The Monty Hall problem is one of the simplest and yet most baffling mathematics puzzles of all: All you have to do is choose between two doors, only one of which has a prize behind it.

Yet everyone gets the answer wrong.

Here is how it works: In the TV show “Let’s Make a Deal,” hosted by Monty Hall, contestants are shown three doors and are told that behind one of them is a Cadillac, and behind the other two are goats. All they have to do is pick one door. After they make their choice, Hall then opens one door, always revealing that the newly opened door has a goat behind it. He then offers the contestant a choice: Stick with your first choice or switch to the other door.

How do you choose in order to maximise your chances of winning?

You don’t need any maths skills to solve it, just a general knowledge of the laws of probability. And yet most people get it wrong. Surely, there are two doors left and one car. It’s a 50-50 chance no matter which door you pick!

Wrong. It is *not* a 50-50 choice, but *the Monty Hall setup biases you to think that it is*.

You should always switch doors to give yourself a 2 out of 3 chance of winning.

Famously, dozens of PhD holders in the US signed their names to the wrong solution when a column in Parade magazine discussed the answer. One of those in the wrong was Paul Erdos, the prolific maths professor who taught at both Manchester and Princeton. Erdos didn’t believe the solution until he was shown a computer simulation of it.

The Monty Hall problem is appealing in large part because even when you understand the correct answer, it still “feels” wrong and it can take a long time to accept that the obvious (incorrect) solution is probably false.

So here is a way of explaining the problem that ought to make sense to everyone. It’s the way I finally admitted that the 50-50 choice in the puzzle is obviously not a 50-50 choice.

This explanation comes from Marilyn Vos Savant, the Parade columnist who initially popularised the Monty Hall Problem. Basically, you need to consider how much information you have about the doors, because the problem delivers different amounts of information for each door, and if you maximise the info then you maximise your chance of picking the door with the car.

Here is how that info adds up:

From the beginning, the contestant only has a 1 in 3 chance of picking the right door. When the host reveals one of the goats, it doesn’t actually change the odds that the contestant initially picked correctly — their first choice remains a 1 in 3 shot.

However, *after* one of the goats is revealed, the contestant has a lot more information than the initial 1 in 3 random choice.

The contestant now effectively has a choice between the “one-of-three” door they chose originally or “two-of-three” remaining doors they initially knew nothing about. That is a choice between sticking with your 1/3 chance or going with a 2/3 chance. And, as you know that the car is not behind one of the 2/3 doors, then the car must be more likely to be behind that other door.

So you must switch to get a 2/3 chance of selecting the car!

This still doesn’t feel right, because the “choice” is confined to two doors.

So here is how to think about that in a way that makes more sense, proposed by Vos Savant back in 1990.

Suppose you have to choose between 1 million doors. Monty Hall then shows you the goats behind 999,998 of the remaining doors. The only two doors left are the one you picked first and the other closed door.

It’s obvious that your first, uneducated, guess was a 1 in a million shot. But now you know that the host knows the car was not behind 999,998 of those doors, and that for some suspicious reason he is leaving that one other door closed. That is an extremely strong clue — a 999,999 in 1 million chance! — that the remaining door has the car.

You’d feel a lot more comfortable about switching, knowing that Hall was careful to avoid opening that one other door out of the 1 million he first offered you.

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