4 ms·

You can distill the disconnect down to one question. Does Monty know where the car is? (The original article says he does, but this often gets lost in the ver

You can distill the disconnect down to one question. Does Monty know where the car is? (The original article says he does, but this often gets lost in the version people read.) Suggest Monty doesn't, and he opens door #3 to reveal a goat only because it wasn't the door you picked (and it was merely luck the car wasn't there). In that case, it is genuinely a 50/50 shot between the remaining two. Now if Monty did know where the car was, and he wouldn't have opened door #3 if the car had been there, then the 2/3 percentage to switch is intact. To many this seems like a minor (or incorrect) distinction but it's little assumptions (Monty knows) that underpin these gotcha questions. That's one reason why Google-style interview questions irritate me so much. In many of them, there's an implicit assumption that is necessary but never stated. Many of the responses Marilyn got seemed to come from this form of irritation, even if some of the people writing in couldn't express a logical or justifiable basis for their irritation. It's an interesting puzzle, but it's too easily rephrased like a con.

MilnerRoute 8y ago
I always start explaining it like this. "There's a one in three chance you guessed right on the first try." And that just doesn't change, even if Monty Hall opens a door. So not only is there a two-in-three chance you guessed wrong on the first try. Monty is helpfully offering you a chance to switch to the correct/winning choice in the second round.
- mchannon 8y ago
  But Monty's only being helpful if I guessed wrong. If it actually was behind door #1, Monty's trying to entice me into picking a goat, and basically being a dick. This is also assuming he could've just opened door #1 right away if it was the correct answer and obviated this whole mess. Monty's free will (an unknown) underpins this entire argument- what choices does he have, and why does he even make me choose a door instead of giving me the car (and the goats) outright? Maybe from a statistical point of view it's better to have Monty pull this shtick, but something that's helpful 2/3 of the time and harmful 1/3 of the time doesn't make the cutoff for helpful.
  - ModernMech 8y ago
    I feel like you're still missing something. The probabilities come out of a table. It's just math. There's no need to bring "free will" into the explanation.
    - username90 8y ago
      No, it is you who are wrong. You can't write down a table if you don't have a proper explanation of the possible event chains, and in this case you don't. Instead when people write down these event chain they add information that isn't there and then think they actually solved the problem. But all you really know is that this one time the host opened a door with a goat, you have no idea if the host always does this since the problem statement doesn't actually say it outright. It might be a good assumption to make, but it isn't in the description so it isn't a mathematically rigorous answer.
      - ModernMech 8y ago
        But once you do have a proper explanation of the possible event chains, free will still doesn't play a part of it. It's a game with rules. You seem to be keyed in on some aspect of the rules being unclear, but even with the rules perfectly clear or not clear at all, the probabilities are the same regardless of the psychological state of the contestant.
  - indigodaddy 8y ago
    But the odds of you guessing wrong are 2/3, so you should switch.
Confusion 8y ago
If Monty would open the door where the car is, the decision whether to switch wouldn't be much of a gamble and there wouldn't be much entertainment in this part of the game show. So of course Monty knows where the car is. There is nothing mysterious or unclear about that. This is just a hard question that is hard to get right and there is no shame in getting it wrong if you have to come up with the answer from scratch, never having heard such a puzzle before. There is no shame in not 'getting' it either. There is shame in allowing not 'getting' it to determine your conclusions.
imh 8y ago
I didn't think what you say is true, but it is. Here's what misled me: Go back to the 100 door version. You start out by opening 1 door, so it's 99/100 that the car is behind another door. If monty just happens to open 98 of those doors and reveal all goats, then that 99% probability that you should switch to one of them combines with new knowledge of which to switch to. Its exceedingly rare, but if it happened, then you should still switch. The above logic is wrong. I simulated and you're correct: n = 10000000 n_doors = 3 first_guess = np.random.randint(n_doors, size=n) car_behind = np.random.randint(n_doors, size=n) should_stay = first_guess == car_behind print('p(should stay)', np.mean(should_stay)) # 0.3333381 monty_opens_all_but = (first_guess + 1 + np.random.randint(n_doors-1, size=n)) % n_doors print('bad setup?: ', np.any(monty_opens_all_but == first_guess)) # False monty_shows_only_goats = np.logical_or(monty_opens_all_but == car_behind, first_guess == car_behind) print('p(monty shows only goats)', np.mean(monty_shows_only_goats)) # 0.6667607 print('p(should stay| monty shows only goats)', np.mean(should_stay[monty_shows_only_goats])) # 0.499936633938 I'm kinda astounded. I do stats for a living, yet without writing out the math, my intuition misled me. I thought I had a framing of the problem that allowed me to use a quick shortcut in my thinking, and that framing was wrong. Two takeaways: 1) Probability is really hard to get the right intuition about. Reasoning by analogy/shortcut problem framing is dangerous. You have to write out the math. 2) The "100 doors" explanation for the usual monty hall problem is correct for subtler reasons than are immediately obvious. You could probably set up a counter monty hall problem to trick people where there are 100 doors and he just happens to show 98 goats.
username90 8y ago
You are right, to make this easier to understand, imagine replacing Monty Hall with a street peddler. You are playing a game with a street peddler. There are 3 cards, two are duds and one is the prize. He lets you pick one card. When you picked a card he flips one of the other two revealing it as a dud, and gives you the option to switch. Do you switch? A person who have heard the naive explanation to the Monty Hall problem would say yes. A smart person would say no. Why? Because it is not in the street peddlers interest to let you win, he has mouths to feed and need the money! So the only reason he reveals another card and asks you to switch is because you picked the right one from the beginning, so staying in this case means 100% chance of winning, and switching is 100% chance of losing. For the naive Monty Hall interpretation to be correct we need two things to be true: Monty Hall always opens a door and the thing behind the door is always a goat. If any of those two are not true then the popular explanation is wrong, and if you look around most explanations of the problem leaves those facts out. For example, in this article they forgot to say that the game host always opens a door and when he opens it it is always a goat. This shows that the author of the article (and almost everyone here at HN) doesn't understand Monty Hall. Instead they talk like non mathematicians like "Why would the game not open a door sometimes?" or "Why would he open a door with not a goat? It said he opened a door with a goat this one time!" etc. The problem with those things is that the explanation only says what happened this one time, so you can't possible write down the event tree without making a lot of extra assumptions about the game.

