# Monty Hall Extended To Two People

Person One enters the Monty Hall problem as usual with usual rules. Of the three doors A, B and C, One chooses B, Monty opens C (goat) and One switches to A calculating that the probability of a car behind A is 2/3.

Before One is allowed to open door A he is asked to sit down and remain silent. He is told that he will get one more chance to switch but not until later. Monty closes door C. The car and goats remain where they are.

Person Two enters the same Monty Hall problem. Of the three doors Two chooses A, Monty opens C (same goat) and Two switches to B calculating that the probability of a car behind B is 2/3.

Before opening door B Monty tells Two he will be allowed one more switch of choice in a few minutes.

Before One and Two have to make their final choice they are allowed to discuss the situation. Both One and Two are brilliant statisticians and 100% honest people and both know that about each other.

One explains what he did and that he reached the conclusion to open A as with the information he had at the time that gave the probability that the car is behind B is 2/3. Two, having been watched by One, does not have to explain that he has calculated the probability that the car is behind A is 2/3.

One and two now get into a discussion to determine the probabilities of the car being behind A or B.

What conclusions do One and Two reach?

My first thoughts are that

P(Car behind A) + P(car behind B) = 2/3 + 2/3 > 1 impossible.

Now too much contradictory information, so now in a situation of a fifty fifty choice.

Any better thoughts?

EDIT 1 Following David K's (quite correct) request for assumptions to be made explicit here they are

1. Monty Hall is 100% honest
2. The goats and car are place behind A, B and C at random.
3. Monty Hall will always open a door with a goat behind it never the car
4. Monty Hall will always open the door that gives away the least information, when equal he will chose a door at random.

EDIT 2 Given these assumptions I believe Alexander Geldhof provided the solution. In order to clarify my thoughts I re-write that as

P(X) means the probability that the car is behind door X.

When One enters he calculates P(A) = P(B) = P(C) = 1/3

Once Monty opens C he knows P(A) = 2/3, P(B) = 1/3 and P(C) = 0;

Two enters and makes his choice and decides P(B) = 2/3.

One provides the following argument to Two

When you entered you calculated that P(B) = 2/3, this was based on what you knew and the logical assumption that P(A) = P(B) = P(c) = 1/3. However my and Monty's previous actions determined that when you entered the P(A) = 2/3, P(B) = 1/3 and P(C) = 0 and I can give you that information.

Once your first choice was A, Monty could only open door C. If the car was behind B he could not open it so had to open C. If the goat was behind B then when he opened it we would know that the goats were behind B and C and so the car was behind A. So he had to open C to give us the least information. In fact this provided no new information since P(C) = 0 is already known.

So given all the information we are currently provided with it remains P(A) = 2/3, P(B) = 1/3 and P(C) = 0. So the best chance of the car is for us to open door A

• There are several assumptions which could be made here - is the host constrained in the second round to open the same door if possible? If the host's choice is constrained then the situation is rather different on the second round from the first. May 11 '19 at 17:24
• @MarkBennet I think he has to be; if he opens a different door then A knows exactly where the car is. May 11 '19 at 17:50
• @IanF1 So what? Monty has no way to guarantee keeping the car's location secret; Two could have made C their first choice. If Monty is willing to "give away the car" that way, why not by opening door B the second time? May 11 '19 at 18:55

At the start, there are three equally likely car-distributions: car behind A, car behind B, car behind C.

Now One comes in and chooses door B. Monty is now forced to open a goat-door which is not B, as by the hidden rule of this problem. In the A-situation, he'd be forced to open C, in the C-situation, he'd be forced to open A, and in the B-case, he can choose doors.

Here's where the trouble starts. By naming Monty's door C, you've merged the A- and C-situations; both are equally likely, so now there's a 2 in 3 chance of the car being behind A, 1 in 3 chance of the car being in B, and ZERO chance of the car being in C.

Two enters. He chooses door A, where the car is going to be two times out of three. What happens after this does not really matter - swapping is only going to net him the car one out of three times. With the extra information of A, this would also be the conclusion of B, being a perfect statistician.

