5 ms·

> 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 impos

by 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.
- JoshuaDavid 5y ago
  The problem is that there are two interpretations of the problem, and is is not obvious that the incorrect one is wrong unless you know that Monty knows what's behind all the doors, and chooses never to open a door with a car behind it. The question isn't "does Monty showing a goat change the odds that it's behind the door that neither you nor Monty picked", it's "does it change that probability from 33% to 50% or from 33% to 66%". Scenario 1: The first, incorrect, interpretation of the problem is "You choose a door, which has either a goat or car behind it. Monty then chooses one of the other two doors, which will also have either a car or goat on it, and opens that door. You then have the choice of whether to switch doors or stay with your original choice". Scenario 2: The second, correct interpretation of the problem is "You choose a door, which has either a goat or a car behind it. Monty then looks behind the other two doors, and chooses the one that has a goat behind it. If both have goats behind them, Monty chooses randomly. Monty opens his chosen door. You then have the choice of whether to switch doors or stay with your original choice". +-----+------+--------+---------+--------+--------+------------+------------+ | Row | Car | Your | Monty's | Result | Result | Frequency | Frequency | | | Door | Choice | Choice | Stay | Switch | Scenario 1 | Scenario 2 | +-----+------+--------+---------+--------+--------+------------+------------+ | 1 | #1 | #1 | #2 | Car | Goat | 1/18 | 1/18 | | 2 | #1 | #1 | #3 | Car | Goat | 1/18 | 1/18 | | 3 | #1 | #2 | #1 | Goat | Goat | 1/18 | 0/18 | | 4 | #1 | #2 | #3 | Goat | Car | 1/18 | 2/18 | | 5 | #1 | #3 | #1 | Goat | Goat | 1/18 | 0/18 | | 6 | #1 | #3 | #2 | Goat | Car | 1/18 | 2/18 | | 7 | #2 | #1 | #2 | Goat | Goat | 1/18 | 0/18 | | 8 | #2 | #1 | #3 | Goat | Car | 1/18 | 2/18 | | 9 | #2 | #2 | #1 | Car | Goat | 1/18 | 1/18 | | 10 | #2 | #2 | #3 | Car | Goat | 1/18 | 1/18 | | 11 | #2 | #3 | #1 | Goat | Car | 1/18 | 2/18 | | 12 | #2 | #3 | #2 | Goat | Goat | 1/18 | 0/18 | | 13 | #3 | #1 | #2 | Goat | Car | 1/18 | 2/18 | | 14 | #3 | #1 | #3 | Goat | Goat | 1/18 | 0/18 | | 15 | #3 | #2 | #1 | Goat | Car | 1/18 | 2/18 | | 16 | #3 | #2 | #3 | Goat | Goat | 1/18 | 0/18 | | 17 | #3 | #3 | #1 | Car | Goat | 1/18 | 1/18 | | 18 | #3 | #3 | #2 | Car | Goat | 1/18 | 1/18 | +-----+------+--------+---------+--------+--------+------------+------------+ In scenario 1, before any door is opened, you chose the correct door in the scenarios corresponding to rows 1, 2, 9, 10, 17, and 18, for an aggregate probability of 6/18 == 1 / 3 of the time having picked the door with a car. Monty then opened a door which happened to have a goat behind it, which eliminates rows 3, 5, 7, 12, 14, and 16. Now you have a 6/12 chance of winning the car if you stay, and a 6/12 chance of winning the car if you switch, and this is how you come to the conclusion that there is no advantage in switching. In scenario 2, you still chose the correct door in the scenarios corresponding to rows 1, 2, 9, 10, 17, and 18, for an aggregate probability of 6/18 == 1 / 3 of the time having picked the door with a car. However, this time Monty's door-opening doesn't eliminate any rows with nonzero probability, since rows 3, 5, 7, 12, 14, and 16 have zero probability to start with. As such, you still have a 6/18 chance of winning the car if you stay, and a 12/18 chance of winning if you switch, so you should switch.
TchoBeer 5y ago
It's great that you're so sure of yourself, just a shame that you're wrong. If the host picks randomly, your chances are in fact 50-50. Think of it this way: imagine every time the hosts picks the car the universe resets. Now imagine you see the host picking a goat. There's a 2/3rds chance in your universe you picked the car, and a 1/3rds chance you picked a goat, and so switching would seem like the bad option. This actually cancels out the effects of the normal Monty Hall problem, and we are left with a 50-50 chance.
- jVinc 5y ago
  It's funny to me that there always seem to be an over representation of rude people among those who get this problem wrong. There is absolutely no need to attack my person. As for the problem, you are just plain wrong. We are talking about are regular old TV-quiz, there is not "universe resetting button" to save your logic. In the regular situation, the host always remove a goat. because otherwise the quiz show is kinda boring. But in the modified version we are considering where he just picks a door at random, then in 2 of the 6 possible outcomes for the door the host picks, he'll be removing the car. Now since the problem has the host showing the doors content, that should be the end of the game show. Who want's to see someone pondering if they should switch between a goat and another goat right? But that doesn't change the logic at all, because the conclusion of "you should always switch" isn't impacted in any way. 0% to 0% is just no change. And in the rest of the cases your'll go from 33% to 66%. Now of cause what would happen in your odd example where the host has a universe resetting button is that you don't have any choice at all, because no matter what the host will reset the universe until they maximize ratings, which might have you get the car or might have you not get the car, but there's no choice to be taken and the outcome that has maximum rating will always happen with 100%. That's why most stats problems avoid introducing "then the host resets the universe" in their problem description, it kind of ruins the whole point of calculating proabilities.
  - listenallyall 5y ago
    I agree with you that "the universe resetting" is a dumb way to visualize or explain the problem. But if Mr. Hall opens the first door randomly (essential that it was a random choice), and it revealed a goat, then no, switching does not provide an advantage. You do not go from 33 to 67%. Instead, you are left with 2 options, both of which originally had a probability of 33% of containing a car, and which, now that the third option has been eliminated, are both currently 50%.