MilnerRoute 8y ago

I always start explaining it like this. "There's a one in three chance you guessed right on the first try." And that just doesn't change, even if Monty Hall opens a door. So not only is there a two-in-three chance you guessed wrong on the first try. Monty is helpfully offering you a chance to switch to the correct/winning choice in the second round.

mchannon 8y ago
But Monty's only being helpful if I guessed wrong. If it actually was behind door #1, Monty's trying to entice me into picking a goat, and basically being a dick. This is also assuming he could've just opened door #1 right away if it was the correct answer and obviated this whole mess. Monty's free will (an unknown) underpins this entire argument- what choices does he have, and why does he even make me choose a door instead of giving me the car (and the goats) outright? Maybe from a statistical point of view it's better to have Monty pull this shtick, but something that's helpful 2/3 of the time and harmful 1/3 of the time doesn't make the cutoff for helpful.
- ModernMech 8y ago
  I feel like you're still missing something. The probabilities come out of a table. It's just math. There's no need to bring "free will" into the explanation.
  - username90 8y ago
    No, it is you who are wrong. You can't write down a table if you don't have a proper explanation of the possible event chains, and in this case you don't. Instead when people write down these event chain they add information that isn't there and then think they actually solved the problem. But all you really know is that this one time the host opened a door with a goat, you have no idea if the host always does this since the problem statement doesn't actually say it outright. It might be a good assumption to make, but it isn't in the description so it isn't a mathematically rigorous answer.
    - ModernMech 8y ago
      But once you do have a proper explanation of the possible event chains, free will still doesn't play a part of it. It's a game with rules. You seem to be keyed in on some aspect of the rules being unclear, but even with the rules perfectly clear or not clear at all, the probabilities are the same regardless of the psychological state of the contestant.
- indigodaddy 8y ago
  But the odds of you guessing wrong are 2/3, so you should switch.

