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Comment by mannykannot | Hacker News Reader

>> It’s important that Monty looked behind the doors before choosing which to open.

> This is often not explicitly stated when the problem is given, which imho is the whole reason this problem has the reputation of being hard to understand.

If that were the only difficulty, why have so many people continued to have trouble accepting it even after this misunderstanding has been cleared up, and even after the correct answer has been explained to them? According to Wikipedia, even Paul Erdős remained unconvinced until he was shown a computer simulation.

I recall mention of an analysis of the responses to Vos Savant's Parade article, concluding that a majority disputing the result were aware of this constraint, and I will post a link if I can find it again (though if a majority did not explain their reasoning, it may not be possible to figure out what assumptions they made. Nevertheless, the question in my first paragraph still stands.)

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I agree that "Monty knowing the answer" isn't the "whole reason" MH is confusing.

IMO, people are confusing MH with a case where the door is opened randomly before your choice.

Let's say Monty shows you three doors, then knowingly opens a wrong door (B) before you choose. Do you want A or C? It's a coin toss, of course.

But suppose you pick door A first and THEN he knowingly opens a wrong door you didn't choose (B). Now the odds of winning are 2/3rds if you switch. But it feels the same as in the previous case. ("How could choosing a door first affect the odds?! It couldn't possibly change anything!!")

The answer is that your choice of a door constrains Monty.

Suppose you choose a door secretly before Monty opens a door. Suppose you choose A, and Monty opens door A, revealing a goat. Well, uh, you can't pick that any more, now, can you? You're forced to choose between B and C, a toss up.

But if you publicly pick door A then Monty can't/won't open A; he has to open one of the other doors. Since you probably chose the wrong door to begin with, and he always opens a wrong door, the other remaining closed door probably has the car.

Extending these types of examples to 100 doors with 99 goats of which Monty opens 98 makes it even easier.

Especially if you walk through what that would look like. You choose door #14. Monty then goes from door to door, opening every single one except door #67. What's special about door #67? Either you picked the door with the car behind it at 1/100 probability, and Monty picked it at random, leaving it closed. Or, with 99/100 probability, Monty is skipping opening that door because it's got the car behind it...

Never thought of it this way, but extending to 100 doors will definitely be my go-to explanation/walkthrough in the future! Thanks for the insight.

I've tried that with people with limited success, though it could be that I've explained it badly.

When you put the games in terms of "you and a friend playing the game simultaneously", most people get it.

You and a friend are playing the game simultaneously; you both decide ahead of time to: a. always pick the same initial door b. no matter what Monty does, you ALWAYS switch, your friend NEVER does.

The possible outcomes before the reveal are: * Monty wins, because he shows you the car. We agree ahead of time that this never happens.

Now, since you know one of you is going to win, since Monty didn't, who has the better chance? Your friend started with a 1/3 chance, and he hasn't switched; does his chance suddenly change because Monty opened a door? Most people (not all!) will agree that it has not.

Since one of you MUST win, and friend has 1/3 chance, what's left?

Woah. That's such a concise way of immediately making the concept clear.

What makes the three doors Monty Hall so counterintuitive is that people tend to correctly reason about the case where the second door is randomly opened and don't understand why it doesn't apply.

I believe that this example makes understanding why people don't get it easier: you are looking for someone with one of your friend. You know they are in one of three rooms. Right before you can open the first one, your friend opens the second one and say: "not there". People assume the Monty Hall problem means that it's more likely your friend is in the third room and not the one you were going to open and think it's silly. And they are right to think that. What they don't get is that the case where your friend opened the correct door is part of the switching choice in the Monty Hall situation.

Your first paragraph looks very plausible, though I am not aware of any data saying it is a prevalent one. Are you suggesting that people still look at this problem as if the choice was random, even when they know it is not? (e.g., if they think the 'random' and 'chosen' cases are equivalent.) That is also plausible, but if it is common, that would imply that the phrasing of the problem is not the root cause of their misunderstanding of the optimal strategy.

Personally, IIRC, my first reaction was to assume the second box opening was not random (perhaps only because the question is not phrased as being conditional on this act revealing a goat) but did not see how this gave any useful information.

Here's another possible way of getting it wrong, regardless of the phrasing of the problem: assuming that, after the reveal (and whether one thinks of it as random or not), one is, as it were, starting over, except with a choice between two boxes rather than three, and no other information.

I definitely don't think it is the only point of confusion, but I definitely think it is a big one. We discussed the problem in my discrete math class and the professor had stated the problem without the constraint. Many in the class, being far too young to even know who Monty Hall is, were confused, and several had their 'aha!' moment once that constraint was mentioned.

> According to Wikipedia, even Paul Erdős remained unconvinced until he was shown a computer simulation.

Interestingly, coding up my own simulation took me from understanding the problem to grokking it. I'd definitely suggest anyone who has programming ability but doesn't grok the problem to write up a simulation.

If anyone reading this uses this example in a class, or is thinking of doing so, perhaps it could be set up as an experiment to see how much the phrasing influences the result?

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