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> Your counter arguments are wrong, but you don't seem to even acknowledge that I am saying that. I do acknowledge that you're saying that I'm wrong. That's wh

by kgwgk 3y ago

> Your counter arguments are wrong, but you don't seem to even acknowledge that I am saying that. I do acknowledge that you're saying that I'm wrong. That's why we're still exchanging arguments! What I don't know exactly is what do you think that it's wrong with my arguments so I try to find where do we agree - and where we don't. Do you think that there is something wrong with the answer I wrote down here https://news.ycombinator.com/item?id=37206445 https://news.ycombinator.com/item?id=37206445 ? It seems that you don't agree that the answer depends on the (relative) value of P(you tell me that you have at least one boy|you have two boys) and P(you tell me that you have at least one boy|you have one boy and one girl). > I'm willing to try to explain why, but not if you don't want to learn and just want to insist you are correct. Well, I could also say that just want to insist that my arguments are wrong but I sincerely hope that you want to learn as much as I do. > Ask yourself how long you would spend trying to explain Monty Hall to someone who kept insisting it was 1/2 to change. As long as needed. Souls are saved one at a time. Here we go. ----- > No matter how you choose the statement "I have at least one (girl/boy)", (prefer one, flip a coins, etc) once you tell me it's always 2/3 boy-girl. Until you tell me it's 1/2. Any algorithm to choose which to say works as long as it's true and you don't convey more information about the children like "my older child is male". That's wrong and I'm going to try to show you that with an example (the mathematical proof is in the link above). Hopefully I'm not misrepresenting your position - please tell me if I do. You are in an auditorium with 600 people. Each of them has two kids. (Let's assume there is no strange thing going on like "meeting of parents with twins" and the sex of the kids is equally probable and independent.) Q: What's the probability that a given person has one boy and one girl? A: 1/2 Q: How many of them do you estimate that have one boy and one girl? A: 300 Each of them write into a paper their name and "I have at least one (girl/boy)" (they never lie and if there is a choice the choose however they want: prefer one, flip a coin, etc.). You have the 600 papers in front of you, but have not read them yet. Q: What's the probability that a given person has one boy and one girl? A: Still 1/2 Q: How many of them do you estimate that have one boy and one girl? A: Still 300 You can win $100 if you guess correctly whether there are more than 350 or less than 350 people with one boy and one girl. Q: What's your guess? A: Less than 350, because my estimate is 300. Q: What will be the probability that a given person has one boy and one girl after you've read the papers? A: 2/3 because once they tell me it's always 2/3 boy-girl. Q: How many of them will you estimate that have one boy and one girl after you've read the papers? A: 400 Q: Do you want you want to change your guess to "more than 350"? A: No, until I read the papers the probability is 1/2 and my estimate is that 300 people have one boy and one girl. Q: So you keep your "less than 350" guess even though you know with certainty that in a few minutes you will estimate that the right answer is around 400 and you will wish you had answered "more than 350"? A: Yes, I'm happy with that. I think I could get the $100 if I answered "more than 350" now but I refuse to do it until I read the papers. You read the papers. Q: What's the probability that a given person has one boy and one girl? A: 2/3 Q: How many of them do you estimate that have one boy and one girl? A: 400 Q: Do you want to change your guess for the $100 prize? A: Yes, now I’d like to answer "More than 350". Thanks for letting me change my guess! Unfortunately you lose, because in a group of 600 pairs of kids we expected around 300 pairs of boy and girl. Writing things on a paper leaves the children unaffected.

