The Monty Hall problem concerns a strategy for a game show. It is loosely based upon an American TV game show called Let’s Make A Deal and is named after the show’s presenter, Monty Hall.

Let’s Make A Deal

The problem was originally posed (in a slightly different form) by Steve Selvin to the scientific journal, The American Statistician in 1975. However, it was made famous by a reader’s letter to Parade magazine in 1990. The letter asked:

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 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 switch your choice?

It’s Decision Time

If you were the contestant, what would your decision be? Stick with the door you originally chose (door 1, in this specific case) or switch to the remaining door (door 2)? Do your odds of winning improve if you switch your choice of door? Or are you more likely to lose the car if you make the switch? Or does it make no difference?

Take a moment to think about your options…

OK. So you’ve made your choice? Was your decision the correct one?

Rules of the Game

Before I answer that, let’s begin by considering some assumptions behind the question:

1. Initially, the car will be behind any one of the three doors, chosen at random, with equal probability. You’re as likely to win by picking Door 1 as you are by picking Door 2 or Door 3. However, Monty knows which door the car is hidden behind.
2. After you have made your initial choice, you then inform Monty of your decision.
3. Monty must then open one of the remaining doors to reveal a goat. (If you picked a goat with your original choice, then Monty has only a single choice of door—that hiding the remaining goat—and must open it. Otherwise, if you picked the car initially, then Monty picks one of the two remaining doors at random.)
4. It’s assumed that you want to win the car, not a goat, and so you’re not going to pick the door that Monty just opened…

Clearly, if those assumptions are incorrect, then the nature of the game changes. For example, if the car is always hidden behind the same door in every game, then you ought to be able to win fairly easily every time.

Possible Strategies

Given that any of the Doors may conceal the car, and that you’re trying to determine whether you should switch your initial selection or not, your initial choice of Door actually isn’t going to make a great deal of difference, so just pick one at random. (I’ll leave it up to you should you wish to explore how your chances of winning are affected by alternative strategies for this decision.)

In the general case, our simplest possible strategies, when offered the opportunity to switch doors, are as follows:

1. Always stick with our original selection.
2. Always switch to the remaining door. (That is, switch to the door that was not our original choice, and that wasn’t opened to reveal a goat.)

It should be noted that, in every game, either Strategy A will win or Strategy B will win: there’s no possible scenario that allows both to win, or neither to win. It should also be obvious that neither strategy guarantees a win. No matter which strategy you choose, sometimes you’ll fail to win the car.

If there’s no difference between the two strategies, then you’ll win the car a roughly equal number of times with each strategy. Otherwise, if one strategy is better than the other, then you’ll likely see more wins with the better strategy.

Why am I using phrases like “roughly equal” and “likely”? Because we’re dealing with games of chance. It’s quite possible (although extremely unlikely) that we could play the game 100 times and get the same outcome (say, winning the car with Strategy A, losing it with Strategy B) each time. Fortunately, the law of averages reassures us that, as the number of games increase, we’ll get closer and closer to the theoretical expected number of wins with each strategy.

What Does Theory Suggest?

When you make your initial choice of door, you have a one-in-three chance of choosing the car, because there is one car and three doors.

So, that being the case, there must be a two-in-three chance that the car is behind one of the other two doors. Agreed?

Now when Monty opens one of the other two doors, you can obviously rule that door out immediately. But the odds that the car is behind one of those two doors hasn’t changed, which means that the probability that the car is concealed by the remaining door must now be two-in-three—all by itself!

Consequently, it is most definitely in your best interests to switch your choice: you’re twice as likely to win the car by switching than by sticking with your original choice.

In particular, we end up with some theoretical results:

• Strategy A has a probability of ⅓ (or 0.333…) of winning the car in a single game. If there are N games, then the expected number of wins is N/3.
• Strategy B has a probability of ⅔ (or 0.666…) of winning the car in a single game. If there are N games, then the expected number of wins is 2N/3.

Does this mean that if we play the game 9 times that Strategy A will win the car 3 times and Strategy B will win it 6 times? No. The expected number is simply our best estimate of the number of times each strategy will win, but there are no guarantees.

I Don’t Believe You!

The theoretical result is highly counter-intuitive, and still attracts a good deal of controversy.

Many—if not most—people are inclined to stick with their original choice. Is this because they feel that Monty is trying to manipulate them away from winning the car, or because of the regret they would feel if it turned out they had selected the car initially, and then switched away from it? Who knows.

But theory is one thing; the proof of the pudding is in the eating.

The Simulation Model

Simulation is a great way to put this to the test. Below is an example that I created using AnyLogic, which is a great general purpose simulation product.

In the simulation, a car and two goats are hidden behind the doors at random at the start of each game. Next, the contestant selects one of the doors, also at random: this turns the door red in color and represents the Strategy A decision.

At this point, Monty opens one of the remaining doors to reveal a goat. The final door then turns green in color and represents the Strategy B decision.

Finally all the doors are opened and we discover which strategy won that particular game. Then a new game begins…

Incidentally, the model is available to download for free, with full source code, from the AnyLogic Cloud, along with many others.

The simulation provides a good deal of evidence for the theoretical results.

(To be completely sure, we need to perform more formal hypothesis testing, using a goodness-of-fit test to compare the observed and expected win frequencies for each strategy. However, that’s a more advanced statistical topic which I’ll leave for another time.)

How did you do? Did your instincts prove correct? Or was the true outcome a little hard to swallow? Don’t worry. Tools like simulation allow us to examine complex situations, like the Monty Hall problem, in detail and determine how our decisions affect the outcomes.

If you have a complex system that exhibits unpredictable, counter-intuitive results, and you’d like some assistance to understand it better, feel free to call me, Mike Allen, on +1 (313) 451-4001 for a free initial consultation.