When I deal with probabilities and get confused, I revert to counting. What are all the possible outcomes? What subset of those outcomes am I looking at? So let’s look at all the posible states.

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats.

Let’s translate that to a simple diagram. We have three doors, let’s label them \(A\), \(B\), and \(C\). We also know that there is a car behind one of the doors, it can be behind door \(A\), \(B\), or \(C\). So we have three possible configurations of the world:

The door with the car has been colored red. The possible states of the world are \(Sa\) (meaning the car is behind door \(A\)), \(Sb\) (car behind door \(B\)), and \(Sc\) (car behind door \(C\)).

Quick, what is the probability of the car being behind door \(A\)? Or to put it another way, what is the probability that the world is in state \(Sa\)? Note that these two questions are the same, but the latter reminds us of the possible states. It is easy then to see that

\[P(S_a) = \frac{1}{3} \\ P(S_b) = \frac{1}{3} \\ P(S_c) = \frac{1}{3} \\\]

Now I’m going to ask you a very different question. Assume that instead of playing, you are observing your friend play the game. What is the probability that he will pick door \(A\)? Did you say 1/3? Of course you did, there are three doors and he must pick one. However, and this is really important, these probabilities are completely independent of the probabilities above!

Think of it this way. If two doors had cars behind them instead of just one, would that change the number of states the world can be in? Would it change the probability of your friend choosing door \(A\)? What if all doors had cars? Now there is just a single state for the world, but your friend can still choose door \(A\) with 1/3 probability!

If we now say that \(A\) is the event of your friend opening door \(A\), we can easily see that

\[P(A) = \frac{1}{3} \\ P(B) = \frac{1}{3} \\ P(C) = \frac{1}{3} \\\]

which is just saying each door has an equal probability of being opened.

The trick to solving this problem is that you are dealing with two different classes of probabilities. One is the probability that the world is in a given state (a priori), the other is that a participant (you or Monty) chooses any given door. If you can keep these two classes of probabilities separate, you will be able to easily solve this problem.