# Making the Monte Hall problem weirder but obvious

post by dynomight · 2020-09-17T12:10:23.472Z · LW · GW · 7 commentsThis is a link post for https://dyno-might.github.io/2020/09/17/making-the-monty-hall-problem-weirder-but-obvious/

The Monty Hall problem is famously unintuitive. This post starts with an extreme version where the solution is blindingly obvious. We then go through a series of small changes. It will be clear that these don’t affect the solution.

## 7 comments

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## comment by Dagon · 2020-09-17T17:04:15.755Z · LW(p) · GW(p)

An easier illustration is "Monty doesn't open any doors, he just gives the player to stay with their chosen door, or switch to ALL of the other doors". This is isomorphic to showing losers for all but one of the other doors.

Like the original, it's somewhat ambiguous if you don't specify that Monty will do this regardless of whether the player guessed right the first time.

## comment by dynomight · 2020-09-17T17:13:52.941Z · LW(p) · GW(p)

That's definitely the central insight! However, experimentally, I found that explanation alone was only useful for people who already understood Monty Hall pretty well. The extra steps (the "10 doors" step and the "Monty promising") seem to lose fewer people.

That being said, my guess is that most lesswrong-ites probably fall into the "already understood Monty Hall" category, so...

## comment by frontier64 · 2020-09-17T23:10:51.123Z · LW(p) · GW(p)

A few months ago I tried a similar process to this with my dad who's pretty smart but like most does not know the Monty Hall Problem.

I put three cards down, showed him one ace which is the winner, shuffled the cards so that only I knew where the ace was and told him to pick a card, after which I would flip over one of the other loser cards. We went through it and he said that it didn't matter whether he switched or not, 50-50. Luckily he did not pick the ace the first time so there was a bit of a uh huh moment.

I repeated the process except using 10 total cards. As I was revealing the loser cards one by one he started to understand that his chances were improving. But he still thought that at the end it's a 50-50 between the card he chose and the remaining card although his resolve was wavering at that point.

I hinted, "What was your chance of selecting the ace the first time", he said, "1 out of 10", and then I gave him the last hint he needed saying, "And if you selected a loser what is that other card there?"

A few seconds later it clicked for him and he understood his odds were 9/10 to switch with the 10 cards and 2/3 to switch with the 2 cards.

He ended up giving me additional insight when he asked what would happen if I didn't know which card was the ace, I flipped cards at random, and we discarded all the worldlines where I flipped over an ace. We worked on that situation for a while and discovered that the choice to switch at the end really is a 50-50. I did not expect that.

## comment by Dach · 2020-09-18T00:45:28.514Z · LW(p) · GW(p)

Amusing anecdote: I once tried to give my mother intuition behind Monte Hall with a process similar to this. She didn't quite get it, so I played the game with her a few times. Unfortunately, she won more often when she stayed than when she switched (n ~= 10), and decided that I was misremembering. A lesson was learned, but not by the person I had intended.

## comment by glagidse · 2020-09-18T08:56:40.874Z · LW(p) · GW(p)

The way I understood the intuition behind the Monty hall problem is so:

1. You've got a million doors

2. Only one door has a prize

3. Imagine what the probability is that you pick one at random (one in a million)

4. Pick one at random

5. Randomly open 99998 doors that you did not pick and do not contain prizes.

6. 2 doors are remaining, one of which you picked.

7. Has the probability that you picked the correct door changed?

8. If yes, why yes? and if no why no? And what is the new probability?