HWR_14 3y ago
I think now I understand. "I will make you make exactly one of two statements and each results in a 2/3 chance of BG" doesn't make sense. "You freely made one of two statements and each results in a 2/3 chance of BG" does. I interpreted "it depends on randomness" as "speak up if you have at least one boy" or "speak up if you have at least one girl", each of which would result in 450 saying something in your example and the math works. So it comes down to if we decide on the question (however we do that) before we look at the kids or after.
- kgwgk 3y ago
  > I interpreted "it depends on randomness" as "speak up if you have at least one boy" or "speak up if you have at least one girl", each of which would result in 450 saying something in your example and the math works. The point is that original problem says "I tell you I have two children and that (at least) one of them is a boy". It doesn't say "I tell you I have two children and [when you ask me to confirm whether (at least) one of them is a boy] that (at least) one of them is a boy". Reasoning from the cases "BB", "BG", "GB", "GG" - and discarding the last one to get p(BB)=1/3 - is implicitly using the cases "BB and I tell you that at least one of them is a boy", "BG and I tell you that at least one of them is a boy", "GB and I tell you that at least one of them is a boy", "GG and I tell you something else like at least one of them is a girl". That breaks the symmetry between the "I tell you that at least one of them is a boy" and the "I tell you that at least one of them is a girl" problems. Using the cases in the previous paragraph the answer for the former is p(BB)=1/3 but the answer to the latter is p(GG)=1. If you want to have the same solution when you switch girl <-> boy ["one of them is a boy, what is the probability that I have two boys" <-> "one of them is a girl, what is the probability that I have two girls"] you should treat them equally. A quite natural way to do so would be to consider the eight equiprobable cases (some of them repeated) BB and I tell you that at least one of them is a boy BB and I tell you that at least one of them is a boy BG and I tell you that at least one of them is a boy BG and I tell you that at least one of them is a girl GB and I tell you that at least one of them is a boy GB and I tell you that at least one of them is a girl GG and I tell you that at least one of them is a girl GG and I tell you that at least one of them is a girl but then the answer to both problems is 1/2. You can make the answer to both problems 1/3 but the eight cases that you would need to consider for that are quite unnatural: BB and [... discard this line somehow ...] BB and I tell you that at least one of them is a boy BG and I tell you that at least one of them is a boy BG and I tell you that at least one of them is a girl GB and I tell you that at least one of them is a boy GB and I tell you that at least one of them is a girl GG and I tell you that at least one of them is a girl GG and [... discard this line somehow ...]

by kgwgk 3y ago

HWR_14 3y ago

I think now I understand. "I will make you make exactly one of two statements and each results in a 2/3 chance of BG" doesn't make sense. "You freely made one of two statements and each results in a 2/3 chance of BG" does. I interpreted "it depends on randomness" as "speak up if you have at least one boy" or "speak up if you have at least one girl", each of which would result in 450 saying something in your example and the math works. So it comes down to if we decide on the question (however we do that) before we look at the kids or after.

kgwgk 3y ago
> I interpreted "it depends on randomness" as "speak up if you have at least one boy" or "speak up if you have at least one girl", each of which would result in 450 saying something in your example and the math works. The point is that original problem says "I tell you I have two children and that (at least) one of them is a boy". It doesn't say "I tell you I have two children and [when you ask me to confirm whether (at least) one of them is a boy] that (at least) one of them is a boy". Reasoning from the cases "BB", "BG", "GB", "GG" - and discarding the last one to get p(BB)=1/3 - is implicitly using the cases "BB and I tell you that at least one of them is a boy", "BG and I tell you that at least one of them is a boy", "GB and I tell you that at least one of them is a boy", "GG and I tell you something else like at least one of them is a girl". That breaks the symmetry between the "I tell you that at least one of them is a boy" and the "I tell you that at least one of them is a girl" problems. Using the cases in the previous paragraph the answer for the former is p(BB)=1/3 but the answer to the latter is p(GG)=1. If you want to have the same solution when you switch girl <-> boy ["one of them is a boy, what is the probability that I have two boys" <-> "one of them is a girl, what is the probability that I have two girls"] you should treat them equally. A quite natural way to do so would be to consider the eight equiprobable cases (some of them repeated) BB and I tell you that at least one of them is a boy BB and I tell you that at least one of them is a boy BG and I tell you that at least one of them is a boy BG and I tell you that at least one of them is a girl GB and I tell you that at least one of them is a boy GB and I tell you that at least one of them is a girl GG and I tell you that at least one of them is a girl GG and I tell you that at least one of them is a girl but then the answer to both problems is 1/2. You can make the answer to both problems 1/3 but the eight cases that you would need to consider for that are quite unnatural: BB and [... discard this line somehow ...] BB and I tell you that at least one of them is a boy BG and I tell you that at least one of them is a boy BG and I tell you that at least one of them is a girl GB and I tell you that at least one of them is a boy GB and I tell you that at least one of them is a girl GG and I tell you that at least one of them is a girl GG and [... discard this line somehow ...]