by jVinc 5y ago

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.
JoshuaDavid 5y ago
The problem is that there are two interpretations of the problem, and is is not obvious that the incorrect one is wrong unless you know that Monty knows what's behind all the doors, and chooses never to open a door with a car behind it. The question isn't "does Monty showing a goat change the odds that it's behind the door that neither you nor Monty picked", it's "does it change that probability from 33% to 50% or from 33% to 66%". Scenario 1: The first, incorrect, interpretation of the problem is "You choose a door, which has either a goat or car behind it. Monty then chooses one of the other two doors, which will also have either a car or goat on it, and opens that door. You then have the choice of whether to switch doors or stay with your original choice". Scenario 2: The second, correct interpretation of the problem is "You choose a door, which has either a goat or a car behind it. Monty then looks behind the other two doors, and chooses the one that has a goat behind it. If both have goats behind them, Monty chooses randomly. Monty opens his chosen door. You then have the choice of whether to switch doors or stay with your original choice". +-----+------+--------+---------+--------+--------+------------+------------+ | Row | Car | Your | Monty's | Result | Result | Frequency | Frequency | | | Door | Choice | Choice | Stay | Switch | Scenario 1 | Scenario 2 | +-----+------+--------+---------+--------+--------+------------+------------+ | 1 | #1 | #1 | #2 | Car | Goat | 1/18 | 1/18 | | 2 | #1 | #1 | #3 | Car | Goat | 1/18 | 1/18 | | 3 | #1 | #2 | #1 | Goat | Goat | 1/18 | 0/18 | | 4 | #1 | #2 | #3 | Goat | Car | 1/18 | 2/18 | | 5 | #1 | #3 | #1 | Goat | Goat | 1/18 | 0/18 | | 6 | #1 | #3 | #2 | Goat | Car | 1/18 | 2/18 | | 7 | #2 | #1 | #2 | Goat | Goat | 1/18 | 0/18 | | 8 | #2 | #1 | #3 | Goat | Car | 1/18 | 2/18 | | 9 | #2 | #2 | #1 | Car | Goat | 1/18 | 1/18 | | 10 | #2 | #2 | #3 | Car | Goat | 1/18 | 1/18 | | 11 | #2 | #3 | #1 | Goat | Car | 1/18 | 2/18 | | 12 | #2 | #3 | #2 | Goat | Goat | 1/18 | 0/18 | | 13 | #3 | #1 | #2 | Goat | Car | 1/18 | 2/18 | | 14 | #3 | #1 | #3 | Goat | Goat | 1/18 | 0/18 | | 15 | #3 | #2 | #1 | Goat | Car | 1/18 | 2/18 | | 16 | #3 | #2 | #3 | Goat | Goat | 1/18 | 0/18 | | 17 | #3 | #3 | #1 | Car | Goat | 1/18 | 1/18 | | 18 | #3 | #3 | #2 | Car | Goat | 1/18 | 1/18 | +-----+------+--------+---------+--------+--------+------------+------------+ In scenario 1, before any door is opened, you chose the correct door in the scenarios corresponding to rows 1, 2, 9, 10, 17, and 18, for an aggregate probability of 6/18 == 1 / 3 of the time having picked the door with a car. Monty then opened a door which happened to have a goat behind it, which eliminates rows 3, 5, 7, 12, 14, and 16. Now you have a 6/12 chance of winning the car if you stay, and a 6/12 chance of winning the car if you switch, and this is how you come to the conclusion that there is no advantage in switching. In scenario 2, you still chose the correct door in the scenarios corresponding to rows 1, 2, 9, 10, 17, and 18, for an aggregate probability of 6/18 == 1 / 3 of the time having picked the door with a car. However, this time Monty's door-opening doesn't eliminate any rows with nonzero probability, since rows 3, 5, 7, 12, 14, and 16 have zero probability to start with. As such, you still have a 6/18 chance of winning the car if you stay, and a 12/18 chance of winning if you switch, so you should switch.