Note that if Two would choose a random door, and Monty would again be forced to open a goat-door, without extra information, swapping would give him the car two times out of three! You just hand-picked the worst option for Two (door A), and by doing so fudged the answer

• No, this is wrong. There are hidden assumptions in the problem statement by OP (as there are hidden assumptions in the original problem). Monty cannot open A in the problem as stated, as then he would have opened both goat-doors and the choice would be rather trivial. May 11 '19 at 16:16
• To recap: One enters and chooses door B randomly. Now Monty opens a goat-door which is NOT B. There's a 2/3 chance that the car is behind A or C, so Monty is forced to pick whatever he chooses; leaving a 2/3 chance of the remaining door having the car. Let's assume he picks C. Now Two enters and chooses a door which is NOT door B and NOT Monty's door C. There's obviously only one pick left - that's where this problem statement has a hidden assumption. Monty now has to open a door which is NOT B, and NOT the door Two picked, A - obviously he can only open C. One knows this and gains no info. May 11 '19 at 16:24
• You are correct. Please edit your answer a little so I can reverse my downvote. May 11 '19 at 16:30
• I have edited my answer. May 11 '19 at 16:32
• I've changed my mind again. You are making the assumption that Monty will never open door B for Two, but that isn't given in the problem. I think that the OP means that the game is played as usual with Two, regardless of what happened in the game with One. In that case, the probability that the car is behind door A is $\frac12$ as the OP said. I think you are correct in pointing out that this issue needs to be addressed in the problem statement. May 11 '19 at 16:59

When the single-round Monty Hall question is asked with "usual" rules, the question often makes the unspoken assumption, which in my opinion really ought to be made an explicit assumption whenever the standard Monty Hall problem is posed, that there is some kind of contract between Monty and the contestants that says that no matter which door the contestant chooses, Monty will always reveal a goat behind a different door and then offer the chance to switch. So let's please be very explicit about this. We assume Monty always follows this rule.

(Without that assumption you can have a "trickster Monty" scenario in which Monty only does the show-goat-offer-switch thing when the contestant's first choice was the car.)

The second unspoken assumption of the standard Monty Hall problem, which I think can be left unspoken without too much harm in that context, is that the contestant's first choice is the car, Monty is equally likely to reveal either of the goats. This seems a reasonable thing to assume while solving the problem, because in the standard one-contestant Monty Hall there is complete symmetry between those two doors; for every reason why Monty might choose one door, there is an equally compelling reason why he might choose the other.

In the two-contestant scenario this symmetry is destroyed. We know the car is either behind door A or door B, so we know that on at least one of the occasions when Monty opened door C, he was forced to do so and it was not simply a random choice between two doors.

So now we have to make some assumptions about how Monty manages this asymmetric behavior. Does Monty run the second round of the game the exact same way he ran the first? Then in the case where the car actually was behind door A, with probability $$\frac12$$ Monty shows the second contestant the goat behind door B rather than door C. The fact that he did not do so is Bayesian evidence that the car is behind door B. It's just exactly enough evidence to cancel out the evidence from the first round. (In two thirds of all games with the initial B-C combination, the car is behind door A, but in only half of those games we see door C opened a second time, whereas all of the other games see door C opened again since the car is behind door B.)

On the other hand, perhaps Monty likes to give away as little information as possible. After all, if he did exactly as he did in the question except that he randomly opened door B after the second contestant's guess, after the contestants compared stories they would know exactly where the car is. So perhaps Monty plays the first round in the standard way, but in the second round he tries to open the same door if possible. Then the only way the second round would give any new evidence is if the second contestant happened to choose door C, forcing Monty to reveal the second goat. Since that did not happen, a good estimate of the probability at the end of the second round is the same as at the end of the first round.

Yet another possibility is that when Monty has a choice, he chooses the door that is later in alphabetical order. In the standard game it is just as plausible that Monty always chooses the door that is earlier in alphabetical order when he has a choice, but we know he had a choice in at least one of the rounds and still chose C, so that preference is ruled out. If it were known that Monty preferred to open the "later" door, then the only thing either contestant could conclude (before comparing notes) is that the car is equally likely to be behind door A or door B, and after comparing notes that conclusion is unchanged.

Since we don't know why Monty opened door C both times, we don't know whether it was just a chance occurrence from playing two rounds by the standard rules, whether it was a deliberate "same as the last time if I can" choice, or whether it was evidence of a bias Monty has toward choosing door C. Those are three very different reasons, and I see no obvious way to assign credence to each of those possibilities. (Even assigning equal probability seems unjustified, since there is none of the symmetry among the reasons that we usually use when we assign equal probabilities.)

So until you tell the contestants (and us) how Monty decides which door to open each time, I would expect the contestants to assign indeterminate probability to whether the car is behind door A. However, assuming Monty could not predict which doors the contestants would choose, I would say the probability of A is at least as good as the probability of B, so I would advise both contestants to make A their final choice.

Update: Based on the edits to the question, in particular the explicit assumption that "Monty Hall will always open the door that gives away the least information," we are in the case I described as "Monty likes to give away as little information as possible," and the conclusion by Persons One and Two should be that the car as probability $$\frac23$$ to be behind door A. This was the proper conclusion at the end of Person One's interaction with Monty, and the only thing Person Two could have done to find out any new information would have been to choose door C.

• Have edited question to make the assumptions I made explicit. May 12 '19 at 9:50