Further and deeper exploration of paradoxes and challenges of intuition and logic can be found in my recently published book, Probability, Choice and Reason.

The Monty Hall Problem is a famous, perhaps the most famous, probability puzzle ever to have been posed. It is based on an American game show, Let’s Make a Deal, hosted by Monty Hall, and came to public prominence as a question quoted in an ‘Ask Marilyn’ column in Parade magazine in 1990.

‘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 all the doors, opens another door, say No. 3, which reveals a goat. He then says to you, “Do you want to switch to door No. 2?” This is not a strategic decision on his part based on knowing that you chose the car, in that he always opens one of the doors concealing a goat and offers the contestant the chance to switch. It is part of the rules of the game.

So should you switch doors?

Consider the probability that you chose the correct door the first time, i.e. No 1 is the door to a car. What is that probability? Well, clearly it is 1/3 in that you have three doors to choose from, all equally likely.

But what happens to the probability that Door No. 1 is the key to the car once Monty has opened one of the other doors?

This again seems quite straightforward. There are now two doors left unopened, and there is no way to tell behind which of these two doors lies the car. So the probability that Door 1 offers the star prize now that Door 2 (or else Door 3) has been opened would seem to be 1/2. So should you switch? Since the two remaining doors would seem to be equally likely paths to the car, it would seem to make no difference whether you stick with your original choice of Door 1 or switch to the only other door that is unopened.

But is this so? Marilyn Vos Savant, in her ‘Ask Marilyn’ column, declared that you should switch doors to boost your chances of winning the car. This answer was howled down by the great majority of the readers who wrote in, and was rejected by even such as Paul Erdos, one of the most prolific mathematicians of all time. Their reasoning was that once the door was opened, only two doors remained closed, so the chance that the car was behind either of the doors was identical, i.e. ½. For that reason, switching or sticking by the contestant should make no difference to the chance of winning the car. Vos Savant argued, in contrast, that switching doubled the chance of winning the car.

Let’s think it through.

When you choose Door 1, there is a 1 in 3 chance that you have won your way to the car if you stick with it. There is a 2 in 3 chance that Door 1 leads to a goat. On the other hand, if you have chosen Door 1, and it is the lucky door, the host must open one of the two doors concealing a goat. He knows that. You know that. So he is introducing useful new information into the game.

Before he opened a door, there was a 2 in 3 chance that the lucky door was EITHER Door 2 or Door 3 (as there was a 1 in 3 chance it was Door 1). Now he is telling you that there is a 2 in 3 chance that the lucky door is EITHER Door 2 or Door 3 BUT it is not the door he just opened. So there is a 2 in 3 chance that it is the door he didn’t open. So, if he opened Door 2, there is a 2 in 3 chance that Door 3 leads to the car. Likewise, if he opened Door 3, it is a 2 in 3 chance that Door 2 leads to the car. Either way, you are doubling your chance of winning the car by switching from Door 1 (probability of car = 1/3) to whichever of the other doors he does not open (probability of car = 2/3).

It is because the host knows what is behind the doors that his actions, which are constrained by the fact that he can’t open the door to the car, introduce valuable new information. Because he can’t open the door to the car, he is obliged to point to a door that isn’t concealing the car, increasing the probability that the door he doesn’t open is the lucky one (from 1/3 to 2/3).

If this is not intuitively clear, there is a way of making it more so. Let’s say there were 20 doors, with a car behind one of them and goats behind 19 of them. Now say we choose Door 1. This means that the probability that this is the winning door is 1 in 20. There is a 19 in 20 probability that one of the other doors conceals the car. Now Monty starts opening one door at a time, taking care not to reveal the car each time. After opening a carefully chosen 18 doors (chosen because they didn’t conceal a car), just one door remains. This could be the door to the car or your original choice of Door 1 could be the path to the car. But your original choice had an original probability of 1/20 of being the winning door. Nothing has changed that, because every time he opens a door he is sure to avoid opening a door leading to a car. So the chance that the door he leaves unopened points to the car is 19/20. So, by switching, you multiply the probability that you have won the car from 1/20 to 19/20.

