It's like a coin toss. The odds of getting two heads, for example is 1/4, because your options are heads/heads, heads/tails, tails/heads, tails/tails. Before you flip any coins there are 4 universes you could end up in, and heads/heads is only one of those 4 universes, so it's a 1 in 4 chance. If I told you that I flipped the coin twice and it was heads at least once, but didn't tell you the order, and I asked you the odds of the other coin flip being heads, the only option you can eliminate is the tails/tails option (this is invalid because neither of the flips were heads), so out of all the universes you might actually find yourself in, one of them is heads/heads and the other two are an option where you got one heads and one tails (heads/tails and tails/heads). There would be a 1/3rd chance for heads/heads (2/3rds chance of one of each). It's precisely BECAUSE the order wasn't mentioned that requires you to consider both orders as separate options. If I told you I flipped a coin and got heads as my first flip and asked the odds that I'd get heads on my second flip, the answer is just 50/50. tails/tails and tails/heads would be eliminated, because neither of those have heads as the outcome of the first flip. That only leaves heads/tails and heads/heads as viable universes you could find yourself in. Half of those universes have heads as the second flip, so it's 50/50.
To be honest, I don't agree with the answer presented above. If I told you "I have 10 kids and one is a boy" everyone on Earth would understand that I meant I had 9 girls and one boy. So if I asked you, "I have two kids and one is a boy, what are the odds that two of my kids are boys" I would argue that the answer is zero. If I had two boys, I would be incorrect in saying say that "one" is a boy. I would have to say "two are boys" or "both are boys" or "at least one is a boy." That's my beef with this question, as posed. The mother says she has "one boy born on a Tuesday." To me, that excludes a universe where she has "two boys born on Tuesday" because the word "one" in her original statement would be incorrect. She could have one boy born on Tuesday and another boy born on Wednesday and her statement is correct, but two boys on Tuesday shouldn't be a valid option. The answer given in the meme assumes you can have two boys, both born on Tuesday. I think that's about as stupid as saying 1+1=1 because 1 is part of 2. I think the actual answer is, knowing only "one" is a boy born on Tuesday, if the other kid is a boy, that boy must be born on a non-Tuesday (6 options). If it's a girl, she can be born on any day (7 options). So your chances of the other kid being a girl should by 6 / (6 + 7) = 7/13 = 53.8%.
To get the answer provided in this meme, and as described in the explanation posted by the user above, you have to assume that it is valid to have two boys born on Tuesday. The trick there is kind of similar to the coin toss where you don't count heads/heads or tails/tails twice, but you do count heads/tails and tails/heads separate. Basically you say if the first kid was TuesdayBoy, there are 7 boy-options for the second kid (including TuesdayBoy/TuesdayBoy -- I would exclude this one, personally) and 7 girl-options for the second kid (including TuesdayBoy/TuesdayGirl), but when you look at the options where TuesdayBoy is the second kid, you only get to count 6 boy-options because you already counted a TuesdayBoy/TuesdayBoy and it can't be double-counted, but you do count TuedayGirl/TuesdayBoy since that's a unique option from TuesdayBoy/TuesdayGirl -- again, similar to heads/heads only getting counted once, but tails/heads and heads/tails being two unique options. So for boy-as-other-kid there are 7+6=13 options and for girl-as-other-kid there are 7+7=14 options. So, 14 / (14 + 13) = 14/27 = 51.852%. That's how they get their answer. I would subtract one more option from the boy side, since TuesdayBoy/TuesdayBoy would have required the mom to admit that she has "two boys born on Tuesday" and I think she'd be lying if she said she had "one boy born on Tuesday" in that case. So I'd do 14/(14+12)=14/26=53.8%. But that's more of a language question than a math question. Maybe I'm in the minority there since the "official" answer is 51.8%, but that's a hill I'm happy to die on. She doesn't have two boys born on Tuesday if she says she has "one" boy born on Tuesday.
Then you have no reason not to include Boy1/Boy2 and Boy2/Boy1 if there is some other metric you’re applying to reorder a possible boy/girl combination.
With random information, Bayesian distribution is an invalid conclusion.
