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Comment by z3t4 | Hacker News Reader

In high school our math teacher introduced a problem that could not be solved, which involved a TV-show game where the contender would choose one of 3 doors, and if there was something behind it, the contender would win it. But the show host would remove one of the doors without something behind it, and ask the contender if he/she would like to switch door.

I simply simulated the game on a computer one million times which revealed that changing the door after the host removed a empty one had a 66% probability of winning. (rather then 50%) but the teacher wouldn't believe me...

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Yeah, the trick with this one is the game show host knows where the prize is. In the case where you picked correctly first (1:3) the switching to the other door will be a loss. But in the case where you picked wrong (2:3) the host will eliminate the non-prize door, so a win is guaranteed if you switch. So, 1/3 * 0 + 2/3 * 1 = 66% of winning for the always switch strategy. Shame on your teacher!

This is a famous problem and the solution is not contested.

I'd rather say that the solution is constantly contested by people who can't shake off their intuition.

To be fair, the problem is often badly posed. The host always reveals a goat, and this is crucial information, but people often fail to make it clear. If the host chose randomly and just happened to reveal a goat, the solution would be the 'intuitive' one.

No, I don't think that's accurate. In the scenario you pose, sticking and switching have an equal reward, so both are valid solutions.

> sticking and switching have an equal reward

Right, and this is the intuitive solution. Nobody argues that switching reduces your chance of success; the split is over whether it increases it to 2/3 or leaves it at 1/3.

(The problem statement ends with 'Is it to your advantage to switch your choice?' The right answer under the intended assumption about the host is 'yes', but the intuitive answer is 'no, it makes no difference'.)

Sorry, I shouldn't have said 'increases it to 2/3 or leaves it at 1/3'. If he picks randomly and reveals a goat, the probability that your original choice of door is a winner is now 1/2, as is the probability that the remaining door is a winner.

I think the reason that people can't shake off their intuition is that the intuition is based on the idea that the host (Monty Hall) has agency.

The specific posing of the problem is that the game is you are presented with three doors, you choose one, and the host opens a door showing a goat, and you have to decide if you want to switch.

The problem is that people imagine a game where you choose a door and the host decides whether to open a second door to convince you to switch OR they open the door you chose to show the goat. So given that they CHOSE to open a goat-door, is it the right move to switch? That problem is a game theory problem and requires that you understand the strategy that the host is employing; the intuition is that the host WANTS you to lose, so will only open the goat-door if you chose right (or has a significantly higher probability of doing so) so switching is "falling for" the trick.

Re-imagining the problem as a different scenario, where you can ask to see which of two doors contains a goat, and then deciding which door to take avoids this trap. Then the solution is not obvious, but not counter-intuitive (maybe; depends on your intuition).

The Monty hall problem (what you’ve described) doesn’t require simulations.

What are the odds you guessed it right on your first pick? (1/3) That’s the only time switching answers will not reveal the prize. The rest of the time there is a losing door and a winning door and the host is forced to reveal the losing door so switching wins 2/3 of the time. It’s that simple.

It's known at the Montey Hall Problem[0] or another variation is The Three Prisoners Problem[1] (IMO [1] is an easier way to conceptualize the probability of different outcomes).

It can be understood that the contestant choosing to switch doors has a probability of winning equivalent to that them being allowed to choose two of the three doors in the first place. By not switching they are choosing one door (1/3) but by switching (since one of the empty doors is removed) they are effectively choosing both the other two doors (2/3).

[0]https://en.m.wikipedia.org/wiki/Monty_Hall_problem

[1]https://en.m.wikipedia.org/wiki/Three_Prisoners_problem

This is the Monty Hall Problem, and though counter-intuitive, is well understood: https://en.wikipedia.org/wiki/Monty_Hall_problem

It's very unclear from your description, but I suspect "could not be solved" means "could not be proved", in which case of course a simulation is not a proof.

This and similar math puzzles are trivial to prove compared to the math problems like the one discussed in the article. It is only that they are counter intuitive that makes it hard for people to believe them. If you change the three door to hundred door problem and have the host open 98 doors, it is rather obvious that you should switch to the one door he has not opened besides the door that you did choose at the start.

I have no dispute with that.

I'm saying that since the OP appears to believe that a simulation proves it (which of course it doesn't), there's a decent chance other parts of the story also are incorrect.

Because yes, this puzzle is quite well known for confounding intuition. Presumably a math teacher knows that when they introduced the question.

> "a problem that could not be solved"

The solution is counter-intuitive, but the problem can be solved with some elementary probability theory.

https://en.wikipedia.org/wiki/Monty_Hall_problem?wprov=sfti1

This problem is tricky with three doors but if you imagine playing the same game but with 100 doors instead of three, then the result is very clear.

You used Monte Carlo method to solve the Monty Hall problem.

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