3 ms·

The explanation in the article is fine as far as it goes, but I think it's much more helpful to emphasize one key fact, which is that MH has to show a goat, and

by topaz0 3y ago

The explanation in the article is fine as far as it goes, but I think it's much more helpful to emphasize one key fact, which is that MH has to show a goat, and can't open your first choice door, even if it's a goat. That means that if you did pick a goat the first time around, he can only have revealed the other goat, in which case switching will necessarily give you the car. That happens with probability 2/3. This is implicit in the "100 doors, MH opens 98 of them" construction discussed in the article but making it explicit makes the idea quite clear and intuitive, even with just 3 doors. I'm moderately sympathetic to the idea that this constraint is not immediately obvious from how the scenario is described, and could technically be ambiguous in some versions. I do think most people will agree that it makes sense to infer this constraint once you think about it. Obviously if he opened the door that you initially picked and it was a goat, you would know to switch to one of the other two (and have 1/2 chance of finding the car). (Or, more absurdly, he could just open the door with the car and you could choose it directly). Or you could imagine a game show where MH can choose to open a door or not, in which case the problem is not well-posed, because you would need to know how he makes the choice.

ghaff 3y ago
There are probably a lot of people at this point who have never watched one of these types of game shows. Maybe if they thought about it, they'd realize that MH probably will always show a goat but it wasn't explicit in the initial question that Marilyn answered. Certainly, that constraint is important to the final solution.
- topaz0 3y ago
  I guess my point is that in the inevitable arguments, whether they are in advice columns or HN comment threads or around dinner tables, sometimes seem to their participants to be about how math works, when they're really about the problem description. Sometimes the disagreement amounts to "I agree that those constraints make sense and I wasn't properly accounting for them" and other times it amounts to "I was actually picturing a scenario where MH might have shown you a car". Either is ok, and there might be room for more discussion after getting there, but it feels like a big waste for either of these to get confused with a fruitless argument where both sides think the other is just doing the math wrong.
- andrewla 3y ago
  No, he wouldn't always open a door -- sometimes he'd offer a trade; like "you can take $1,000 cash from me now, or have your door". I wonder if anyone has gone through every episode of "Let's Make a Deal" where this scenario occurs and look at the empirical distribution of switch/not-switch.
empyrrhicist 3y ago
nvm
- topaz0 3y ago
  Actually it does matter. If you lose automatically when a car is shown, that will happen in 1/3 of games: none of the times your first pick was a car (0 * 1/3 = 0), half of the times your first pick was a goat (1/2 * 2/3 = 1/3). The remaining 2/3 are equally split between the times you chose a car first (which are guaranteed to complete) and the times you chose a goat first and MH randomly chose the other goat rather than the car. (Note: this analysis does depend on the constraint that MH doesn't open your first pick door! Edit: actually the answer also becomes equal odds if he can pick any door, regardless of whether it was your first pick or the car. In that case, 1/3 of games will not complete, and the rest will be 50/50 between the unopened doors.)
- graeme 3y ago
  Monty never shows the car. That was stated in the original reply from Marilyn, and it should be obvious to anyone who's watched a game show. Why on earth would monty open a door, show you the prize (which you now can't get) and then ask you if you want to switch your door? No one would design a game show like that, so it means the question only has sense if interpreted as "Monty always opens a goat door".
SeanLuke 3y ago
> The explanation in the article is fine as far as it goes, but I think it's much more helpful to emphasize one key fact, which is that MH has to show a goat, and can't open your first choice door, even if it's a goat. The second part I agree with, but not the first part; because it sounds like you're saying MH is required by the rules to reveal a goat. But the problem works fine if he always opens a remaining door at random and it happens to be a goat. I'd say it as: once you had chosen a door, Monte was required to open some other door. And the door he happened to open, for whatever reason, had a goat. Wouldn't that work?
- topaz0 3y ago
  This has been addressed several times in the comments on this post, and the answer is no, it does matter that he has to select a goat. If he might have selected a car and just got lucky in selecting the goat, then all of the following had equal probability: 1. you picked car first, he picked the first goat 2. you picked car first, he picked the other goat 3. you picked first goat first, he picked the car 4. you picked first goat first, he picked the other goat 5. you picked other goat first, he picked the car 6. you picked other goat first, he picked the first goat You happen to know that he picked a goat, so you can eliminate 3 and 5, but the remaining four possibilities are still equally likely. Half of them let you win by switching. This is also the "MH stumbles into a door on accident" variant that others have mentioned. Another way of getting intuition about your version is to realize that, if he might have shown you a car, then the fact that he has opened a goat is evidence against the hypothesis that your first pick was a goat, because half of the times you choose a goat he would open a car, whereas if you chose a car he would have to show you a goat every time. It eliminates some of the ways things could have turned out if you started on a goat without eliminating any of the ways things could have turned out if you started on a car, so the former is relatively less likely in retrospect than it was before MH opened the goat. More obvious in the 100-door case: if your first pick was a goat, then the fact that he hasn't shown you a car after opening 98 doors at random (or equivalently leaving one door closed at random) either means that you were very lucky in the first pick and got the can (in which case it's easy to open 98 doors without a car) or he got very lucky in his random choices. Most games played in this way (98%) would end with MH revealing the car, and the fact that you don't happen to be in one of them only tells you that one or the other of you got very lucky. The probabilities work out that the two options are equally likely.

