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The trick that finally got me to accept the conclusion was to imagine there were 100 doors. You pick one, then Monty opens 98 other doors. Then you obviously sw

The trick that finally got me to accept the conclusion was to imagine there were 100 doors. You pick one, then Monty opens 98 other doors. Then you obviously switch, as the original door has a 1% chance of being right, and the other door has a 99% of being right.

sdfhdhjdw3 4y ago
I've hear that before many times, but caution. You're just leaning on your intuition, probably not understanding why you got to the wrong answer in the first place.
aaronjl 4y ago
What helped me is the fact that the only way you WON'T win by switching, is if you correctly choose it the first time. And that only happens 1/3 of the time
- injb 4y ago
  The problem I always had with this (and still have) is that you're not asked about the probability (i.e. what trend you expect to see in a string of repeated experiments), but rather you're asked: "Is it to your advantage to switch your choice?" Well that depends on where the prize is. If the prize is behind the other door, then yes it's to your advantage to switch. Otherwise, it isn't. How can there be any more to it than that? You're not going to be given the opportunity to play the game 100 times. This is your one shot and the location of the prize is already fixed.
  - jjbinx007 4y ago
    Imagine there are 100 doors. You pick one at random. You have a 1 in 100 chance of getting the right answer. If the host opens all the doors that contain the booby prizes bar one (in this case 98 goats) and leaves one door remaining, plus your original guess, it's pretty obvious you are unlikely to have guessed correctly on your first try. Scale it down to 3 doors and the host always revealing a goat and you have the same situation except there's a 1 in 3 chance you'll win by sticking with your first guess and a 2 in 3 chance if you swap.
  - nkrisc 4y ago
    Well if you don't switch then your odds of winning are a simple 1 out of 3. Switching inverts your initial guess, so instead of losing 2 out of 3 times because your initial pick was wrong, switching makes you win 2 out of 3 times when your initial pick is wrong.
  - Jtsummers 4y ago
    Right, it's your one shot. So what gives you better odds (that is, is to your advantage). If you don't switch, you're selecting the lower odds chance of winning, putting yourself at a disadvantage. Since you don't know where it is, but do have a chance to double your chance of winning, then switching is advantageous. Expected value is the thing to consider.
thehappypm 4y ago
The problem I have with this is that the original Monty Hall problem has him opening one door out of 2. With the 100 doors analogy, wouldn’t Monty be opening just one of the remaining doors? Or half? Why all 98?
- injb 4y ago
  Then the question of "switching" wouldn't make sense. Which of the remaining 98 would you be asked to switch to? The whole point is that you end up with a choice of sticking with the door you picked, or switching to the one remaining door.
  - colmanhumphrey 4y ago
    Seems fine to me. Do you want to stick with your original or switch to one of the remaining 98? It’s still worth switching too, just a smaller increase in probability.
- waynesonfire 4y ago
  i think the idea is to demonstrate more easily the information leak by exaggerating the amount of information being leaked. with 3 doors, there are only so many the host can chose to open. with a 100, there is up to 98. sure, it can be 1 or 2... or 98.
- jasonswett 4y ago
  The rule is "after the player has guessed a door, open ALL the other doors except the door that was guessed plus one other door". I think your confusion is because if there are 100 doors then the rule is self-evident. But when there are only 3 doors, the rule of "open all doors except the door that was guessed plus one other" is indistinguishable from "open just one door". The key is knowing that the rule is the "open all..." rule.
  - colmanhumphrey 4y ago
    That’s not the rule from the question in the link nor in the original game (the game’s rule isn’t quite the same as the classic problem, it’s even messier). Your proposed rule doesn’t uniquely extend the original, it’s just one of the possible generalizations.
dahfizz 4y ago
> Then you obviously switch, as the original door has a 1% chance of being right, and the other door has a 99% of being right. It's not obvious to me that this is true. Why do we update the chance of the "new" door but not the one we already selected? Intuitively, I would expect the odds of both doors to increase to 50% as the other 98 doors are opened. I know that you're correct, I just don't find this to be an intuitive explanation.
- mcphage 4y ago
  The odds of your original door being right are 1%—that never changes. It’s determined by the options when the game was set up and when you picked it. If your original door was wrong—99% chance—then it’s in one of the 99 other doors. But the host opens 98 of them, so if it was one of the 99 other doors—again, 99% chance—then the only option left is the last door.

