[montyhallstuff]

Consider 3 doors: A, B, C. If you pick the right one you win the prize. Say you pick door A. Monty Hall opens B or C and reveals that there's nothing behind. Now he asks if you want to stay with A or switch to the closed door. What should you do to maximize your chances of winning?

Right before any of the doors is opened we have no idea what to expect, so there are 3 equaly likely possibilities:

1. prize is behind A 2. prize is behind B 3. prize is behind C.

In case (1) you win by staying with your choice of A. In case (2) Monty Hall must open C (by the rules), so you'll win if you switch to B and in (3) the host will open B(by the rules) and you should switch to C.

In 2 cases out of 3 (2/3) you win by switching and in only one case out of 3 (1/3) you win by staying with your first choice. Since 2/3 > 1/3, you should switch.

This problem is easy if you know the defintions of sample space, event and the Equally Likely Probability Formula.

[russellspitzer]

I prefer thinking about the 100 doors case.

You are shown 100 doors 1 ... 100. Imagine you pick 1 door.

Of the other 99 doors Monty tells you which 98 of them don't have the prize.

He now asks if you want to switch to the 99th door that he didn't cancel out.

Intuitively that seems to make much more sense to me.

[posterboy]

Yes, indeed it did for me, as well, but then I'm not trusting my intuition that failed me for the three door version, and I'm desperately trying to prove it wrong, because I'm biased by my first commitment to an answer.

I think that commitment bias or what it's called is also an explanation for the misunderstanding in the first place. The argument against the increased chance, as I would use, works both ways and that is much more apparent the smaller the initial chance is. But I only intuitively used it to defend my position.

[montyhallstuff]

Just in case....

A sample space is the set of all possible outcomes of a random experiment. An event is a subset of a sample space.

Let S be the finite sample space with equally likely outcomes in it. Let E be an event. Then the probability of E is the number of outcomes in E over the total number of outcomes in S.

Now consider 52 cards in a deck. The sample space of outcomes here are the 52 cards. What is the event that the chosen card is a black face card? E = {J-club,Q-club,K-club,J-spade,Q-spade,K-spade}. What is the probability that the chosen card is a black face card? 6/52 by the formula above.

Monty Hall problem can be solved by using the above as a guide.

Sample space here : {switch, switch, stay-with-your-original-choice}. By the formula above, the probability of switching is 2/3 and the probability of staying put is 1/3.

If you're doubting the formula, I can provide you with a proof.

[retsibsi]

As I mentioned in another comment, it's important to note that Monty Hall is guaranteed to open a door with a goat behind it (a losing door). If he chose randomly and just happened to reveal a goat, the answer would be different.

[disgruntledphd2]

Indeed. This is the major reason I don't like the Monty Hall problem. When using stats and probability (at least in science), the universe is not your adversary.