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Well don't I feel like an idiot for saying any of that. Why were you running a "computer simulation" of the problem? Were you, even? >"The two" ?? What "two"?

Well don't I feel like an idiot for saying any of that. Why were you running a "computer simulation" of the problem? Were you, even? >"The two" ?? What "two"? Your "explanations" are very unclear. The only two there are: the two children. The sex of one is independent from the sex of the other. >You will find many things difficult if you don't learn how to slow down, take time, be precise, and explain clearly. Be more like Feynman - take time to explain things clearly and precisely. Okay, I'll try, good buddy. Monty Hall is different because it precludes upfront the possibility of there being any ratio other than 1:2. If we were, however, to say that any door could have either a goat or a car, then Monty's elimination of some other door would not inform you on the odds of your door YOUR door. Yea or nay? The only way for Monty to give you any information at all about whether to switch is if the rule binds him to tell you something about the odds of the unchosen doors. In the original problem, this happens, because the ratio is bound 1:2. In our revised problem, this does not happen - unless he tells you he can't remove a door because neither is a goat. But! Using this as an analogue to our children problem, we've already precluded that possibility upfront by eliminating GG, thus Monty can't provide us any information because his B-G choice is unbound by any rule.

ColinWright 13y ago
> Why were you running a "computer simulation" of the > problem? Were you, even? I was reconstructing what happened when Erdős was first told of the Monty Hall problem. He answered 1/2, and was very frustrated for a time. In the end he said something like "But you're not telling me the why!" Moving on ... >> "The two" ?? What "two"? Your "explanations" are >> very unclear. > The only two there are: the two children. The sex > of one is independent from the sex of the other. So we are assuming that this chap has exactly two children, and that their sexes are independent with probability 1/2. > Monty Hall is different because it precludes upfront > the possibility of there being any ratio other than > 1:2. If we were, however, to say that any door could > have either a goat or a car, then it wouldn't matter > whether you stayed or switched. Monty's elimination > of some other door would not inform you on the odds > of your door YOUR door. Yea or nay? Yes, agreed. > The only way for Monty to give you any information at > all about whether to switch is if the rule binds him > to tell you something about the odds of the unchosen > doors. Yes. > In the original problem, this happens, because the > ratio is bound 1:2. OK. > In our revised problem, this does not happen - unless > he tells you he can't remove a door because neither > is a goat. Yes. > Using this as an analogue to our children problem, > we've already precluded that possibility upfront > by eliminating GG, thus Monty can't provide us any > information because his B-G choice is unbound by > any rule. I fail to see how that is in any way relevant. There are four equally likely possibilities for the children. If we label them older first then we have BB, BG, GB, GG. Only by labelling them in some consistent order to we get equally likely possibilities. The father says that at least one child is a boy. That eliminates the option of GG. Are you somehow suggesting that the probability distribution of the other options is changed by this? If so, how? Just as Monty's choice gives no information about the door we have selected (and which therefore remains as chance 1/3 of being the winning door), telling us that it's not the case that both children are girls gives us no further information about the other three equally likely possibilities. So in https://news.ycombinator.com/item?id=7022615 https://news.ycombinator.com/item?id=7022615 I asked you this ... A: So now let's take another experiment. I tell you I'm going to flip two coins until at least one of them is heads. I do that. Now I ask you to bet on whether or not there is a tail. What do you think are fair odds? B: ... C: ... There are three options. 1: Your answers to A, B, and C are not all the same; 2: They are all the same, and not 1/3; 3: They are all the same, and all 1/3. In https://news.ycombinator.com/item?id=7022707 https://news.ycombinator.com/item?id=7022707 you answered: > The answer to all 3 is 1/2. That turns out not to be the case. The answer to A is 1/3. Write a computer program to simulate it and then post it here. Seriously, don't just theorize about it, try it.
- gametheoretic 13y ago
  >That turns out not to be the case. The answer to A is 1/3. Write a computer program to simulate it and then post it here. Seriously, don't just theorize about it, try it. I misread A. I agree that it's 1/3 if you flip pairs until at least one is heads. I disagree that living in a world where pairs of tails don't exist and living in a world where pairs of tails do exist but are ignored are the same thing. >Are you somehow suggesting that the probability distribution of the other options is changed by this? Yes. I'm saying it's an unwitting double Monty Hall. The fact that he can't give you 3 cars changes the odds. Going (invisibly) from 4 possible worlds (including GG) to 3 changes the prob dist of the remaining 3, by which I mean 2. If Monty must guarantee you at least one boy, then there aren't really 3 remaining possibilities. The unguaranteed can be a boy, or it can be a girl. What is the third?
  - ColinWright 13y ago
    >> The father says that at least one child is a boy. >> That eliminates the option of GG. Are you somehow >> suggesting that the probability distribution of >> the other options is changed by this? > Yes. ... Going (invisibly) from 4 possible worlds > (including GG) to 3 changes the prob dist of the > remaining 3, by which I mean 2. If Monty must > guarantee you at least one boy, then there aren't > really 3 remaining possibilities. The unguaranteed > can be a boy, or it can be a girl. What is the > third? The standard reasoning is that there are two options, but they are not equally likely. Equally likely options are, labelling oder child first, BB, BG, GB. These were equally probable before we gained the extra information, and the standard interpretation is that they remain equally probable after. In the Monty Hall case you choose a door, and it has probability 1/3 of being right. It remains 1/3 probable after he chooses a door. The other two doors together have probability 2/3 of concealing the car. It remains 2/3 after he has chosen and opened a door. The probabilities of the branches don't change in any of these cases. Before you are told anything about the children we have four equally likely possibilities: BB, BG, GB, GG. After one is eliminated, why should the others still not be equi-probable? > A: So now let's take another experiment. > I tell you I'm going to flip two coins > until at least one of them is heads. I > do that. Now I ask you to bet on whether > or not there is a tail. What do you > think are fair odds? And you say: > I misread A. I agree that it's 1/3 if you > flip pairs until at least one is heads. OK. Now B: > B: Now I hunt among couples with two children > until I find one that doesn't have two girls, > and I ask: what are the odds they have two > boys? Do you agree that it's 1/3?

