# The Monty Hall Problem Explained

Tags: Math

The Monty Hall Problem is a probability problem that goes as follows:

Consider a game show in which you (the contestant) are placed in front of three doors-- A, B, and C. The host (named Monty Hall after the original host of the real game show Let’s Make a Deal) tells you that there is a prize behind one of these three doors. The other two doors have nothing behind them. Your goal is to select the door with the prize behind it. That’s all-- you just pick a door. Let’s say you picked Door A.

Monty knows which door has the prize. Once you’ve made your pick, Monty opens one of the other doors, let’s say Door B, and shows that there’s nothing behind it. Monty asks if you want to switch your choice over to Door C.

I’m sure you’re on the edge of your seat.

The two questions are: what is the probability that you will win the prize if you stay with Door A? What about if you switch to Door C?

At first this sounds like a prolonged trick question-- surely the probability is ⅓ for both options, right?

Not so fast.

The true odds contradict the intuitions of nearly everyone who reads this, including thousands of the readers of the 1990 Parade magazine column that popularized this puzzle. Thousands of reader letters (including over 1,000 from PhDs) were sent to the author Marilyn vos Savant disagreeing with her assessment. Some of the most established mathematicians of our time went to their deathbeds stubbornly unconvinced of the real answer.

What’s this answer that supposedly causes an existential crisis in statisticians and laypeople alike? Here it is: You would have a ⅓ probability of winning the prize if you stuck with Door A, but a ⅔ probability of winning if you switched to Door C.

You play the game again and again and again-- the prize is randomly placed behind a door, you select a door, and then Monty opens another door and shows that it’s empty. He asks if you want to switch to the remaining door that neither of you picked. If you switched every time you would win 66% of the games.

How is this possible? How is it not 33% everytime for every door? How is it different between doors? Even if we tentatively accept that the probability can change, how does it go from 33% to 66%? If it changes at all, and the choice is narrowed from 1 in 3 to 1 in 2, shouldn’t it change from 33% to 50%? And even that’s a hard sell-- the staunch intuition of nearly all who read this is that the probability should not budge one iota in either direction, let alone reveal a strategy that allows you to win two-thirds of the games. This isn’t just a quirk found in computer simulations or the results of some convoluted math-- you can play this game in real life and discover the paradoxical probabilities yourself. The Mythbusters tested this with a mock game show set. The results? Myth confirmed. What the hell’s going on here? How do the odds double in a game so full of random choice?

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Here’s how: the rules of true random probability cease to apply to ⅔ of the rounds because the host makes a subtle non-random choice.

Let’s go through this again and identify where randomness is and isn’t: First, the prize was placed behind a random door. Then, you randomly pick any door that takes your fancy. You have a ⅓ chance of selecting the correct door at this stage-- so far so good.

Let’s consider what happens when you do select the correct door. The host must open an empty door and he is greeted with a choice about which door he wants to open here because both of the doors you didn’t select are empty. He could flip a coin for all he cares, it’s all random. Chance rules the day so far. If you switched doors you would lose. This scenario happens ⅓ of the time.

The key here is to recognize that in ⅔ of the rounds you select the wrong door at first, and to then consider the situation the host is in once this is the case. To say that you selected the “wrong door” is to say that you selected one of the two empty doors, leaving only one empty door left. Due to the mandate the host must follow, the host is forced to open that one remaining empty door. This means that ⅔ of the time the prize is behind the door that neither you nor the host selected. If you switch doors you will win.

In other words, when you get it wrong, the host picks the other door that would’ve also been wrong. So when you switch doors to the only one left that neither of you picked, it has the prize behind it. This happens ⅔ of the time. The other ⅓ of the time you picked the correct door at first, so you’ll lose those rounds after switching. Voila-- 66% success rate by switching every time.

If it’s hard to digest that the host is making a non-random choice, consider that if he truly were making a random choice during his turn he would open the door with the prize behind it at least some of the time. But he never does. That’s his one rule-- always open an empty door. ⅔ of the time there is only one left when it’s his turn. The prize was placed behind a random door, and you selected a random door, but the host does not select a random door ⅔ of the time.

© Joseph Bowen