In September of 1990, columnist Marilyn vos Savant was asked the infamous “Monty Hall” problem. Inspired by the game show Let’s Make a Deal, the concept of the game is fairly straightforward:
“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. You pick a door, say No. 1, and the host, who knows what’s behind the other doors, opens another door, say No. 3, which has a goat. He then says to you, ‘Do you want to pick door No. 2?’ Is it to your advantage to take the switch?”
Savant observed that switching doubles your chance at winning the car (from 1/3 to 2/3), much to the dismay of many of her readers. She was met with outrage. She notes that she received 10,000—mostly negative—letters, including many from Math and Statistics Ph.Ds (the featured letters are really worth reading). While intuition says that the reveal should not alter the probability of winning, this is wrong. Why is this problem so counter-intuitive?
The Bayesian Way
If you still don’t believe that switching doubles your chance at winning, let’s see if a 250-year old probability equation—Bayes Theorem—can change your mind. It’s written as follows.
\[ \underbrace{P(A | B)}_{\text{Posterior}} = \underbrace{P(A)}_{\text{Prior}} \times \frac{\underbrace{P(B | A)}_{\text{Likelihood}}}{\underbrace{P(B)}_{\text{Marginal}}} \]
Let’s solve the Monty Hall problem with this equation.
Case 1: Car Behind Door i
There are three doors, which we will call doors i, j, and k. We always begin on door i, which may or may not have the goat. This leaves doors j and k for the reveal, and one of these will always have a goat behind it. Let’s begin with finding the probability that the goat is behind door i.
Let \( A_i \) be the event that there is a car behind door \( i \).
Let \( B_k \) be the event that there is a goat revealed behind door \( k \).
\[ P(A_i | B_k) = P(A_i) \times \frac{P(B_k | A_i)}{P(B_k)} \]
Using the Law of Total Probability, we calculate the marginal, \( P(B_k) \), as:
\[ P(B_k) = P(B_k | A_i) P(A_i) + P(B_k | A_j) P(A_j) + P(B_k | A_k) P(A_k) \] \[ P(B_k) = \left(\frac{1}{2} \times \frac{1}{3} \right) + \left(1 \times \frac{1}{3} \right) + \left(0 \times \frac{1}{3} \right) \] \[ P(B_k) = \frac{1}{6} + \frac{1}{3} + 0 = \frac{1}{2} \]
This gives:
\[ P(A_i | B_k) = P(A_i) \times \frac{P(B_k | A_i)}{\frac{1}{2}} \]We are interested in the probability of the car being behind door i, given that we have revealed a goat behind door k (our posterior). Above, I calculated the marginal—the probability that the goat is revealed behind door k. This leaves just two unknown values – our prior and our likelihood. Our prior is the probability that the car is behind door i. This is fairly intuitive—it is 1/3. There are three doors behind which there could be a car, and door i is one of them. The likelihood is the probability that door k is revealed with the goat, given that the car is behind door i. Knowing that the car is behind the chosen door, the revealed door could be either j or k. This gives 0.5 as the likelihood. We plug in and get the following:
\[ P(A_i | B_k) = \frac{1}{3} \times \frac{\frac{1}{2}}{\frac{1}{2}} = \frac{1}{3} \]
This is our posterior. Given that door k is revealed with the goat, the probability of the car being behind door i is 1/3. The additional information about the goat behind door k does not increase our chance of the car being behind our door, despite the removal of one of the doors. Our posterior is the same as our prior. This can be written mathematically as:
\[ P(A_i | B_k) = P(A_i) \]
Case 2: Car behind Door j
While it may be obvious that the posterior in this case is 2/3, it is worth writing out. In this case, we will still choose door i initially, leaving doors j and k for the reveal (one of these will always have a goat behind it). The math is similar.
Let \( A_j \) be the event that there is a car behind door \( j \).
Let \( B_k \) be the event that there is a goat revealed behind door \( k \).
\[ P(A_j | B_k) = P(A_j) \times \frac{P(B_k | A_j)}{P(B_k)} \]
The marginal, \( P(B_k) \), is the same as above:
\[ P(B_k) = \frac{1}{2} \]
This gives:
\[ P(A_j | B_k) = P(A_j) \times \frac{P(B_k | A_j)}{\frac{1}{2}} \]Our prior is the same as above—the probability of the door j having the car is also 1/3. The change that doubles our probability of winning is the likelihood. Rather than 1/2, the likelihood is 1: given that the car is behind door j, the probability that we reveal the goat behind the third door is 1 (remember, door i is chosen by the player and thus cannot be revealed). This gives
\[ P(A_j | B_k) = \frac{1}{3} \times \frac{1}{\frac{1}{2}} = \frac{2}{3} \]
So switching is twice as likely to win you the car!
Even with this math, the answer still may be hard to believe. Why is this so unintuitive?
What’s the trick?
The statistical slight of hand that makes this problem so difficult has everything to do with the reveal of the goat. The probabilities reflect the sneaky fact that the revealed door requires a goat behind it. This leads to a disconnect from the correct answer (to switch) and what we want to believe (it doesn’t matter). By door j not being chosen to reveal the goat, we gain some evidence in favor of switching to door j. The effect is subtle: there’s a reason why it tripped up so many people. To illustrate this point, I will draw from Judea Pearl’s Book of Why. In this book he proposes an altered version of the game where the revealed door could either have a goat or car behind it. If the door (door k) has the goat, then you proceed as usual. If it is revealed that the car is behind door k, you lose. This modified version of the game does not condition on the reveal of a goat behind door k—just the reveal of door k. In this case, the probability of winning whether you stay or switch is the same.
The Monty Hall problem speaks to the challenges of probabilistic thinking. It also speaks to a quality of causality that is hard to wrap one’s head around: collider bias. There are two variables that are causing door k to be revealed with the goat: the door you chose initially (door i) and the location of the car. By the rules of the game, you are inadvertently conditioning on the goat being shown behind door k, causing this weird, counterintuitive phenomenon. The instance of collider bias (i.e., conditioning on the reveal of a goat) makes it extremely difficult to alter our original intuition—that switching and staying are equally likely to win you the car.
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