TchoBeer 5y ago

It's great that you're so sure of yourself, just a shame that you're wrong. If the host picks randomly, your chances are in fact 50-50. Think of it this way: imagine every time the hosts picks the car the universe resets. Now imagine you see the host picking a goat. There's a 2/3rds chance in your universe you picked the car, and a 1/3rds chance you picked a goat, and so switching would seem like the bad option. This actually cancels out the effects of the normal Monty Hall problem, and we are left with a 50-50 chance.

jVinc 5y ago
It's funny to me that there always seem to be an over representation of rude people among those who get this problem wrong. There is absolutely no need to attack my person. As for the problem, you are just plain wrong. We are talking about are regular old TV-quiz, there is not "universe resetting button" to save your logic. In the regular situation, the host always remove a goat. because otherwise the quiz show is kinda boring. But in the modified version we are considering where he just picks a door at random, then in 2 of the 6 possible outcomes for the door the host picks, he'll be removing the car. Now since the problem has the host showing the doors content, that should be the end of the game show. Who want's to see someone pondering if they should switch between a goat and another goat right? But that doesn't change the logic at all, because the conclusion of "you should always switch" isn't impacted in any way. 0% to 0% is just no change. And in the rest of the cases your'll go from 33% to 66%. Now of cause what would happen in your odd example where the host has a universe resetting button is that you don't have any choice at all, because no matter what the host will reset the universe until they maximize ratings, which might have you get the car or might have you not get the car, but there's no choice to be taken and the outcome that has maximum rating will always happen with 100%. That's why most stats problems avoid introducing "then the host resets the universe" in their problem description, it kind of ruins the whole point of calculating proabilities.
- listenallyall 5y ago
  I agree with you that "the universe resetting" is a dumb way to visualize or explain the problem. But if Mr. Hall opens the first door randomly (essential that it was a random choice), and it revealed a goat, then no, switching does not provide an advantage. You do not go from 33 to 67%. Instead, you are left with 2 options, both of which originally had a probability of 33% of containing a car, and which, now that the third option has been eliminated, are both currently 50%.