Conditional Probability and the Monty Hall Problem
You’ve been selected from the audience of a game show to come up and play a game. The host walks you up to the stage, where you find three doors labelled 1, 2, and 3. He says, “Behind one of these doors is a brand new car. If you pick the door correctly, you get the car. If you guess wrong, you get a goat.” Having no way of knowing which door hides the car, you pick door 1. The host then walks over to door 3, opening it to reveal a goat. After doing so, he gives you the chance to change your pick to door 2.
The question: should you stay with door 1 or switch to door 2?
Luckily, you’re enrolled in INFO 2040 and are an expert in game theory and conditional probability. You think for a few moments, and then say, “I’d like to switch to door 2.” The host walks over to the second door and opens it, revealing yet another goat. Despite the fact that you didn’t win a car, you can console yourself with the knowledge that switching doors is the dominant strategy for this game.
This game is known as the Monty Hall problem. At first glance, it might seem that both door 1 and door 2 have the same chance of holding the car, so switching doors shouldn’t affect your chances of winning the car. However, as the BetterExplained article describes, when the host opens door 3 to reveal a goat, it improves the chance that door 2 holds the car. Intuitively, there is a 1/3 chance that the car is behind our original choice, and a 2/3 chance that the car is behind one of the other two doors. When the host reveals one of these doors to have a goat, it’s as if the 2/3 probability is given entirely to the other door. Swapping to this other door is thus the best option.
We can compute the conditional probabilities of winning if we stay or switch. The probability of winning if we stay is P(car is behind door 1 | host reveals door 3) and the probability of winning if we switch is P(car is behind door 2 | host reveals door 3). We will assume that you originally choose door 1. If the car is behind door 1, then the host reveals door 3 50% of the time (he could reveal either of 2 or 3). If the car is behind door 2, then the host reveals door 3 100% of the time (he is forced to reveal it).
The car is behind each door with probability 1/3, so P(host reveals door 3) = (1/3)(1/2) + (1/3)(1) = 1/2. If the car is behind door 2, then the host is forced to reveal door 3, so P(car is behind door 2 and host reveals door 3) = P(car is behind door 2) = 1/3. Computing the conditional probability, P(car is behind door 2 | host reveals door 3) = P(car is behind door 2 and host reveals door 3) / P(host reveals door 3) = (1/3)/(1/2) = 2/3. On the other hand, P(car is behind door 1 and host reveals door 3) = 1/6, so P(car is behind door 1 | host reveals door 3) = (1/6)/(1/2) = 1/3. The linked Wikipedia article goes through this computation in a couple of different ways and also includes helpful images for visualizing these probabilities.
We can see that there is a 2/3 chance of winning if we switch, and only a 1/3 chance if we stay, so it is always better to switch doors than to stay!
https://betterexplained.com/articles/understanding-the-monty-hall-problem/
https://en.wikipedia.org/wiki/Monty_Hall_problem#Conditional_probability_by_direct_calculation