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brooksmosesChris Hallbeck posted a cute little short that involves the Monty Hall Paradox at https://www.youtube.com/shorts/At_LNDO1eq0, and it includes an explanation of the result that has a lot of people saying "I finally understand it!" in comments. And, after watching it and reading some comments, I think I have an even more intuitive explanation.
The paradox is this: In front of you is a game show host, and three doors. Behind one of the doors is a Shiny New Car, or some other great prize that you may win. Behind the other two doors are goats. You select a door. The host then opens one of the doors that you didn't select, revealing a goat, and then offers you an choice: Do you keep the door you selected first, or do you switch to the other door that they didn't open? If the door you select (either by keeping your first selection or switching to the other one) is the one with the prize, you win the prize!
If your door contains the goat, my understanding is that you do not actually get to keep the goat.
(A key datapoint -- often omitted from the descriptions! -- is that this is how the process always goes, and you know that fact. The host will always open a door with a goat, and will always offer the opportunity to switch. This is not a case where the host is being devious and only trying to get you to switch away if you start out choosing the prize.)
The paradoxical result is that switching will lead to the prize twice as often as not-switching, even though it looks like a random choice between two doors that you have no information about.
Explanation behind cut....
First, notice that the host is being devious in how they ask the questions, and the door you pick first isn't a choice of a door to open. It's a choice about how the second half of the problem will be set up. And that means the second half of the problem isn't set up randomly, so you pretty immediately have an intuitive reason to expect it's not going to be 50/50.
It's a little cleaner to think about this as a collaborative way to select the final two doors. You pick a door, and then the host picks a door to include. The host indicates their pick by opening the door that isn't included, revealing its goat. (The video explanation extends this to a case that starts with 1000 doors, and then the host opens 998 of them.)
So, let's suppose you know ahead of time that door 2 has a goat. Then, when the host first asks you for a door, you can pick door 2 to include. That means, when the host picks a door to include in the final choice, they have to pick the one with the prize. You have forced their choice, and you know you forced it -- and, since you know they had to pick the door with the prize, you "switch" and open the door they picked, and you get the prize. There's no randomness there. If you know that door 2 has a goat, and you pick it as your first choice, you can always win. 100% of the time.
But, in reality, you don't know. But you can make a more-likely-than-not guess! Suppose that you guess that door 2 has a goat, and then act accordingly. 2/3 of the time, you're right in that guess, so following the "I know door 2 is bad" logic means you win. And 1/3 of the time, your guess about door 2 is wrong, so following the logic leads you to lose.
(Of course, you could pick one of the other doors to guess about; since you don't actually have any knowledge ahead of time, it works out the same.)
And that's why "switching" will be correct 2/3 of the time.
The paradox is this: In front of you is a game show host, and three doors. Behind one of the doors is a Shiny New Car, or some other great prize that you may win. Behind the other two doors are goats. You select a door. The host then opens one of the doors that you didn't select, revealing a goat, and then offers you an choice: Do you keep the door you selected first, or do you switch to the other door that they didn't open? If the door you select (either by keeping your first selection or switching to the other one) is the one with the prize, you win the prize!
If your door contains the goat, my understanding is that you do not actually get to keep the goat.
(A key datapoint -- often omitted from the descriptions! -- is that this is how the process always goes, and you know that fact. The host will always open a door with a goat, and will always offer the opportunity to switch. This is not a case where the host is being devious and only trying to get you to switch away if you start out choosing the prize.)
The paradoxical result is that switching will lead to the prize twice as often as not-switching, even though it looks like a random choice between two doors that you have no information about.
Explanation behind cut....
First, notice that the host is being devious in how they ask the questions, and the door you pick first isn't a choice of a door to open. It's a choice about how the second half of the problem will be set up. And that means the second half of the problem isn't set up randomly, so you pretty immediately have an intuitive reason to expect it's not going to be 50/50.
It's a little cleaner to think about this as a collaborative way to select the final two doors. You pick a door, and then the host picks a door to include. The host indicates their pick by opening the door that isn't included, revealing its goat. (The video explanation extends this to a case that starts with 1000 doors, and then the host opens 998 of them.)
So, let's suppose you know ahead of time that door 2 has a goat. Then, when the host first asks you for a door, you can pick door 2 to include. That means, when the host picks a door to include in the final choice, they have to pick the one with the prize. You have forced their choice, and you know you forced it -- and, since you know they had to pick the door with the prize, you "switch" and open the door they picked, and you get the prize. There's no randomness there. If you know that door 2 has a goat, and you pick it as your first choice, you can always win. 100% of the time.
But, in reality, you don't know. But you can make a more-likely-than-not guess! Suppose that you guess that door 2 has a goat, and then act accordingly. 2/3 of the time, you're right in that guess, so following the "I know door 2 is bad" logic means you win. And 1/3 of the time, your guess about door 2 is wrong, so following the logic leads you to lose.
(Of course, you could pick one of the other doors to guess about; since you don't actually have any knowledge ahead of time, it works out the same.)
And that's why "switching" will be correct 2/3 of the time.
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Date: 2025-10-09 01:28 pm (UTC)If the host doesn't always open a losing door and offer the contestant the chance to switch, a host who wanted you to win the car could behave as described in this explanation, and a host who wanted you to lose could reveal a losing door only when he knew that you had already chosen the winning door. The third possibility in that scenario is a host who was instructed to be unpredictable, to increase suspense for the audience, and before each show was told something like "open door 2 (which contains a goat) if the contestant picks door 1 or 3, but don't offer a switch if they pick door 2" or "show the person a losing door if they first pick door 1, and otherwise just open the first door they chose."