ColinWright 13y ago

> Why were you running a "computer simulation" of the > problem? Were you, even? I was reconstructing what happened when Erdős was first told of the Monty Hall problem. He answered 1/2, and was very frustrated for a time. In the end he said something like "But you're not telling me the why!" Moving on ... >> "The two" ?? What "two"? Your "explanations" are >> very unclear. > The only two there are: the two children. The sex > of one is independent from the sex of the other. So we are assuming that this chap has exactly two children, and that their sexes are independent with probability 1/2. > Monty Hall is different because it precludes upfront > the possibility of there being any ratio other than > 1:2. If we were, however, to say that any door could > have either a goat or a car, then it wouldn't matter > whether you stayed or switched. Monty's elimination > of some other door would not inform you on the odds > of your door YOUR door. Yea or nay? Yes, agreed. > The only way for Monty to give you any information at > all about whether to switch is if the rule binds him > to tell you something about the odds of the unchosen > doors. Yes. > In the original problem, this happens, because the > ratio is bound 1:2. OK. > In our revised problem, this does not happen - unless > he tells you he can't remove a door because neither > is a goat. Yes. > Using this as an analogue to our children problem, > we've already precluded that possibility upfront > by eliminating GG, thus Monty can't provide us any > information because his B-G choice is unbound by > any rule. I fail to see how that is in any way relevant. There are four equally likely possibilities for the children. If we label them older first then we have BB, BG, GB, GG. Only by labelling them in some consistent order to we get equally likely possibilities. The father says that at least one child is a boy. That eliminates the option of GG. Are you somehow suggesting that the probability distribution of the other options is changed by this? If so, how? Just as Monty's choice gives no information about the door we have selected (and which therefore remains as chance 1/3 of being the winning door), telling us that it's not the case that both children are girls gives us no further information about the other three equally likely possibilities. So in https://news.ycombinator.com/item?id=7022615 https://news.ycombinator.com/item?id=7022615 I asked you this ... A: So now let's take another experiment. I tell you I'm going to flip two coins until at least one of them is heads. I do that. Now I ask you to bet on whether or not there is a tail. What do you think are fair odds? B: ... C: ... There are three options. 1: Your answers to A, B, and C are not all the same; 2: They are all the same, and not 1/3; 3: They are all the same, and all 1/3. In https://news.ycombinator.com/item?id=7022707 https://news.ycombinator.com/item?id=7022707 you answered: > The answer to all 3 is 1/2. That turns out not to be the case. The answer to A is 1/3. Write a computer program to simulate it and then post it here. Seriously, don't just theorize about it, try it.

gametheoretic 13y ago
>That turns out not to be the case. The answer to A is 1/3. Write a computer program to simulate it and then post it here. Seriously, don't just theorize about it, try it. I misread A. I agree that it's 1/3 if you flip pairs until at least one is heads. I disagree that living in a world where pairs of tails don't exist and living in a world where pairs of tails do exist but are ignored are the same thing. >Are you somehow suggesting that the probability distribution of the other options is changed by this? Yes. I'm saying it's an unwitting double Monty Hall. The fact that he can't give you 3 cars changes the odds. Going (invisibly) from 4 possible worlds (including GG) to 3 changes the prob dist of the remaining 3, by which I mean 2. If Monty must guarantee you at least one boy, then there aren't really 3 remaining possibilities. The unguaranteed can be a boy, or it can be a girl. What is the third?
- ColinWright 13y ago
  >> The father says that at least one child is a boy. >> That eliminates the option of GG. Are you somehow >> suggesting that the probability distribution of >> the other options is changed by this? > Yes. ... Going (invisibly) from 4 possible worlds > (including GG) to 3 changes the prob dist of the > remaining 3, by which I mean 2. If Monty must > guarantee you at least one boy, then there aren't > really 3 remaining possibilities. The unguaranteed > can be a boy, or it can be a girl. What is the > third? The standard reasoning is that there are two options, but they are not equally likely. Equally likely options are, labelling oder child first, BB, BG, GB. These were equally probable before we gained the extra information, and the standard interpretation is that they remain equally probable after. In the Monty Hall case you choose a door, and it has probability 1/3 of being right. It remains 1/3 probable after he chooses a door. The other two doors together have probability 2/3 of concealing the car. It remains 2/3 after he has chosen and opened a door. The probabilities of the branches don't change in any of these cases. Before you are told anything about the children we have four equally likely possibilities: BB, BG, GB, GG. After one is eliminated, why should the others still not be equi-probable? > A: So now let's take another experiment. > I tell you I'm going to flip two coins > until at least one of them is heads. I > do that. Now I ask you to bet on whether > or not there is a tail. What do you > think are fair odds? And you say: > I misread A. I agree that it's 1/3 if you > flip pairs until at least one is heads. OK. Now B: > B: Now I hunt among couples with two children > until I find one that doesn't have two girls, > and I ask: what are the odds they have two > boys? Do you agree that it's 1/3?