It's kind of like me reading the wikipedia page on the Monty Hall problem.

I read an explanation about why it makes sense to change doors. But no, my gut tells me there's a 50/50 chance. I scroll down, repeat...

That gut feeling approach is very human like. You have a bias and even when the facts say that you are wrong you think that there must be a mistake, because your original bias is so strong.

Maybe we need a dozen LLMs with different biases. Let them try to convince the main reasoning LLM that it’s wrong in various ways.

Or just have an LLM that is trained on some kind of critical thinking dataset where instead of focusing on facts it focuses on identifying assumptions.

That would be a true Mixture of Experts.

I sometimes put the 4 biggest models like this to converge on an optimal solution

1/3 chance you picked the door with the car, 2/3 chance it's behind one of the other two doors.

These probabilities don't change just because you subsequently open any of the doors.

So, Monty now opens one of the other 2 doors and car isn't there, but there is still a 2/3 chance that it's behind ONE of those 2 other doors, and having eliminated one of them this means there's a 2/3 chance it's behind the other one!!

So, do you stick with your initial 1/3 chance of being right, or go with the other closed door that you NOW know (new information!) has a 2/3 chance of being right ?!

The other way to see it is by just looking at the different outcomes of car behind door A, B or C.

Let's call the door you initially pick A.

car initial monty stick swap

A A B A C -- or Monty picks C, and you swap to B

B A C A B

C A B A C

So, if you stick, get it right 1/3, but swap get it right 2/3.

it's easier to think about it with 100 doors.

if you get to pick one and he opens 98 of the remaining ones, obviously you would switch to the remaining one you didnt pick, since 99/100 times the winning door will be in his set.

Is it though? Instinctively the initial choice and the last remaining door have the same odds of 1/100.

On the initial choice yes. But on the second choice, that other door is a single door that is the sum of the odds of the other 99 doors. So you're second choice would be to keep the door you initially chose (1/100) or select the other door (99/100).

Remember, the host always knows which is the correct door, and if you selected incorrectly on the initial choice they will ALWAYS select the correct door for the second choice.

I thought it would be obvious that I’m not arguing the statistical facts, but the idea that “it is easier to think about” the 100 doors scenario. There is simply no straightforward explanation that works for laypeople.

I think the issue most lay people have is that the host opening a door changes the odds of winning, because he knows where the prize is.

I think the easiest way to demonstrate that this is true is to play the same game with two doors, except the host doesn't open the other door if it has the prize behind it. This makes it obvious that the act of opening the door changes the probability of winning, because if the host opens the other door, you now have 100% chance of winning if you don't switch. Similarly, if they don't open the other door, you have a 0% chance of winning, and should switch. It's the fact that the host knows and chooses that is important.

It's only once you get over that initial hurdle that the 100 door game becomes "obvious". You know from the two door example that the answer isn't 50/50, and so the only answer that makes sense is that the probability mass gets concentrated in the other door.

It's probably easier for most people to not think of them as two remaining doors, but two remaining sets. Originally, with one hundred doors, if the goal object is only behind one of them, then there would be a 1/100 probability it would be behind the initially chosen door, which comprises one set, while there's a 99/100 probability that the goal object is behind one of the doors in the set of not originally chosen doors. If 98/99 of the doors in the not originally chosen doors set are excluded as having the goal object, then this does not change that there's a 99/100 probability that the goal object is behind a door in this set, it just means it wasn't one of the other doors in the set.

It is easier to think about this in the case of a very large number of doors. It is unlikely that you could have picked the right door from, say, 1B.

Chasing this tangent a bit -- I have never been happy with the Monty Hall problem as posed.

To me the problem is that it is posed as a one-shot question. If you were in this actual situation, how do you know that Monty is not deliberately trying to make you lose? He could, for example, have just let you open the first door you picked, revealing the goat. But he chose to ask you to switch, then maybe that is a big hint that you picked the right door the first time?

If the game is just "you will pick a door, he will reveal another door, and then you can choose to switch" then clearly the "usual" answer is correct; always switch because the only way you lost is if you guessed correctly the first time (1/3).

But if the game is "try to find the car while the host tries to make you lose" then you should never switch. His ideal behavior is that if you pick the door with the goat then he gives you the goat; if you pick the door with the car then he tries to get you to switch.

His ideal adversarial strategy becomes non-trivial when you know about it.

It is very likely “just flip a coin to turn it back to 50/50” but may be something statistically sophisticated.

If his desire is for the contestant to lose, then he can't really do better (formally) than winning 2/3 of the time by simply opening the door that they choose. In practice, always opening a goat-door and always asking to switch for a car-door can do slightly better than 2/3 because some contestants, unaware of his strategy and objectives, might choose to switch.

If his objective is more subtle -- increasing suspense or entertainment value or getting a kick out of people making a self-destructive choice or just deciding whether he likes a contestant -- then I'm not sure what the metrics are or what an optimal strategy would be in those cases.

Given that his motives are opaque and given no history of games upon which to even inductively reason, I don't think you can reach any conclusion about whether switching is preferable. Given the spread of possibilities I would tend to default to 50/50 for switch/no-switch, but I don't have a formal justification for this.

Yeah I studies Statistics in graduate but still believes that it's wrong. It's mathematically correct but it's wrong! I refuse to believe it!