I want to attempt to explain an elementary problem that has been discussed frequently in the mass public eye. Marilyn Vos Savant (or if you like, Marilyn "Not So" Savant) discussed the problem in a column in Parade magazine. Her solution was correct, but her explanation (I believe) is lacking. Said column generated a huge response. The problem goes as follows...

Suppose we are a contestant on a game show a la "Let's Make a Deal" (hence the name of the problem). The host confronts us with three doors. Behind one door, we are told, is a fabulous prize. There is nothing of value behind the other doors. Suppose, without loss, we choose door #1 (if this bothers you, then just imagine we repaint all the doors so that #1 is the one we chose). After we make our choice, the host will open one of the other doors not containing the prize and will give us the chance to 'switch' doors (i.e. choose the other remaining door).

The question: Should we switch?

Incorrect solution : One's intuition might say "well, at that stage, there's a 50/50 chance that the prize is behind the door we chose (or the other door, for that matter), so we have an equal chance either way. So there's really no reason to switch.

Unfortunately, intuition proves to be wrong. We will show that we have a 2/3 chance of winning by switching doors when asked. We will try to give a coherent analysis of why this is the case.

Correct solution:

There are three cases.  Assumingthe prize is randomly placed (or at least we have no knowledge of where it is placed), each of these cases will occur with probability 1/3.

Case 1: Prize behind door 1. (Remember, we selected door 1 initially).

prize---empty---empty

--1---------2--------3---

In this case, let's assume the host will randomly open either door 2 or door 3. Regardless, in this case, if we switch doors we will lose.

Case 2: Prize behind door 2.

empty---prize---empty

--1---------2-------3---

In this case, assuming the host will never show us where the prize actually is before giving us the choice to switch (that wouldn't be much of a game, of course), the host is forced to open door 3 here. Thus, in this case, if we switch doors (i.e. switch from door 1 to door 2, which remains), we win.

Case 3: Prize behind door 3.

Hopefully, it's clear enough that this case is, for all practical purposes, identical to case 2. If we switch we win.

The punchline: Observe we may be in cases 1,2, or 3 each with equal probability of 1/3 (assuming the prize was randomly placed). Note that, since switching in cases 2 and 3 lead to a win, this says the probability of winning by switching doors is 2/3, while the probability of winning by "standing pat" is only 1/3.

Comment #1: Many people are very skeptical of this result. The mathematics here doesn't lie, but for the skeptic, it is possible to observe the fruits of this result by simulation. Get a friend and play this game many times, or write a computer program to simulate this game. You will see that players who switch will win close to 2/3 of the time (by the law of large numbers, as the number of trials approaches infinity, the relative frequency of trials where a 'switching' player wins will approach 2/3).

Comment #2: For the mathematician, we can make all of this precise and use the language of conditional probability, but for sake of ease of understanding we have not done so. The author believes that this treatment of the problem is more cogent than most previous attempts to explain it.

The moral of the story: When conditioning on events, one must be careful to ensure that conditioning really does incorporate all pertinent 'information'. For example, suppose I have a bucket containing 10 white balls and 10 red balls. I will draw and remove balls from the bucket, and allow myself the luxury of thinking I can do this in a random manner. If a draw (and remove) a red ball, then I can safely say that the probability of drawing a red ball on the next trial is 9/19 (9 reds left/19 balls total left). In this case, all the 'information' in the situation is clearly understood. However, in the Monty Hall problem, the naive/incorrect approach neglects the fact that the host will behave in a certain manner depending on the player's choice. The host has information about the game that the player does not have access to , (in particular, the door that the host chooses to open is a piece of 'information', but its evidential value is somewhat disguised to the player) and thus the confusion may arise.

'Updating' probabilities (i.e. how to condition properly) is a big problem in Bayesian philosophy. For more information, read about Bayesian philosophy and The Ravens' Paradox .