH problem, the contestant is given a second "choice" with the host-reveal but still only keeps the original (or switches). One of my goals in creating this alternative was to eliminate the host variable which is a clear source of confusion (and trickery) in the original, leading to such misassumptions as

• The host's knowledge of what is behind all three doors creates a mathematical bias since he won't ever choose the car at random. If MH only presents the option when the car wasn't selected (to throw the contestant off), this would not change the math when testing for when the contestant chooses the car first. If the host always chooses a goat, it's because he always has at least one goat to choose from and is supposed to reveal a goat, not choose a door at random.

• Showing the contestant that one of the other 2 doors has a goat gives the contestant new information that affects the outcome. it is not eliminating the goat (seeing the goat) that makes switching more likely to reveal the car, it is eliminating the door.

If my variation has the pick-2 parameter but has the "only one door allowed" rule reinstated , switching is still the better option. Here is the final version:

• A host and contestant stand before 3 doors. The host advises the contestant that 1 of the 3 doors hides a car while the other doors each hide a goat.

• The host advises the contestant to choose 2 of the 3 doors to check for the car.

• The contestant chooses door 1 and door 3.

• The host asks the contestant to choose between opening 1 and 3 or switch to opening door 2.

In this scenario, the host has done nothing to interfere and the contestant knows that at least one of the two selected doors has a goat, but must still risk choosing the wrong door of his selected 2 or switching. While the odds may still seem 1/2 at first, the possibility that the contestant chose both goats and thus may have no chance with his current subset makes the better odds in switching clearer.

Last question: What would be the actual probability of choosing the car if we don't know if the car exists in the subset 1 or 0 times? Just a comment mentioning a concept or wiki page is fine. Just curious on the math and have no idea what search for.

Answer 1

If the host is under no obligations except not to lie, "behind one of the doors is a goat" reveals absolutely nothing. There is no conditional probability here. The chances of winning are 2/3 if the contestant stays, and 1/3 if he switches.

Also, if he wins, the goat should ride shotgun; they're notoriously bad drivers.

Edit: To answer your last comment, this version is more intuitive than the original. In the original, one has to interpret the information provided by the host's big reveal; in your variant, it is easy to see that the information is useless and that we can ignore the host.

Edit 2: I want to express the solution in an absolutely clear way that is not unique to this problem, because I think this mode of thought will be helpful to people who are perennially confused by these kinds of puzzles.

First, let's take the original Monty Hall problem. There are 3 doors, 2 hiding goats and one hiding a car. You choose a door uniformly at random. Now, there is a 1/3 chance that you've chosen the car, and 2/3 chance that you've chosen a goat. The host reveals a goat behind a door other than the one you've chosen. Now what?

2/3 of the time, you will be in this situation:

• you are standing in front of a door with a goat
• the other unopened door has a car
• if you switch, you will win
• if you stay, you will lose

1/3 of the time, you will be in this situation:

• you are standing in front of a door with a car
• the other unopened door has a goat
• if you switch, you will lose
• if you stay, you will win

2/3 of the time, switching is correct. Having no other information, you should switch.

Now, let's look at your problem in exactly the same way. There are 3 doors, 2 hiding goats and one hiding a car. You choose two doors uniformly at random. Now, there is a 2/3 chance that you've chosen the car and a goat, and 1/3 chance that you've chosen two goats. Regardless of what the host does:

2/3 of the time, you will be in this situation:

• you have chosen two doors, one with a car and one with a goat
• the unchosen door has a goat
• if you switch, you will lose
• if you stay, you will win

1/3 of the time, you will be in this situation:

• you have chosen two doors, both with goats
• the unchosen door has a car
• if you switch, you will win
• if you stay, you will lose

2/3 of the time, staying is correct. Having no other information, you should stay.

Answer 2

The contestant has a higher probability of winning the car if he stays. Let's label the items as $G_1$, $G_2$, and $C$. When the contestant chooses, he will have chosen $(G_1, G_2)$, $(C, G_1)$, or $(C, G_2)$, so the probability of choosing exactly two goats is $\frac{1}{3}$, and the probability of choosing a goat and a car is $\frac{2}{3}$. Switching is only beneficial when the contestant has two goats, but he probably has a goat and a car, so he shouldn't switch.

Perhaps you'll find the following convincing.

Suppose we had $1000$ doors, with a car behind one of the doors and $999$ goats in the rest. The host let's you choose 999 doors. Do you think it's a good idea to change your option to that one remaining door?

It's the same idea for $3$ doors.

Answer 3

Let me take your version with marbles, and let me show that you can transform the two problems into each other step by step so that it is obvious that in your scenario, it is better to stay.

OK, let's start by re-stating the marble version of Monty Hall again:

You take one marble out without looking at it. The host looks at the remaining marbles and removes a grey one. Now he offers you to switch. It is of advantage to you to switch.

So far, so good. But it is clear that it doesn't matter where the two marbles are when the show master removes one. Therefore it is obvious that the following is equivalent to the original:

You take one marble out with your right hand without looking. You then take out the two other marbles with your left hand without looking. The host then looks at the marbles in your left hand and removes a grey one. You then put the remaining marble of the left hand back. Now the host offers you to switch. Of course it's still better to switch.

But now it's obvious that putting the left hand marble back in is pointless. Therefore we can simplify it as follows:

You take a marble out with your right hand without looking. You then take out the two other marbles with your left hand without looking. The host looks at the marbles in the left hand and removes a grey one, so you now have a single marble in each hand. Then he offers you to choose whether you want the marble in your left hand, or the marble in your right hand. It is of advantage to take the left hand.

But it is clearly not relevant in which order you take out the marbles. Therefore the following version is again equivalent:

You take out two marbles with your left hand without looking. You then take the remaining marble out with your right hand without looking. The host looks at the marbles in the left hand and removes a grey one, so you now have a single marble in each hand. Then he offers you to choose whether you want the marble in your left hand, or the marble in your right hand. It is of advantage to take the marble in the left hand.

But again, the exact location of your marble doesn't matter. Therefore you can just omit taking the right-hand marble altogether, so you get:

You take out two marbles without looking. The host looks at the marbles in your hand and removes a grey one. Then he offers you to choose whether you want the marble in your hand, or the marble in the bag, in other words, he offers you to switch. It is of advantage to take the marble in your hand, that is, not to switch.

Now as a final step we observe that it doesn't matter whether the show master removes the grey marble or just tells you it is there, because after all, after looking at the marbles you'll be able to figure out which one is the green anyway. Therefore we arrive at your scenario:

You take out two marbles without looking. The host tells you that one of the marbles in your hand is grey. Then he offers you to switch. It is of advantage not to switch.

Indeed, we can go even a step further, by observing that the show master doesn't tell you anything you didn't already know: If there are two marbles in your hand, but only one is green, then there must be a grey marble in your hand. So we arrive at:

You take out two marbles hand without looking. Then the host offers you to switch. It is of advantage not to switch.

Now of course, you don't get any additional information after you took out the two marbles, therefore it doesn't change anything if you make the decision right from the beginning:

The host offers you the choice to either draw two marbles, or to first draw two marbles and then to switch. It is of advantage to just draw two marbles.

Of course drawing two marbles and then switching is equivalent to just drawing one marble, we arrive at the following situation:

The host offers you the choice to either draw two marbles, or just one marble. It is of advantage to draw two marbles.

Indeed, it is quite obvious to everyone that drawing two marbles is an advantage over drawing just one marble. Therefore the whole sequence in reverse can be used to demonstrate that in the original Monty Hall problem, it is of advantage to switch.