Hi guys,
Well, I am sure everyone knows the problem of three doors where the host of the show opens one of them after the player had already chosen one. In case you do not know the game yet, check its wikipedia entry: http://en.wikipedia.org/wiki/Monty_Hall_problem
It is an old problem I know. Chances are whoever is reading this is familiar with the answers already. But something that may go unnoticed is the fact of what causes confusion. And from where does this confusion actually comes from?
So I gonna introduce 2 visions. First, what I call the host's vision. And later the player's perspective. But first, some definitions:
I am the player, so I decided to open door number 1. Suppose there are 3 possible hypotheses: H1, H2, H3. All of them are respective to the case where the door with the prize is number 1,2 or 3. Then let D be the data available, which is whether the host decided to open door 2 (D2) or door 3 (D3). This is pretty usual, I know =)
P(H1)=P(H2)=P(H3)=1/3
The Host
From the host's perspective, there is (probability of choosing door 2 or 3, given that some hypothesis is true):
P(D2|H1) = 1/2 P(D3|H1)= 1/2
P(D2|H2)= 0 P(D3|H2)= 1
P(D2|H3) = 1 P(D3|H3)= 0
And then, we calculate using Bayes, the generic equation:
P(Hi|Dd)= P(Dd|Hi)P(Hi) / P(Dd)
Now, just substituting the probabilities. We have the possibility of the hypotheses being correct, given that the host opens door 2 or door 3:
P(H1|D2)= 1/3
P(H2|D2)= 0
P(H3|D2)= 2/3
P(H1|D3)= 1/3
P(H2|D3)= 2/3
P(H3|D3)= 0
That is, changing from door 1 is always nice. Right?
Yes, right. But lets see from a different point of view.
The Player
If we think from the player perspective, there are only two doors left. And the chance now is 1/2 for either one. So it does not matter changing.
Actually, if we get the last answers and just pick a door at random, we get the 1/2. For example, suppose the host opens door 2:
1/2*P(H1|D2) + 1/2*P(H3|D2) = 1/2*1/3 + 1/2*2/3= 1/2
So, if we choose at random. We get in fact the 1/2. And that is exactly the same probability that we think intuitively.
This is where the confusion arises. We think that either way should be correct and the other false. But actually, both are correct in this case. The player is thinking without priors (that is, he is not giving a different weight for either door). What is very plausible in choosing things that are the same.
But... in this case there are priors! When the host opens door 2, the door 3 increases its chance of being the correct, much more than the door 1 (chosen by the player)!!!
Why this, you may ask??
The game's rules create this. The door chosen by the player is static, it will not change by the host actions. The other doors, however, will kind of evolve. They will be selected wisely to left only the "most interesting available". Because the host can not remove the prize's door!
This happens, because the remaining doors must have the prize! And simply because the player's door are not being affected by the host's tip (the door removal process). The player's door remains with its probability while the other door rise its probability.
We human beings tend to give equal possibility for everything which should be the same. But sometimes, they are more differences than we can easily perceive.
Thus, there are different dynamics between the chosen door and the rest of the doors is crucial. And this is hidden by non trivial rules, which purposely don't make this explicit.
