5 ms·

From reading the problem as posed in the article, the confusion seems similar to the "plane on a treadmill" thing, where people are interpreting the premise of

by jml7c5 5y ago

From reading the problem as posed in the article, the confusion seems similar to the "plane on a treadmill" thing, where people are interpreting the premise of the problem differently. (In plane on a treadmill: "the treadmill moves backwards @ -1 * plane airspeed" vs "the treadmill moves backwards to keep plane airspeed at 0". In this: "the host will open the door at random and in this example it happen to have a goat" vs "the host will never open a door with a car behind it".)

dllthomas 5y ago
The presentation of the problem included the phrase "the host, who knows what's behind the doors, opens another door", so I think in that case the interpretation that Monty deliberately reveals a goat is more appropriate (in addition, obviously, to being what was intended), although I agree there is wiggle room enough to admit both possibilities and clarity about which case you are discussing is important as it changes the answer.
- jVinc 5y ago
  > as it changes the answer. It doesn't. If the host randomly reveals a car, then you have 0% chance to win. If he doesn't you have 66% chance to win by switching.
  - dllthomas 5y ago
    I once thought that had to be the case. I wrote the simulation to demonstrate that I was correct. I learned that I wasn't. I encourage you to do likewise. I think I now have an understanding of why the right answer is the right answer, but I thought I had that before; I am more confident in the answer than my reasoning.
    - deleted 5y ago
      [deleted]
    - jVinc 5y ago
      I owe you an apology, you're correct, I was wrong. I can't pinpoint where my reasoning failed me exactly, but my confidence certainly was misplaced, and that's from someone who did do the simulation originally to understand the outcomes, but not to understand the distinction between the host deliberately or randomly revealing the door. I'm sorry, but thank you for you persistence in correcting me. It was a fruitful learning experience to dig into this problem again.
- jml7c5 5y ago
  It does, but it doesn't emphasize it enough to allay confusion, particularly for those unfamiliar with the "gotcha" in the problem. It's easy to overlook in that description unless you know that it is the important clue to understanding. Kind of like the classic riddle "John's mother has four children. They are named March, April, and May. What's the last child's name?" which uses underemphasis for almost comedic effect.
rsa25519 5y ago
> In this: "the host will open the door at random and in this example it happen to have a goat" vs "the host will never open a door with a car behind it".) If the host opens a door with a goat, then it doesn't matter whether or not it was intentional.
- dllthomas 5y ago
  Yes it does. Or more precisely, it matters whether the host could be counted on to do so reliably; the mechanism for that doesn't matter. There's a difference here, that our language obscures, between procedure and hypothetical.
  - tpush 5y ago
    The intention of the host only matters if the contestant would have to choose the subsequent action (switching or not) before the hosts opens a door. If the host has revealed a goat door, and the contestant then has to decide what to do, the intentions of the host for having chosen the door are irrelevant.
    - dllthomas 5y ago
      (Noting again that the mechanism doesn't matter, what matters is the odds of various behavior by the host, but - I think reasonably - using "intentions of the host" as a proxy for that.) The intentions of the host do matter. Imagine the host picks the correct door by the following procedure: 1) picks an available door at random; 2) if that door has a goat, opens it; 3) if that door has the car, opens the other door. I hope you will agree that this is equivalent to the problem as originally intended - Monty can be relied on to reveal a goat, and exactly why doesn't matter. Breaking it down into equally likely cases, assuming the contestant picks door 3: A) The car is behind door 1, Monty picks door 1, Monty corrects. B) The car is behind door 1, Monty picks door 2 C) The car is behind door 2, Monty picks door 1 D) The car is behind door 2, Monty picks door 2, Monty corrects E) The car is behind door 3, Monty picks door 1 F) The car is behind door 3, Monty picks door 2 When Monty reveals the goat behind (say) door 2, we know we're in case A, B, or F. All remain equally likely, and switching wins in A and B. If Monty would not have corrected, then revealing the goat behind door 2 eliminates (the new) A as well, leaving us with only B and F, again equally likely. If all of this remains unconvincing, I encourage you to write a simple simulation of the problem.
jVinc 5y ago
> vs "the treadmill moves backwards to keep plane airspeed at 0". You might be correct in thinking people misunderstand it as that, but that's physically impossible, so I don't think you can argue that people have some sort of alternate understanding under which they are actually correct. That's just one type of wrong reasoning people might apply to the problem. Add to that, in your hypothetical understanding of Monty Hall, that actually still doesn't change anything. Because even if the host picks a door completely at random, you still should always switch, it would just sometimes be the case that the host goes "ohh, that's to bad, I revealed the car and you can't win it now", but obviously that doesn't do anything to change your odds, because you're always better off or the same by switching, it's just in the unfortunate cases you'r switching from 0% change to 0% chance. But that doesn't do anything to change the fact that if he didn't reveal the car you're still going from 33% to 66%.
- npinsker 5y ago
  It does change your odds -- you're now living in a world in which the car didn't get revealed, and by Bayesian reasoning that means it's more likely that you live in a world where you picked the correct door.
  - jVinc 5y ago
    I didn't state that it doesn't change your odds. Obviously going from 33% to 0% because the car isn't possible to win anymore will be a change of your odds. I stated it doesn't change anything, because in one situation you go from 33% to 66% by changing and in the other you go from 0% to 0% by changing. So there is no difference to the logic of always switching because it either improves your odds or keeps them the same.
    - PebblesRox 5y ago
      If the host reveals a goat at random rather than by special knowledge, your odds don't go from 33% to 66% by switching. The case where the host reveals a goat is only two out of three, not three out of three because one third of the time, the host will reveal the car. In one of those two cases, you picked a goat and in the other you picked a car. So you're at 50% odds whether you switch or not, if the host doesn't know where the car is. If the host does know, then there are three out of three cases where the host reveals a goat. In one out of three cases you picked the car but in the other two cases you picked the goat. So that's why your odds go up if you switch.

