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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 li

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.

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.