This is an archive of past discussions about Monty Hall problem. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page.
Q: There are clearly two doors, how can it not be 50/50?
A: The simplest explanation may be that any player's initial probability of not picking the car is 2/3. Monty's action of revealing a goat behind a door doesn't change that. Therefore, the player should accept the offer to switch. It doubles his likelihood of winning the car.
The subsequent Q & A explains the difference between the "conditional" approach and the simpler unconditional approaches.
Q: Why do some references insist an unconditional solution (like the one immediately above) does not address the problem "as stated"?
A: As usually stated, the problem has two steps. Step 1 is when the player initially picks a door. Step 2 is when the host opens a door. According to Morgan et al., Gillman, and Grinstead and Snell the problem is therefore about the conditional probability of winning by switching following step 2, at which point the player can see which of the remaining two doors the host opens. It is at this point that the player is given the opportunity to switch doors. These sources say the question pertains not to all players, but only to players who have picked a door (for example door 1) and then have seen the host open a different door (for example door 3). This distinction concerns the difference between
and
is the unconditional probability of winning by switching. It is roughly what fraction of all players who play the game will win by switching. Gillman suggests this would be the probability of interest in a revised version of the problem where "you need to announce before a door has been opened whether you plan to switch." [italics in the original!]
is the conditional probability of winning by switching given that the player initially picked door 1 and the host has opened door 3. It is roughly what fraction of all players who pick door 1 and see the host open door 3 will win by switching.
These are different questions which may or may not have the same numeric answer. Morgan et al., Gillman, and Grinstead and Snell all say the Monty Hall problem, as usually stated, asks the conditional question not the unconditional one.
An unconditional solution such as "a player has a 2/3 chance of initially picking a goat, all of these players who switch win the car, so 2/3 of all players who switch will win the car" correctly answers the unconditional question, but doesn't necessarily say anything about the conditional problem. For example, this solution says the answer for both the Parade version and the fully explicit Krauss and Wang version of the Monty Hall problem is 2/3. Morgan et al. and Gillman both introduce a different variant, identical to the Krauss and Wang version except instead of
If both remaining doors have goats behind them, he chooses one randomly
they specify
If both remaining doors have goats behind them, he chooses the leftmost door
In this variant, an unconditional solution also correctly answers the unconditional question. The player still has a 2/3 chance of initially picking a goat, and in this case the host must reveal the other goat, so the solution above still applies and 2/3 of all players who switch will win the car. However, in this variant, if the player initially picks door 1 and the host opens door 3 the player knows the car is behind door 2 and has a 100% probability of winning by switching. If the player initially picks door 1 and the host opens door 2, either the player's initial pick was the car (a 1/3 chance) or the car is behind door 3 (also a 1/3 chance). This means, given that the host has opened door 2, the player's chance of winning by switching is 1/2. The point of introducing this variant is to show the difference between the unconditional and conditional questions. In this variant, these questions have different answers exposing the difference between unconditional and conditional solutions.
Using an unconditional solution produces the correct numeric answer for the expicit Monty Hall problem as stated by Krauss and Wang, but it is not answering the conditional question these sources say is asked by the problem. The issue with the approach is hidden because in this version of the problem the conditional and unconditional answers are the same. Applying the same solution to a variant where the unconditional and conditional solutions are different, like the "leftmost door variant", reveals the issue.
Q: Why do most references ignore this "conditional" problem and address the puzzle more simply, equivalent to what Morgan et al. refer to as an "unconditional" solution?
A: The problem as stated is usually interpreted as having a certain symmetry, such that the naming, numbering, or positioning of the doors (1, 2, and 3, or left, middle, right) and the particular initial choice of the contestant and the choice of the host of which door to open carry no information that would distinguish a particular game state from the generic game state. This forces the "conditional" answer to be the same in all cases and also the same as the "unconditional" answer. As Selvin says "The basis to my solution is that Monty Hall knows which box contains the keys and when he can open either of two boxes without exposing the keys, he chooses between them at random" (Selvin 1975).
Since the door numbers are exchangeable, the player's initial choice might as well have been random. Suppose that this is the case, and that he happens to choose Door 1. We can now completely solve MHP by first using symmetry to argue that the relationship between the doors when identified only by their roles - door hiding car; door chosen by player, door opened by host, door left closed by host - is independent of the numbering of the doors. The identities of door chosen, door opened, door left closed could equally well be any of the six possibilities (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1). By symmetry of our initial state of knowledge, which possibility is actually realized is independent of the question whether the door first chosen or the door left closed hides the car. This implies that the player's decision whether to stay with door-chosen or to switch to door-left-closed can just as well be made in advance ignoring the door numbers of a particular case (in Vos Savant's words, *say*, (1,3,2)). They give him no further information. They are irrelevant.
This is an archive of past discussions about Monty Hall problem. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page.