by jml7c5 5y ago

dllthomas 5y ago

The presentation of the problem included the phrase "the host, who knows what's behind the doors, opens another door", so I think in that case the interpretation that Monty deliberately reveals a goat is more appropriate (in addition, obviously, to being what was intended), although I agree there is wiggle room enough to admit both possibilities and clarity about which case you are discussing is important as it changes the answer.

jVinc 5y ago
> as it changes the answer. It doesn't. If the host randomly reveals a car, then you have 0% chance to win. If he doesn't you have 66% chance to win by switching.
- dllthomas 5y ago
  I once thought that had to be the case. I wrote the simulation to demonstrate that I was correct. I learned that I wasn't. I encourage you to do likewise. I think I now have an understanding of why the right answer is the right answer, but I thought I had that before; I am more confident in the answer than my reasoning.
  - deleted 5y ago
    [deleted]
  - jVinc 5y ago
    I owe you an apology, you're correct, I was wrong. I can't pinpoint where my reasoning failed me exactly, but my confidence certainly was misplaced, and that's from someone who did do the simulation originally to understand the outcomes, but not to understand the distinction between the host deliberately or randomly revealing the door. I'm sorry, but thank you for you persistence in correcting me. It was a fruitful learning experience to dig into this problem again.
jml7c5 5y ago
It does, but it doesn't emphasize it enough to allay confusion, particularly for those unfamiliar with the "gotcha" in the problem. It's easy to overlook in that description unless you know that it is the important clue to understanding. Kind of like the classic riddle "John's mother has four children. They are named March, April, and May. What's the last child's name?" which uses underemphasis for almost comedic effect.

rsa25519 5y ago

> In this: "the host will open the door at random and in this example it happen to have a goat" vs "the host will never open a door with a car behind it".) If the host opens a door with a goat, then it doesn't matter whether or not it was intentional.

dllthomas 5y ago
Yes it does. Or more precisely, it matters whether the host could be counted on to do so reliably; the mechanism for that doesn't matter. There's a difference here, that our language obscures, between procedure and hypothetical.
- tpush 5y ago
  The intention of the host only matters if the contestant would have to choose the subsequent action (switching or not) before the hosts opens a door. If the host has revealed a goat door, and the contestant then has to decide what to do, the intentions of the host for having chosen the door are irrelevant.
  - dllthomas 5y ago
    (Noting again that the mechanism doesn't matter, what matters is the odds of various behavior by the host, but - I think reasonably - using "intentions of the host" as a proxy for that.) The intentions of the host do matter. Imagine the host picks the correct door by the following procedure: 1) picks an available door at random; 2) if that door has a goat, opens it; 3) if that door has the car, opens the other door. I hope you will agree that this is equivalent to the problem as originally intended - Monty can be relied on to reveal a goat, and exactly why doesn't matter. Breaking it down into equally likely cases, assuming the contestant picks door 3: A) The car is behind door 1, Monty picks door 1, Monty corrects. B) The car is behind door 1, Monty picks door 2 C) The car is behind door 2, Monty picks door 1 D) The car is behind door 2, Monty picks door 2, Monty corrects E) The car is behind door 3, Monty picks door 1 F) The car is behind door 3, Monty picks door 2 When Monty reveals the goat behind (say) door 2, we know we're in case A, B, or F. All remain equally likely, and switching wins in A and B. If Monty would not have corrected, then revealing the goat behind door 2 eliminates (the new) A as well, leaving us with only B and F, again equally likely. If all of this remains unconvincing, I encourage you to write a simple simulation of the problem.

jVinc 5y ago

> vs "the treadmill moves backwards to keep plane airspeed at 0". You might be correct in thinking people misunderstand it as that, but that's physically impossible, so I don't think you can argue that people have some sort of alternate understanding under which they are actually correct. That's just one type of wrong reasoning people might apply to the problem. Add to that, in your hypothetical understanding of Monty Hall, that actually still doesn't change anything. Because even if the host picks a door completely at random, you still should always switch, it would just sometimes be the case that the host goes "ohh, that's to bad, I revealed the car and you can't win it now", but obviously that doesn't do anything to change your odds, because you're always better off or the same by switching, it's just in the unfortunate cases you'r switching from 0% change to 0% chance. But that doesn't do anything to change the fact that if he didn't reveal the car you're still going from 33% to 66%.

npinsker 5y ago
It does change your odds -- you're now living in a world in which the car didn't get revealed, and by Bayesian reasoning that means it's more likely that you live in a world where you picked the correct door.
- jVinc 5y ago
  I didn't state that it doesn't change your odds. Obviously going from 33% to 0% because the car isn't possible to win anymore will be a change of your odds. I stated it doesn't change anything, because in one situation you go from 33% to 66% by changing and in the other you go from 0% to 0% by changing. So there is no difference to the logic of always switching because it either improves your odds or keeps them the same.
  - PebblesRox 5y ago
    If the host reveals a goat at random rather than by special knowledge, your odds don't go from 33% to 66% by switching. The case where the host reveals a goat is only two out of three, not three out of three because one third of the time, the host will reveal the car. In one of those two cases, you picked a goat and in the other you picked a car. So you're at 50% odds whether you switch or not, if the host doesn't know where the car is. If the host does know, then there are three out of three cases where the host reveals a goat. In one out of three cases you picked the car but in the other two cases you picked the goat. So that's why your odds go up if you switch.