There are some fun and tricky puzzles with probability, like the famous Monty Hall problem, where sometimes even professional mathematicians get confused. But there is one easy "hack" to quickly solving even the most bizarre such conundrums, the ensemble technique. An ensemble is just a fancy word for a collection of very many items. I explain the logic of the method along with some justifications for it in this article but here I want to do something more fun and easy: I want to show how to use the method in practice to solve a couple of probability puzzles that are normally considered tricky.

The Monty Hall problem.

There are many great explanations of this famous puzzle. Here's one from Better Explained, whose excellent article also provides a few ways to understand the correct solution:

The Monty Hall problem is a counter-intuitive statistics puzzle:

There are 3 doors, behind which are two goats and a car.

You pick a door (call it door A). You’re hoping for the car of course.

Monty Hall, the game show host, examines the other doors (B & C) and opens one with a goat. (If both doors have goats, he picks randomly.)

Here’s the game: Do you stick with door A (original guess) or switch to the unopened door? Does it matter?

Surprisingly, the odds aren’t 50-50. If you switch doors you’ll win 2/3 of the time!

The article then explains many different approaches to making the correct answer more intuitive. Let's see how the ensemble method handles this problem. You don't need to have already read the article where I laid out the foundations of the method, I will try to explain how it applies to this problem in a self-contained way.

First, let's pose the question: suppose you switch doors, what's the chance you win the car? The main idea of the ensemble method is simple - just imagine the whole situation has happened, independently, a gazillion (G) times. Perhaps there are G parallel universes and each one has a version of you participating in the game and switching from door A to the other unopened door. The method then just tells you to count different possibilities.

1. Before the game show host reveals one door with a goat.

As far as you know, the car is equally likely to be behind any of the three doors, so you expect:

1/3G universes with the car behind door A

1/3G universes with the car behind door B

1/3G universes with the car behind door C

2. After the host reveals one door with a goat.

Now the puzzle says the host checks both unpicked doors B and C, finds one with a goat and shows it to you (if both have goats he chooses randomly). So what happens to the three groups of universes above? For the first group, where the car is behind door A, the host will choose randomly to open B or C, so the 1/3G universes will split into two groups:

1/6G universes with the car behind A and door B revealed.

1/6G universes with the car behind A and door C revealed.

Notice that in either of these groups when you switch doors from your original choice of A you will lose. For the other two cases, it's even simpler. If the car is behind B then according to the rules the host can only open C and vice versa, so we'll have:

1/3G universes with the car behind B and therefore door C revealed.

1/3G universes with the car behind C and therefore door B revealed.

Notice that in either of these groups when you switch doors from your original choice of A you will win. If for example door C is revealed you will see it has a goat, will pick B and get the car. So we have:

1/6G + 1/6G universes where you lose, 1/3G in total

1/3G + 1/3G universes where you win, 2/3G in total

Out of the G total universes you could be in you would then lose in 1/3 of them and win in 2/3 of them. Of course you don't know which of those G universes you actually are, you are equally likely to be in any of them, so your probability of being in a losing one is 1/3 and in a winning one is 2/3.

I hope it's clear how the method gets you the probabilities. If it's a little disorienting to think of parallel universes you can also think about you, in this universe, participating in G separate independent games. Pretty much all other tricky probability puzzles can be solved by introducing the ensemble and doing some simple counting, without the need to learn and remember more sophisticated formulas like the Bayes theorem. In the next part, I will show how the ensemble handles another famous puzzle, the two children problem.

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