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But if the car door is picked, there's no further game to play. Surely the only interesting thing to ask is, conditioned on seeing a goat, what's the probabilit

But if the car door is picked, there's no further game to play. Surely the only interesting thing to ask is, conditioned on seeing a goat, what's the probability the third door contains a car. The cases of observing a car are irrelevant since that's not the scenario. I'm still not convinced it's any different whether Monty Hall knows or not so long as the goat door is opened. Edit: having written this, thinking about the 100 door case. If 98 random doors open (that aren't the one I picked) then the fact they all contain goats is pretty suggestive that I have the car, or at least 50/50. I'm not convinced by the code example though.

kragen 3y ago
your first paragraph is what i thought before running the simulation, but it turned out to be wrong. if my simulation isn't convincing to you, try writing one that is
- hgomersall 3y ago
  I'm convinced i think, just not initially by the simulator! I needed to reason my way separately but I think I'm happy with the sim now.
topaz0 3y ago
Yes, it's exactly 50/50. It's not hard to work out the details, as I did in another thread elsewhere on this article. Here goes: -There's a 1/100 chance my door is a car, in which case it doesn't matter which of the others stays closed, it will always be a goat. The game will proceed, and switching will lose. -There's a 99/100 chance my door is a goat, in which case the car is behind some other door. Choosing 98 out of 99 doors to open at random is the same as choosing 1 out of 99 doors to leave closed at random. So the chance that the car stays hidden in this case (so that switching will win) is 1/99, and 98/99 that the game ends early because the car is revealed. -Adding it up, the game ends without the chance to make a choice 99/100 * 98/99 = 98/100 of the time. Of the remaining 2%, 1/100 comes from the first case (switching loses) and 99/100 * 1/99 = 1/100 comes from the second case (switching wins). The strategies are equally effective.
tsimionescu 3y ago
> the fact they all contain goats is pretty suggestive that I have the car, or at least 50/50 That's exactly the point. If he doesn't know, then it's exactly 50/50 and there is no reason to switch. If he does know, then it's 1/NUM_DOORS versus NUM_DOORS-1/NUM_DOORS, so you'd be crazy not to switch. The point is, if he picks at random, in the 100 door case, the vast majority of the time he will open the car door while opening those 98 doors. The case where you get to pick again would be exceedingly rare. Conversely, if he only opens goat doors, you will get your second pick 100% of the time.
- hgomersall 3y ago
  This is why one can't ignore the car picked cases, so the simulator is correct.
  - EGPRC 3y ago
    I believe the confusion with "ignoring the car picked cases" comes due to thinking that they are additional games to the original ones, instead of being part of those that constitute the 2/3 in which the player starts picking wrong. So you thought that eliminating them you were left with the original 1/3 vs 2/3, when you actually removed half of the 2/3.

kragen 3y ago

your first paragraph is what i thought before running the simulation, but it turned out to be wrong. if my simulation isn't convincing to you, try writing one that is

hgomersall 3y ago
I'm convinced i think, just not initially by the simulator! I needed to reason my way separately but I think I'm happy with the sim now.

topaz0 3y ago

Yes, it's exactly 50/50. It's not hard to work out the details, as I did in another thread elsewhere on this article. Here goes: -There's a 1/100 chance my door is a car, in which case it doesn't matter which of the others stays closed, it will always be a goat. The game will proceed, and switching will lose. -There's a 99/100 chance my door is a goat, in which case the car is behind some other door. Choosing 98 out of 99 doors to open at random is the same as choosing 1 out of 99 doors to leave closed at random. So the chance that the car stays hidden in this case (so that switching will win) is 1/99, and 98/99 that the game ends early because the car is revealed. -Adding it up, the game ends without the chance to make a choice 99/100 * 98/99 = 98/100 of the time. Of the remaining 2%, 1/100 comes from the first case (switching loses) and 99/100 * 1/99 = 1/100 comes from the second case (switching wins). The strategies are equally effective.

tsimionescu 3y ago

> the fact they all contain goats is pretty suggestive that I have the car, or at least 50/50 That's exactly the point. If he doesn't know, then it's exactly 50/50 and there is no reason to switch. If he does know, then it's 1/NUM_DOORS versus NUM_DOORS-1/NUM_DOORS, so you'd be crazy not to switch. The point is, if he picks at random, in the 100 door case, the vast majority of the time he will open the car door while opening those 98 doors. The case where you get to pick again would be exceedingly rare. Conversely, if he only opens goat doors, you will get your second pick 100% of the time.

hgomersall 3y ago
This is why one can't ignore the car picked cases, so the simulator is correct.
- EGPRC 3y ago
  I believe the confusion with "ignoring the car picked cases" comes due to thinking that they are additional games to the original ones, instead of being part of those that constitute the 2/3 in which the player starts picking wrong. So you thought that eliminating them you were left with the original 1/3 vs 2/3, when you actually removed half of the 2/3.