sdfhdhjdw3 4y ago

I've hear that before many times, but caution. You're just leaning on your intuition, probably not understanding why you got to the wrong answer in the first place.

aaronjl 4y ago

What helped me is the fact that the only way you WON'T win by switching, is if you correctly choose it the first time. And that only happens 1/3 of the time

injb 4y ago
The problem I always had with this (and still have) is that you're not asked about the probability (i.e. what trend you expect to see in a string of repeated experiments), but rather you're asked: "Is it to your advantage to switch your choice?" Well that depends on where the prize is. If the prize is behind the other door, then yes it's to your advantage to switch. Otherwise, it isn't. How can there be any more to it than that? You're not going to be given the opportunity to play the game 100 times. This is your one shot and the location of the prize is already fixed.
- jjbinx007 4y ago
  Imagine there are 100 doors. You pick one at random. You have a 1 in 100 chance of getting the right answer. If the host opens all the doors that contain the booby prizes bar one (in this case 98 goats) and leaves one door remaining, plus your original guess, it's pretty obvious you are unlikely to have guessed correctly on your first try. Scale it down to 3 doors and the host always revealing a goat and you have the same situation except there's a 1 in 3 chance you'll win by sticking with your first guess and a 2 in 3 chance if you swap.
- nkrisc 4y ago
  Well if you don't switch then your odds of winning are a simple 1 out of 3. Switching inverts your initial guess, so instead of losing 2 out of 3 times because your initial pick was wrong, switching makes you win 2 out of 3 times when your initial pick is wrong.
- Jtsummers 4y ago
  Right, it's your one shot. So what gives you better odds (that is, is to your advantage). If you don't switch, you're selecting the lower odds chance of winning, putting yourself at a disadvantage. Since you don't know where it is, but do have a chance to double your chance of winning, then switching is advantageous. Expected value is the thing to consider.

thehappypm 4y ago

The problem I have with this is that the original Monty Hall problem has him opening one door out of 2. With the 100 doors analogy, wouldn’t Monty be opening just one of the remaining doors? Or half? Why all 98?

injb 4y ago
Then the question of "switching" wouldn't make sense. Which of the remaining 98 would you be asked to switch to? The whole point is that you end up with a choice of sticking with the door you picked, or switching to the one remaining door.
- colmanhumphrey 4y ago
  Seems fine to me. Do you want to stick with your original or switch to one of the remaining 98? It’s still worth switching too, just a smaller increase in probability.
waynesonfire 4y ago
i think the idea is to demonstrate more easily the information leak by exaggerating the amount of information being leaked. with 3 doors, there are only so many the host can chose to open. with a 100, there is up to 98. sure, it can be 1 or 2... or 98.
jasonswett 4y ago
The rule is "after the player has guessed a door, open ALL the other doors except the door that was guessed plus one other door". I think your confusion is because if there are 100 doors then the rule is self-evident. But when there are only 3 doors, the rule of "open all doors except the door that was guessed plus one other" is indistinguishable from "open just one door". The key is knowing that the rule is the "open all..." rule.
- colmanhumphrey 4y ago
  That’s not the rule from the question in the link nor in the original game (the game’s rule isn’t quite the same as the classic problem, it’s even messier). Your proposed rule doesn’t uniquely extend the original, it’s just one of the possible generalizations.

dahfizz 4y ago

> Then you obviously switch, as the original door has a 1% chance of being right, and the other door has a 99% of being right. It's not obvious to me that this is true. Why do we update the chance of the "new" door but not the one we already selected? Intuitively, I would expect the odds of both doors to increase to 50% as the other 98 doors are opened. I know that you're correct, I just don't find this to be an intuitive explanation.

mcphage 4y ago
The odds of your original door being right are 1%—that never changes. It’s determined by the options when the game was set up and when you picked it. If your original door was wrong—99% chance—then it’s in one of the 99 other doors. But the host opens 98 of them, so if it was one of the 99 other doors—again, 99% chance—then the only option left is the last door.