# Monty Hall Problem Intuition

This summer a friend came over with their 13 years old nephew for a visit. Their nephew mentioned the Monty Hall problem and we talked about it for hours. Eventually I think we found a good intuition to explain the best strategy.

## The problem

A TV show. 3 closed doors. 1 prize.

The flow is:

- The candidate picks a door. That door remains closed.
- The host, who knows where the prize is, opens another door which doesn’t have the prize behind it.
- The candidate is asked which one of the remaining closed doors they want to open. “Stay or switch”.

## The misunderstanding

At the last step, we usually assume that each remaining closed door has an equal chance of having the prize behind it. And probably because of the status quo bias, we usually stay.

But this is wrong. The remaining closed doors don’t have an equal chance. Which is counter-intuitive.

## The intuition

Since it was difficult to convince each other using statistic counting, as we disagreed on what or how to count, I proposed a pragmatic experimental approach.

“Let’s just play say 20 times and count the results, win or loss. For the first
10 rounds we’ll adopt the *stay* strategy. For the subsequent other 10 ones the
*switch* strategy.”

Additionally, in order to distort the probability space and emphasize patterns, let’s also use playing cards and pretend the prize is the ace of spades.

Ok so let’s play! First round. Strategy is going to be “always stay”.

- The candidate picks a card out of 52 and they keeps it covered.
- The host, who knows where the ace of spades is, would then uncover all cards except one.
- But since your strategy is “always stay”, it actually doesn’t matter what the host does, and you can uncover your card right away.

It appears clearly here that the “always stay” strategy leads to a 1/52 chance of picking the ace of spades.

How about the “always switch” strategy?

- The candidate picks a card out of 52 and keeps it covered.
- The host, who knows where the ace of spades is, uncovers all the remaining cards except one.
- Since our strategy is “always switch”, The candidate gives up their initial pick, and the host uncovers the last card.

With this strategy, what is the chance of the candidate picking the ace of spades? This is more tricky to answer.

One way to look at it is this: since there are 2 uncovered cards, one being the ace of spades, the probability is 1/2.

Another way to look at it is this:

- Step 1: The candidate has a 1/52 chance to win at this step. And
**there is a 51/52 probability that the ace of spade is in the host’s heap**. - Step 2: After reducing the host’s heap, there is thus a much higher probability (51/52) that the last uncovered card is the ace of spades.

Yes, we didn’t end up actually counting the results and playing all 20 rounds. But the couple of runs corroborated the theory.

## Interpretation

So there is this illusion that, after the second step (after the space reduction), there would be a 1/2 chance of winning. Meaning that it doesn’t matter which card the candidate picks.

Whereas, in fact, the candidate is not picking a second time! The question at step 2 really is: which card is most probably the ace of spades: the one the candidate picked initially (“stay”) or the last one in the host’s heap (“switch”)?

This post was kindly reviewed by Léon.