4 ms·

>Do you just mean the subset of "he chose randomly and happened to draw a goat out of one of the two you did not choose"? In which case switching isn't benefici

by basch 6y ago

>Do you just mean the subset of "he chose randomly and happened to draw a goat out of one of the two you did not choose"? In which case switching isn't beneficial because Yes switching is. Switching is beneficial IF he shows you a goat out of the doors you didnt choose.

kgwgk 6y ago
https://news.ycombinator.com/item?id=24713352 https://news.ycombinator.com/item?id=24713352
majormajor 6y ago
If the host opens a door truly at random... - you choose door 1, goat -- host chooses door 1, goat --- switch: 1/2 chance of being right --- stay: you're wrong -- host chooses door 2, goat --- switch: you are right --- stay: you are wrong -- host chooses door 3, car --- switch: you are right --- stay: you are wrong - you choose door 2, goat -- host chooses door 1, goat --- switch: you are right --- stay: you're wrong -- host chooses door 2, goat --- switch: 1/2 chance of being right --- stay: you are wrong -- host chooses door 3, car --- switch: you are right --- stay: you are wrong - you choose door 3, car -- host chooses door 1, goat --- switch: you are wrong --- stay: you are right -- host chooses door 2, goat --- switch: you are wrong --- stay: you are right -- host chooses door 3, car --- switch: you are wrong --- stay: you are right so there are 6 times out of 9 where the host shows you a goat total, and you are right if you switch 3 out of those 6 times. if you stay each of those 6 times, you're right 2 out of the 6 times. there are 4 times the host shows you a goat that you didn't choose, and if you switch you are right 2 out of those times. but in this scenario, you are actually right 2 out of those 4 times if you stay, too. this seems opposite of what i expected. I guess if the host is choosing at random, and you conditionalize on "being shown a goat AND a door you didn't choose" then you're reducing that particular sample space down to be more heavy in the "you got it right originally" scenario, because the "you got it wrong originally" scenarios rule out half the goat doors from the hosts choices we're considering.
- basch 6y ago
  > so there are 6 times out of 9 where the host shows you a goat total. No. He picks goat 100% of the time. Choosing not goat is outside the bounds/parameters of the initial constraints.
klmadfejno 6y ago
Nope. Assume you choose the car (1/3). He will always choose a goat. If you switch, you have a 100% chance of losing. (1/3) * (1/1) = 1/3 chance of losing if you switch. Assume you choose a goat (2/3). There is a 50% chance that he opens a goat if he chooses randomly. If this happens, there's a 100% chance that you win if you switch. (2/3) * (1/2) * 100% = 1/3 chance that you win if you switch. The remaining 1/3 chance is voided because we've conditioned the game on him randomly choosing a goat, and not a car. So the probability of winning if you switch is (1/3) / ((1/3) + (1/3)) = (1/2) exactly 50/50. In the traditional monty hall, switching gives a 2/3 chance of winning. This is still true in the random variant but only if you're allowed to choose the car door in monty opens that one by chance.

by basch 6y ago

kgwgk 6y ago

https://news.ycombinator.com/item?id=24713352 https://news.ycombinator.com/item?id=24713352

majormajor 6y ago

If the host opens a door truly at random... - you choose door 1, goat -- host chooses door 1, goat --- switch: 1/2 chance of being right --- stay: you're wrong -- host chooses door 2, goat --- switch: you are right --- stay: you are wrong -- host chooses door 3, car --- switch: you are right --- stay: you are wrong - you choose door 2, goat -- host chooses door 1, goat --- switch: you are right --- stay: you're wrong -- host chooses door 2, goat --- switch: 1/2 chance of being right --- stay: you are wrong -- host chooses door 3, car --- switch: you are right --- stay: you are wrong - you choose door 3, car -- host chooses door 1, goat --- switch: you are wrong --- stay: you are right -- host chooses door 2, goat --- switch: you are wrong --- stay: you are right -- host chooses door 3, car --- switch: you are wrong --- stay: you are right so there are 6 times out of 9 where the host shows you a goat total, and you are right if you switch 3 out of those 6 times. if you stay each of those 6 times, you're right 2 out of the 6 times. there are 4 times the host shows you a goat that you didn't choose, and if you switch you are right 2 out of those times. but in this scenario, you are actually right 2 out of those 4 times if you stay, too. this seems opposite of what i expected. I guess if the host is choosing at random, and you conditionalize on "being shown a goat AND a door you didn't choose" then you're reducing that particular sample space down to be more heavy in the "you got it right originally" scenario, because the "you got it wrong originally" scenarios rule out half the goat doors from the hosts choices we're considering.

basch 6y ago
> so there are 6 times out of 9 where the host shows you a goat total. No. He picks goat 100% of the time. Choosing not goat is outside the bounds/parameters of the initial constraints.

klmadfejno 6y ago

Nope. Assume you choose the car (1/3). He will always choose a goat. If you switch, you have a 100% chance of losing. (1/3) * (1/1) = 1/3 chance of losing if you switch. Assume you choose a goat (2/3). There is a 50% chance that he opens a goat if he chooses randomly. If this happens, there's a 100% chance that you win if you switch. (2/3) * (1/2) * 100% = 1/3 chance that you win if you switch. The remaining 1/3 chance is voided because we've conditioned the game on him randomly choosing a goat, and not a car. So the probability of winning if you switch is (1/3) / ((1/3) + (1/3)) = (1/2) exactly 50/50. In the traditional monty hall, switching gives a 2/3 chance of winning. This is still true in the random variant but only if you're allowed to choose the car door in monty opens that one by chance.