Confusion 8y ago

If Monty would open the door where the car is, the decision whether to switch wouldn't be much of a gamble and there wouldn't be much entertainment in this part of the game show. So of course Monty knows where the car is. There is nothing mysterious or unclear about that. This is just a hard question that is hard to get right and there is no shame in getting it wrong if you have to come up with the answer from scratch, never having heard such a puzzle before. There is no shame in not 'getting' it either. There is shame in allowing not 'getting' it to determine your conclusions.

imh 8y ago

I didn't think what you say is true, but it is. Here's what misled me: Go back to the 100 door version. You start out by opening 1 door, so it's 99/100 that the car is behind another door. If monty just happens to open 98 of those doors and reveal all goats, then that 99% probability that you should switch to one of them combines with new knowledge of which to switch to. Its exceedingly rare, but if it happened, then you should still switch. The above logic is wrong. I simulated and you're correct: n = 10000000 n_doors = 3 first_guess = np.random.randint(n_doors, size=n) car_behind = np.random.randint(n_doors, size=n) should_stay = first_guess == car_behind print('p(should stay)', np.mean(should_stay)) # 0.3333381 monty_opens_all_but = (first_guess + 1 + np.random.randint(n_doors-1, size=n)) % n_doors print('bad setup?: ', np.any(monty_opens_all_but == first_guess)) # False monty_shows_only_goats = np.logical_or(monty_opens_all_but == car_behind, first_guess == car_behind) print('p(monty shows only goats)', np.mean(monty_shows_only_goats)) # 0.6667607 print('p(should stay| monty shows only goats)', np.mean(should_stay[monty_shows_only_goats])) # 0.499936633938 I'm kinda astounded. I do stats for a living, yet without writing out the math, my intuition misled me. I thought I had a framing of the problem that allowed me to use a quick shortcut in my thinking, and that framing was wrong. Two takeaways: 1) Probability is really hard to get the right intuition about. Reasoning by analogy/shortcut problem framing is dangerous. You have to write out the math. 2) The "100 doors" explanation for the usual monty hall problem is correct for subtler reasons than are immediately obvious. You could probably set up a counter monty hall problem to trick people where there are 100 doors and he just happens to show 98 goats.

username90 8y ago

You are right, to make this easier to understand, imagine replacing Monty Hall with a street peddler. You are playing a game with a street peddler. There are 3 cards, two are duds and one is the prize. He lets you pick one card. When you picked a card he flips one of the other two revealing it as a dud, and gives you the option to switch. Do you switch? A person who have heard the naive explanation to the Monty Hall problem would say yes. A smart person would say no. Why? Because it is not in the street peddlers interest to let you win, he has mouths to feed and need the money! So the only reason he reveals another card and asks you to switch is because you picked the right one from the beginning, so staying in this case means 100% chance of winning, and switching is 100% chance of losing. For the naive Monty Hall interpretation to be correct we need two things to be true: Monty Hall always opens a door and the thing behind the door is always a goat. If any of those two are not true then the popular explanation is wrong, and if you look around most explanations of the problem leaves those facts out. For example, in this article they forgot to say that the game host always opens a door and when he opens it it is always a goat. This shows that the author of the article (and almost everyone here at HN) doesn't understand Monty Hall. Instead they talk like non mathematicians like "Why would the game not open a door sometimes?" or "Why would he open a door with not a goat? It said he opened a door with a goat this one time!" etc. The problem with those things is that the explanation only says what happened this one time, so you can't possible write down the event tree without making a lot of extra assumptions about the game.