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Personally I don't think you are given any new information in the Monty Hall problem that's worth anything. You always knew there was a 100% chance that one of the doors in that set has a goat behind it, which door that is is pretty much irrelevant.

Or to put it another way how many people would stick with their chosen door if instead of opening a door and showing a goat they were simply asked "would you like to stick with your current choice of door, or switch and take the best result from the other doors?"

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> switch and take the best result from the other doors

That is, effectively, what the choice boils down to. But that being said, since the problem is stated as only getting to pick one "other" door, the "extra information" received is knowing which of the other 2 doors has the best result.

I didn’t question the math, but what made me intuitively believe the Monty Hall problem was this: Me: I am thinking of a number between 1 and a million. If you guess correctly, you get a prize.

you make your guess Me: Ok, I will reveal that the number is either $YOUR_GUESS or 452,871. You may keep your original guess or switch to 452,871.

Of course, you would switch, because the chances you guessed correctly originally were 1 in a million.

Wow. I don't think it's ever been explained to me this well before.

Try thinking of it this way: The information is in which door the host does not open.

2/3rds of the time, your initial choice will be a goat. In those cases the host deliberately avoids opening the remaining door with a car - thereby telling you where it is.

If you chose the car door initially, then indeed you get no new information. But that only happens 1/3rd of the time.

I am curious if that changes your opinion.

Read their second paragraph. That is exactly how they think of it, and they do think it makes it obvious you should switch.

Ah! Apologies, I did not read that properly.

When Monty reveals a goat, your odds increase from 1/3 to 1/2 if Monty is picking randomly and just happened not to reveal a goat (and if he revealed a car we start from the top, or you automatically lose, or you automatically win, or...), but to 2/3 if Monty will reliably reveal a goat. I think it's hard to say that you have no extra information in that latter case.

It's true that we can add other options (pick both doors) of equal likelihood, but it's because of our increased information that the reduced option of picking just one door isn't actually worse off.

The GP knows all this, and agrees it's better to switch (look at the second paragraph). They just dislike the common formulation (that is given in terms of the extra information from the boot being opened), and find another formulation more compelling (your original choice is one door, while the switch is actually to the best prize from the other two doors, not simply to another one door).

The Monty Hall problem is easily resolved by imagining there are a million doors instead of just three. The host opens 9,999,998 doors, revealing goats behind each. You'd obviously switch doors to the one the host left closed, and the reasoning you followed will be just as valid for any number of doors including 3.

This problem is completely different, since the number of degrees of freedom remains fixed throughout the decision process. As someone else pointed out, it is trivially resolvable by realizing that no matter how many times you switch envelopes, you're really only making one choice, and doing so with 50-50 odds of picking the more profitable envelope. The offer to switch amounts to meaningless magician's patter, whether the offer is made once or a million times.

I always think of the Monty Hall problem with 100 doors. And with the game show host revealing 98 'goat' doors thus leaving two doors left (the original guess, and another one). You'd switch then right? :)

Your first paragraph makes it seem like you don't accept the common solution to the Monty Hall problem, which is why people have responded explaining it to you, even though the second paragraph makes it clear that you do, and just dislike the way it is explained.

Personally, I'm not sure that it would be easier to convince someone that the second formulation is equivalent to the original problem, but next time I find a dis-believer I will give it a shot!

When I was initially exposed to Monty Hall, what confused me the most is that I didn't know that the host would always open a door with a goat, rather than just picking one of the remaining doors at random. In the latter case (i.e. the case where the host can pick a car), even provided that you are in the universe where he picked a goat, switching and staying both yield a 50% chance of success.

It becomes a lot clearer if you increase the number of goats.

I have a deck of a billion cards, one entitles you to the car, the others have a picture of a goat. You pick one at random but do not get to see it.

Of the remaining 999,999,999 cards, I discard 999,999,998 goats.

Do you want your initial card, or the remaining card in my deck?

Or to frame it another way, do you want the car iff you picked it first or iff you did not pick it first?

The GP is not contesting this, they just think that the whole discarding/revealing of the goats is an unnecessarily complicated explanation for why this works.

The way they put it, the strategy becomes obvious if you instead reformulate the problem like so: there are three doors with three different prizes. You get to pick one door, but are then given a choice: stick with your original choice, or choose to get the best prize from the other two doors.

The solution (switch!) is more obvious in the second formulation. I have no idea if it's easier to convince someone else of the reasoning in the first formulation, or to convince them that the second formulation is equivalent to the original.

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