by topaz0 3y ago

ghaff 3y ago

There are probably a lot of people at this point who have never watched one of these types of game shows. Maybe if they thought about it, they'd realize that MH probably will always show a goat but it wasn't explicit in the initial question that Marilyn answered. Certainly, that constraint is important to the final solution.

topaz0 3y ago
I guess my point is that in the inevitable arguments, whether they are in advice columns or HN comment threads or around dinner tables, sometimes seem to their participants to be about how math works, when they're really about the problem description. Sometimes the disagreement amounts to "I agree that those constraints make sense and I wasn't properly accounting for them" and other times it amounts to "I was actually picturing a scenario where MH might have shown you a car". Either is ok, and there might be room for more discussion after getting there, but it feels like a big waste for either of these to get confused with a fruitless argument where both sides think the other is just doing the math wrong.
andrewla 3y ago
No, he wouldn't always open a door -- sometimes he'd offer a trade; like "you can take $1,000 cash from me now, or have your door". I wonder if anyone has gone through every episode of "Let's Make a Deal" where this scenario occurs and look at the empirical distribution of switch/not-switch.

empyrrhicist 3y ago

nvm

topaz0 3y ago
Actually it does matter. If you lose automatically when a car is shown, that will happen in 1/3 of games: none of the times your first pick was a car (0 * 1/3 = 0), half of the times your first pick was a goat (1/2 * 2/3 = 1/3). The remaining 2/3 are equally split between the times you chose a car first (which are guaranteed to complete) and the times you chose a goat first and MH randomly chose the other goat rather than the car. (Note: this analysis does depend on the constraint that MH doesn't open your first pick door! Edit: actually the answer also becomes equal odds if he can pick any door, regardless of whether it was your first pick or the car. In that case, 1/3 of games will not complete, and the rest will be 50/50 between the unopened doors.)
graeme 3y ago
Monty never shows the car. That was stated in the original reply from Marilyn, and it should be obvious to anyone who's watched a game show. Why on earth would monty open a door, show you the prize (which you now can't get) and then ask you if you want to switch your door? No one would design a game show like that, so it means the question only has sense if interpreted as "Monty always opens a goat door".

SeanLuke 3y ago

> The explanation in the article is fine as far as it goes, but I think it's much more helpful to emphasize one key fact, which is that MH has to show a goat, and can't open your first choice door, even if it's a goat. The second part I agree with, but not the first part; because it sounds like you're saying MH is required by the rules to reveal a goat. But the problem works fine if he always opens a remaining door at random and it happens to be a goat. I'd say it as: once you had chosen a door, Monte was required to open some other door. And the door he happened to open, for whatever reason, had a goat. Wouldn't that work?

topaz0 3y ago
This has been addressed several times in the comments on this post, and the answer is no, it does matter that he has to select a goat. If he might have selected a car and just got lucky in selecting the goat, then all of the following had equal probability: 1. you picked car first, he picked the first goat 2. you picked car first, he picked the other goat 3. you picked first goat first, he picked the car 4. you picked first goat first, he picked the other goat 5. you picked other goat first, he picked the car 6. you picked other goat first, he picked the first goat You happen to know that he picked a goat, so you can eliminate 3 and 5, but the remaining four possibilities are still equally likely. Half of them let you win by switching. This is also the "MH stumbles into a door on accident" variant that others have mentioned. Another way of getting intuition about your version is to realize that, if he might have shown you a car, then the fact that he has opened a goat is evidence against the hypothesis that your first pick was a goat, because half of the times you choose a goat he would open a car, whereas if you chose a car he would have to show you a goat every time. It eliminates some of the ways things could have turned out if you started on a goat without eliminating any of the ways things could have turned out if you started on a car, so the former is relatively less likely in retrospect than it was before MH opened the goat. More obvious in the 100-door case: if your first pick was a goat, then the fact that he hasn't shown you a car after opening 98 doors at random (or equivalently leaving one door closed at random) either means that you were very lucky in the first pick and got the can (in which case it's easy to open 98 doors without a car) or he got very lucky in his random choices. Most games played in this way (98%) would end with MH revealing the car, and the fact that you don't happen to be in one of them only tells you that one or the other of you got very lucky. The probabilities work out that the two options are equally likely.