Off topic, but I saw your tweet complaining about the Monty Hall Problem and it’s been in my brain ever since… I can’t see the flaw with concluding that switching doors will win 2/3 of the time. What have I missed??
@The Unstoppable Gravy Express: nothing. I have no idea what MGK’s argument is, but he’s wrong. The explanation that convinced me is that if a stranger wandered on set after the door is opened then he’d have a 50% chance of guessing right, but that’s because he lacks the information you have – which door you picked originally.
@Zyzzyva: After thinking about it all morning, I’ve also come up with this way of explaining it… work out what’s behind the unchosen door (the one you didn’t pick that Monty didn’t open) in each scenario.
If you originally chose correctly, the unchosen door is 100% for sure a goat.
If you originally chose wrong, the unchosen door is 100% for sure a car.
Since there are 2 wrong doors to choose and only 1 correct door, the unchosen door has the car 2/3 of the time, and you should switch.
My problem is that I’m incapable of approaching it as a statistical problem. I’ll always approach it as a psychological problem: why is Monty showing me the goat? Is he trying to shake me? To give me a hint? I have yet to see the problem posed in a fashion which would convince me that Monty’s motivations are irrelevant.
I don’t think that’s just your problem, I think that’s the problem.
If one of the rules of the game is “Monty is required to offer you a switch,” then you should definitely switch. But if Monty is allowed any discretion about when he makes a switch offer, it becomes a lot less clear (and in fact, you could argue that switching is a flat-out bad plan, on the “cider in your ear” principle).
Well, what if Monty is an evil wizard who can transform goats into cars and vice-versa with a twitch of his nose, then you’ll ALWAYS lose… unless you quickly behead him, then storm the stage and take the car by force!
I mean, you could answer “Lady & the Tiger” problems with “I put a slab of meat right outside the door and listen for growling” but that kind of misses the point.
It becomes clearer that switching is the winning move if you imagine that there are 100 doors, and Monty shows you 98 goats. You had a 1/100 chance of picking the car. Monty has left only one door closed, so 99/100, the door that Monty leaves closed must be the car. (This is basically Unstoppable Gravy Express’s argument, but with more doors.)
I wonder: does this “always switch” strategy work with Deal or No Deal, assuming that a high-value briefcase is still in play at the end of the game? I feel like it’s basically the same problem.
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Off topic, but I saw your tweet complaining about the Monty Hall Problem and it’s been in my brain ever since… I can’t see the flaw with concluding that switching doors will win 2/3 of the time. What have I missed??
I do like the Ken Burns documentaries, so of course I had to laugh at your accurate description!
@The Unstoppable Gravy Express: nothing. I have no idea what MGK’s argument is, but he’s wrong. The explanation that convinced me is that if a stranger wandered on set after the door is opened then he’d have a 50% chance of guessing right, but that’s because he lacks the information you have – which door you picked originally.
Or, pragmatically: simulation proves the case.
@Zyzzyva: After thinking about it all morning, I’ve also come up with this way of explaining it… work out what’s behind the unchosen door (the one you didn’t pick that Monty didn’t open) in each scenario.
If you originally chose correctly, the unchosen door is 100% for sure a goat.
If you originally chose wrong, the unchosen door is 100% for sure a car.
Since there are 2 wrong doors to choose and only 1 correct door, the unchosen door has the car 2/3 of the time, and you should switch.
Or in table form:
YOUR CHOICE………………………Goat / Goat / Car
MONTY REVEALS…………………Goat / Goat / Goat
THEREFORE OTHER DOOR HAS: Car / Car / Goat
So in 2 of 3 cases, switching gets you the car.
After playing the simulation 20 times, changing my choice 10 times and keeping it 10 times…I had 20 goats. Dangers of small sample size, I guess. 🙂
@Candlejack: yup. 😀
The “run 1000 times” button grinds out 67-33 pretty consistently.
Why are they showing The Roosevelts: An Intimate History in Canada? Are you guys remotely interested?
My problem is that I’m incapable of approaching it as a statistical problem. I’ll always approach it as a psychological problem: why is Monty showing me the goat? Is he trying to shake me? To give me a hint? I have yet to see the problem posed in a fashion which would convince me that Monty’s motivations are irrelevant.
I don’t think that’s just your problem, I think that’s the problem.
If one of the rules of the game is “Monty is required to offer you a switch,” then you should definitely switch. But if Monty is allowed any discretion about when he makes a switch offer, it becomes a lot less clear (and in fact, you could argue that switching is a flat-out bad plan, on the “cider in your ear” principle).
Shad: Franklin Roosevelt was the first Canadian President of the United States.
Well, what if Monty is an evil wizard who can transform goats into cars and vice-versa with a twitch of his nose, then you’ll ALWAYS lose… unless you quickly behead him, then storm the stage and take the car by force!
I mean, you could answer “Lady & the Tiger” problems with “I put a slab of meat right outside the door and listen for growling” but that kind of misses the point.
Monty always shows you a goat.
It becomes clearer that switching is the winning move if you imagine that there are 100 doors, and Monty shows you 98 goats. You had a 1/100 chance of picking the car. Monty has left only one door closed, so 99/100, the door that Monty leaves closed must be the car. (This is basically Unstoppable Gravy Express’s argument, but with more doors.)
I wonder: does this “always switch” strategy work with Deal or No Deal, assuming that a high-value briefcase is still in play at the end of the game? I feel like it’s basically the same problem.