The Monty Hall Problem Explained

The Monty Hall problem is a famous mathematical "trick" that involves probability, decision-making, and an unintuitive result. It goes something like this:

You are a contestant on Let's Make A Deal. The host, one Mr. Monty Hall, offers you a choice between three doors. Behind two of the doors are goats, but behind the final door is a brand new car. At the end of the game, whichever door you have selected, you keep the prize it hides.

It is assumed, of course, that the player would prefer to win the car.

After your decision is made, Mr. Hall walks to one of the doors you didn't select and opens it, revealing a goat. He turns to you and offers a new choice: You may keep the door you chose originally, or switch to the final remaining door. What should you do? Does it matter?

If you have enough interest in mathematical puzzles to still be reading this page, you're almost certainly already aware that the answer is yes, you should switch to the final door. Staying on your original door gives you a 1/3 chance of winning the car, while switching gives you a 2/3 chance. But why? Well, there are a lot of explanations floating around that I don't particularly like. They make sense if you already intuitively understand the reason. But as explanations? I find them unhelpful. Even if you're able to prove the result in one of these ways, it still feels unsettling. Why should the choice matter? Sure, it does, but... why?

I'd like to offer an explanation that I think makes more intuitive sense. Let's look at the problem based on two cases:

Case 1: You choose the car

The contestant chooses door A, which has the car behind it.

This is the case where your original decision, at the beginning of the game, is the door with the car behind it. It's probably obvious that switching will always make you lose; you can't switch to the car if you've already chosen the car! But let's walk through it, as practice for the second case.

We'll assume arbitrarily that you chose door A. It doesn't really matter, since the same logic applies regardless of the specific configuration of doors and choice.

Monty Hall can open either door B or door C.

Next, Mr. Hall reveals the contents of a door. He has a choice to make: He reveals either B or C. In this case, both doors have a goat behind them, so the choice he makes is irrevelant.

Whichever door he picks, you should not switch. Whether B or C remains, in both cases, there is a goat behind it.

And so, our case 1 result: If you picked the car with your first guess, switching gives you a 0% chance of winning. Sure, that's obvious enough. Let's move on to the second case.

Case 2: You choose a goat

The contestant chooses door A again, but now that door has a goat.

This case is much more interesting. Again, we'll assume you picked door A for simplicity, and we'll assume the car is behind door B.

We again put ourselves in the shoes of Monty Hall. He has to reveal either door B or C. But unlike in the first case, his hand is forced. Remember, he must reveal a goat! Of the two doors he can pick from, only one hides a goat, so he must reveal door C.

Monty Hall can only open door C.

Now, let's note the current situation. We know that our door, door A, has a goat (because we assumed it). We see that door C has a goat (because it's been revealed). There is only one remaining door, and the car has to be somewhere, so that door has to be the winner! When we're offered a chance to switch, we'd be fools to not take it.

We now have our case 2 result: If you picked a goat with your first guess, switching gives you a 100% chance of winning.

That, right there, is the key insight. The fact that starting on a goat and switching always leads to victory is the point that's so easy to miss, and figuring it out is the trick to understanding the whole problem. But just in case, let's tie it all together.

Tying it all together

We've shown that if you choose the car on your first guess, switching will lead to you winning 0% of the time. We've also shown that if you choose a goat on your first guess, switching will lead to you winning 100% of the time. But, of course, you don't know what's behind the door you picked!

To figure out the probability that switching doors is beneficial, remember that the cases where you should switch are the same cases as the ones where you picked a goat. And what are those odds? Well, there are three doors, and two have goats. At the time you make your decision, you have no other information. With nothing to influence your decision, you have a 2/3 chance of picking a goat.

Finally, finally, our answer. We have shown there is a 2/3 chance that switching to the final door is beneficial.