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Has anyone else spent hours on understanding this and just accepted they’ll never accept it? I completely get all the explanations but it just feels too weird.

bemusedthrow75 3y ago
No. I am not a mathematician but I got it once I understood that this is not drawing coloured balls randomly from a bag (which is how all my school probability problems seemed to go). That is: - the setup of the system matters. - The state of the system at the point of the decision to switch matters. - The choices don’t get re-randomised. So the probabilities assigned to the original choice (and the remaining alternative) still count. If the host had closed a curtain over the stage and randomised the remaining doors, then it would be 50:50. But he didn’t. So you’re still in the probabilities of the original choice. One of the goats has been removed. The car and one goat remain: you know this for sure. You are being offered a door knowing that behind it must, necessarily, be the opposite of your original choice, and the probabilities have not been reset. If you originally picked the goat, that door absolutely has a car behind it. And there's a 2/3 chance you picked the goat originally. So by inference there's a 2/3 chance the door has a car behind it. You should switch. I didn’t get it until I had written a simulation to see it for myself though!
full_monty 3y ago
I think I have a simple explanation. There's 1/3 chance of picking the car. 1/3 of the time you pick the car, switch and lose. 2/3 of the time you pick a goat, switch and win. Why? Because 2/3 of the time you picked a goat, Monty shows you the other goat, so if you switch you definitely get the car.
- bemusedthrow75 3y ago
  Yes but in fact Monty always shows you a goat. There's always a goat for him to choose regardless of your choice, he knows where it is, and he's not going show you the car or the goat you already picked. Showing you the goat is the event that, as you say, guarantees that the other door has the opposite of your original choice behind it. Because nobody closed the curtain to shuffle the choices. One of the things I think people struggle with -- and I struggle with -- is that probability isn't about hypothesising about a single event that happened and how it might have happened. It's about encapsulating all the possible ways a single specified scenario can play out in a single expression. Monty shows you a goat this time, but this means Monty always shows you a goat. There's no scenario where he is unable to show you a goat. Just like you always only pick one door. And there's always only two goats and one car. (Full marks for username choice)
Sohcahtoa82 3y ago
Based on other comments, it seems the people that refuse to accept the answer that it's better to switch and not 50/50 are nitpicking how the game is described. The idea is supposed to be that Monty always reveals a goat. He does not pick a door at random to reveal before offering the choice to switch. If you pick door A, and the car is in door B, he will reveal door C and ask if you want to switch. If you pick A, the car is in C, then he will open door B and ask if you want to switch. Expanding it to 100 doors with Monty opening 98 of them and then offering you the opportunity to switch should make it very intuitive. Let's say you pick door 7, and Monty then opens every door except your door and say, door 48, all revealing goats, then it should make you think "Wow...very odd that he didn't reveal door 48...there must be a reason he skipped that specific door". Make it 1,000,000 doors. You choose door 1. Monty opens every door except number 423,901, and all the doors he opened had goats. You're tired and hungry because all you've done for 4 days is watch Monty open doors to goats, but you should certainly be thinking it's odd that he skipped that one specific door. You should probably switch, though maybe take a nap before claiming your car.
NoboruWataya 3y ago
In addition to the "100 door" formulation that helped me a lot, the way I rationalise it is that: * When you choose your first door, the chance the car is behind that door is 1/3, for obvious reasons. * When Monty opens the goat door, nothing actually changes that is relevant to your odds. He is always going to open a goat door, whether your first choice is a car or a goat, so it's basically irrelevant. He hasn't move the car, he hasn't given you any information you didn't already know when you made your first choice. So the odds of your first choice being correct can't have changed. They are still 1/3. * The only possible outcomes are that your first-chosen door has the car, or the other remaining door has the car. The probability if your first-chosen door having the car is (still) 1/3, and the probabilities of all possible outcomes must add up to 1, so the probability of the other door having the car are 1 - 1/3 = 2/3.

bemusedthrow75 3y ago

No. I am not a mathematician but I got it once I understood that this is not drawing coloured balls randomly from a bag (which is how all my school probability problems seemed to go). That is: - the setup of the system matters. - The state of the system at the point of the decision to switch matters. - The choices don’t get re-randomised. So the probabilities assigned to the original choice (and the remaining alternative) still count. If the host had closed a curtain over the stage and randomised the remaining doors, then it would be 50:50. But he didn’t. So you’re still in the probabilities of the original choice. One of the goats has been removed. The car and one goat remain: you know this for sure. You are being offered a door knowing that behind it must, necessarily, be the opposite of your original choice, and the probabilities have not been reset. If you originally picked the goat, that door absolutely has a car behind it. And there's a 2/3 chance you picked the goat originally. So by inference there's a 2/3 chance the door has a car behind it. You should switch. I didn’t get it until I had written a simulation to see it for myself though!

full_monty 3y ago

I think I have a simple explanation. There's 1/3 chance of picking the car. 1/3 of the time you pick the car, switch and lose. 2/3 of the time you pick a goat, switch and win. Why? Because 2/3 of the time you picked a goat, Monty shows you the other goat, so if you switch you definitely get the car.

bemusedthrow75 3y ago
Yes but in fact Monty always shows you a goat. There's always a goat for him to choose regardless of your choice, he knows where it is, and he's not going show you the car or the goat you already picked. Showing you the goat is the event that, as you say, guarantees that the other door has the opposite of your original choice behind it. Because nobody closed the curtain to shuffle the choices. One of the things I think people struggle with -- and I struggle with -- is that probability isn't about hypothesising about a single event that happened and how it might have happened. It's about encapsulating all the possible ways a single specified scenario can play out in a single expression. Monty shows you a goat this time, but this means Monty always shows you a goat. There's no scenario where he is unable to show you a goat. Just like you always only pick one door. And there's always only two goats and one car. (Full marks for username choice)

Sohcahtoa82 3y ago

Based on other comments, it seems the people that refuse to accept the answer that it's better to switch and not 50/50 are nitpicking how the game is described. The idea is supposed to be that Monty always reveals a goat. He does not pick a door at random to reveal before offering the choice to switch. If you pick door A, and the car is in door B, he will reveal door C and ask if you want to switch. If you pick A, the car is in C, then he will open door B and ask if you want to switch. Expanding it to 100 doors with Monty opening 98 of them and then offering you the opportunity to switch should make it very intuitive. Let's say you pick door 7, and Monty then opens every door except your door and say, door 48, all revealing goats, then it should make you think "Wow...very odd that he didn't reveal door 48...there must be a reason he skipped that specific door". Make it 1,000,000 doors. You choose door 1. Monty opens every door except number 423,901, and all the doors he opened had goats. You're tired and hungry because all you've done for 4 days is watch Monty open doors to goats, but you should certainly be thinking it's odd that he skipped that one specific door. You should probably switch, though maybe take a nap before claiming your car.

NoboruWataya 3y ago

In addition to the "100 door" formulation that helped me a lot, the way I rationalise it is that: * When you choose your first door, the chance the car is behind that door is 1/3, for obvious reasons. * When Monty opens the goat door, nothing actually changes that is relevant to your odds. He is always going to open a goat door, whether your first choice is a car or a goat, so it's basically irrelevant. He hasn't move the car, he hasn't given you any information you didn't already know when you made your first choice. So the odds of your first choice being correct can't have changed. They are still 1/3. * The only possible outcomes are that your first-chosen door has the car, or the other remaining door has the car. The probability if your first-chosen door having the car is (still) 1/3, and the probabilities of all possible outcomes must add up to 1, so the probability of the other door having the car are 1 - 1/3 = 2/3.