What is the mistake in the following contradiction to the solution of the Monty Hall problem?

Question

Let us say in a game show there are three doors - 1,2 and 3 - behind which are hidden 1 special prize and 2 useless objects. The host knows which door leads to which object. Let there be two contestants in the game - $A$ and $B$. First, $A$ is asked to play the game. $A$ chooses Door 1. Then the host opens Door 3 revealing that it is not the Door leading to the prize. He then asks if $A$ wants to switch. It is widely known that switching to Door 2 would give you twice the chance of winning, so he switches. Then, $B$, who doesn't know anything about $A$'s game is asked to come and play. He chooses Door 2. Then the host opens Door 3 again and asks $B$ if he wants to switch. $B$ says yes, because it gives him twice the chance of winning. So, for $A$, it is more likely that the prize is behind Door 2 and for $B$, it is more likely that the prize is behind Door 1. How can this happen if there is nothing unique about $A$ or $B$?

Answer 1

Look at it this way - suppose the host has decided he likes door #3, so he'll open it if at all possible. Two contestants play, blind to each other, and choose door #1 and door #2 respectively. There are three possibilities:

• If the prize is behind door #1, the host will open door #3 for both. Both players switch, one of them wins.

• If the prize is behind door #2, the host will open door #3 for both. Both players switch, one of them wins.

• If the prize is behind door #3, the host can't open his favorite door, door #3, so he'll open door #2 for the first player and door #1 for the second. Both players switch to door #3, and both players win.

By saying that the host opened door #3 for both, we rule out that last case, introducing new information not known to either player.

Answer 2

So to clarify the question: $A$ and $B$ both come to play the game and they each have their deterministic strategies--given the action of the host that is. $A$ picks Door 1 first and then after the host uncovers a door, picks the remaining door. Then $B$ picks Door 2 first and then after the host uncovers a door, picks the remaining door.

This means that (no matter whether or not the host favors opening Door 3) that $A$ wins and $B$ loses if the prize is behind Door #2, and $B$ wins and $A$ loses if the prize is behind Door #1, and both $A$ and $B$ win if the prize is behind Door #3.

This is another way of looking at it. Your analysis above didn't take into account that $A$ and $B$ with their strategies will be sometimes be forced into picking Door #3.

Let us now assume that both $A$ and $B$ each take their first picks as they have before, but now they both have secret knowledge that the host will always pick Door #3 whenever he can. Then both $A$ and $B$ know that the probability of being right changing their first picks when the Host picks #3 is $\frac{1}{2}$ (not $\frac{2}{3}$), while the probability of being right switching when the host picks another door besides #3 is actually 1. [make sure you see why]. So neither $A$ nor $B$ sees the prize as being more likely to be behind Door #1 instead of Door #2 or vice versa.

Answer 3

You can look at it this way: in order to ensure that he always opens a non-prize door before offering the player a chance to switch, we ensure that Monty already knows which door the prize is behind.

So in your scenario, you have at least three people who have estimated the probability that the car is behind door $1.$ Their three answers fall into one of the following two cases:

• Player A's estimated probability is $\frac13,$ Player B's estimated probability is $\frac23,$ and Monty's estimated probability is $1.$
• Player A's estimated probability is $\frac13,$ Player B's estimated probability is $\frac23,$ and Monty's estimated probability is $0.$

The "contradiction" you've found is just an extension of the not-always-comfortable application of probability to situations where different people have different knowledge. The probability that one estimates in those case is actually a conditional probability given what one knows.

• Player A knows that the rules prevented Monty from opening door $1,$ that Monty opened door $3,$ and that the car is not behind door $3.$

• Player B knows that the rules prevented Monty from opening door $2,$ that Monty opened door $3,$ and that the car is not behind door $3.$

What's "unique" about each player is that A knows something B does not know, and B knows something A does not know.

Monty, meanwhile, knows a lot more than either player:

• Monty knows that he opened door $3,$ that the car is behind door $x,$ and that he was prevented from opening either door $1$ or door $2$ by the players' choices (and in one case, also because the car was there).

If everyone knew what Monty knew, there would be none of this nonsense about probabilities of $\frac23$ or $\frac12$ that the car was behind a particular door; the probability would be either $1$ or $0.$