If he didn’t know what lay behind the doors, he could inadvertently have opened the door to the car, so when he does so this adds no new information save that he has randomly eliminated one of the doors. If he randomly opens 18 doors, not knowing what is behind them, and two doors now remain, they each offer a 1 in 2 chance of the car. So you might as well just flip a coin – and hope!

Even when it is explained this way, I find that many people find it impossible to grasp the intuition. So here’s the clincher.

Say I have a pack of 52 playing cards, which I lay face down. If you choose the Ace of Spades, you win the car. Every other playing card, you win nothing. Go on, choose one. This is now laid aside from the rest of the deck, still face down. The probability that the card you have chosen is the Ace of Spades is clearly 1/52.

Now I, as the host, know exactly where the Ace of Spades is. There is a 51/52 chance that it must be somewhere in the rest of the deck, and if it is I know where. Now, I carefully turn over the cards in the deck one a time, taking care never to turn over the Ace of Spades, until there is just one card left. What is the chance that the one remaining card from the deck is the Ace of Spades? It is 51/52 because I have carefully sifted out all the losing cards to leave just one card, the Ace of Spades. In other words, I have presented you with the one card out of the remaining deck of 51 that is the Ace of Spades, assuming that it was not the card you chose in the first place. The chance that the card you chose in the first place was the Ace of Spades is 1/52. So the card I have selected for you out of the remaining deck has a probability of 51/52 of being the Ace of Spades. So should you switch when I offer you the chance to give up your original card for the one that I have filtered out of the remaining 51 cards (taking care each time never to reveal the Ace of Spades). Of course you should. And that’s what you should tell Monty Hall every single time. Switch!

Appendix

In the standard description of the Monty Hall Problem, Monty can open door 1 or door 2 or door 3. The car can be behind door 1, door 2 or door 3. The contestant can choose any door.

We can apply Bayes’ Theorem to solve this.

D1: Monty Hall opens Door 1.

D2: Monty Hall opens Door 2.

D3: Monty Hall opens Door 3.

C1: The car is behind Door 1.

C2: The car is behind Door 2.

C3: The car is behind Door 3.

The prior probability of Monty Hall finding a car behind any particular door is P(C#) = 1/3,

where P(C1) = P (C2) = P(C3).

Assume the contestant chooses Door 1 and Monty Hall randomly opens one of the two doors he knows the car is not behind.

The conditional probabilities given the car being behind either Door 1 or Door 2 or Door 3 are as follows.

P(D3 I C1) = 1/2 … as he is free to open Door 2 or Door 3, as he knows the car is behind the contestant’s chosen door, Door 1. He does so randomly.

P(D3 I C3) = 0   … as he cannot open a door that a car is behind (Door 3) or the contestant’s chosen door, so he must choose Door 2.

P (D3 I C2) = 1 … as he cannot open a door that a car is behind (Door 2) or the contestant’s chosen door (Door 1).

These are equally probable, so the probability he will open D3, i.e. P(D3) = ½ + 0 + 1 / 3 = 1/2

So, P (C1 I D3) = P(D3 I C1). P(C1) / P(D3) = 1/2 x 1/3 / 1/2 = 1/3

Therefore, there is a 1/3 chance that the car is behind the door originally chosen by the contestant (Door 1) when Monty opens Door 3.

But P (C2 I D3) = P(D3 I C2).P(C2) / P (D3) = 1 x 1/3 / 1/2 = 2/3

Therefore, there is twice the chance of the contestant winning the car by switching doors after Monty Hall has opened a door.

Exercise

Question 1. You are given the choice of three boxes, coloured red, yellow and blue. Two are empty, a thousand pound prize is inside the third. You are asked to select a box. You select the blue box.

I know which box contains the prize and must open one of the boxes that I know to be empty. I open the yellow box and it is empty.

To maximise your chance of winning the prize, should you switch to the red box, stick with the blue box, or does it not matter? Why?

Question 2. Suppose in a separate experiment, you are given the choice of three boxes, pink, green and violet. Two are empty, a thousand pound prize is inside the third. You are asked to select a box. You select the green box.

I do not know which box contains the prize and must open one of the boxes. I open the pink box and it is empty.

To maximise your chance of winning the prize, should you switch to the violet box, stick with the green box, or does it not matter? Why?