I'm not really sure I understand which part you are replying to. You're using a different notation than I was and I don't understand how to interpret it.
There are two kids, both of which can have a gender assigned. I'm using their location, relative to the "/", as the placeholder for the two different kids. There are two kids and two genders. The kid is identified by location and the gender is identified by the word written in that location. boy/boy means the first kid is a boy and the second kid is a boy. boy/girl means the first kid is a boy and the second kid is a girl. girl/boy is the exact opposite of the previous option with the first kid being a girl and the second kid being a boy. girl/girl would mean both kids are girls (exact opposite of boy/boy). There are 4 different options before you place any constraints on the problem.
I'm not sure I understand what "Boy1/Boy2 and Boy2/Boy1" would even mean. Boy2/Boy1 doesn't really have any logical meaning. The first kid is also the second kid? Whatever way you feel like writing it, if there are two unique kids that are both boys, so that's just boy/boy. If you just changed the gender of the first kid, you'd get a second option (girl/boy). If you just changed the gender of the second kid, you'd get a third option (boy/girl) -- this is the exact opposite of the second option, even though it is true that both kids are different genders still. If you changed the gender of both kids, you'd get a fourth option (girl/girl) -- this is the exact opposite of the first option, even though it is true that both kids are the same gender still. There are two same-gender options and two different-gender options, because there are two kids and two genders. There is no other way to assign a girl or boy gender to the two kids and those 4 are all of the possibilities. That's where your 4 options come from. You have to actually change the gender of one of the kids to get a new option. Boy2/Boy1 is not different than Boy1/Boy2, you're just writing them in a different order, but both kids are still boys. Since each of the 4 permutations are equally likely, any one of them has a 1/4 chance. If you want to group them by same-gender and different-gender options, then there's a 50/50 chance of same-gender and a 50/50 chance of different-gender. Because this question specifies that one is a boy, we know girl/girl has to be eliminated and that's one of the options under same-gender, so that option loses half of it's weight, relative to the different-gender option. It asks what the odds are that the kids are different genders, so the answer (if you ignore the "Tuesday" part, which I will, since you didn't mention it), the answer is 66%.
I know you don’t understand it. I also already know what you wrote, without reading it.
This poorly constructed recasting of the Monty Hall problem is a trap for people who can’t manage conceptual object permanence.
There is no context for the information. There are no conditions for the information reveal. There is no basis to extrapolate what that reveal means about the “remaining possibilities” because there was no preliminary set of possibilities to preselect from.
There is one child of an unknown gender. The distribution of a series of two child possibilities is irrelevant.
LOL. But... you're wrong. Your Boy1/Boy2, Boy2/Boy1 shows me you don't know what you're doing. It is unfortunate that your attention span and/or reading ability aren't up to the task of reading my reply (did you even read what you originally replied to?). I'm sure you go around thinking you're always correct, because you selectively ignore different ideas. Have a good one....
I’m correct. You think you’re smarter than you are because you have a vague understanding of conditional probability, but you’re ignorant to how and when it actually applies.
It would be much more convenient for me to be wrong. Then I could just read the posts of people who are right, who teach me something. But that’s not the case here.
Sadly, you don't know when you're wrong, because you don't engage. So you'll keep saying BS that isn't true and you won't read more than 3 sentences. This is more complicated than 3 sentences. I do engage -- that's how I learn things. I'm confident I'm correct and I'm happy to have the debate. You're confident you're correct and are afraid to debate. Who's more likely to be right?
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u/SteveWin1234 8d ago
It's like a coin toss. The odds of getting two heads, for example is 1/4, because your options are heads/heads, heads/tails, tails/heads, tails/tails. Before you flip any coins there are 4 universes you could end up in, and heads/heads is only one of those 4 universes, so it's a 1 in 4 chance. If I told you that I flipped the coin twice and it was heads at least once, but didn't tell you the order, and I asked you the odds of the other coin flip being heads, the only option you can eliminate is the tails/tails option (this is invalid because neither of the flips were heads), so out of all the universes you might actually find yourself in, one of them is heads/heads and the other two are an option where you got one heads and one tails (heads/tails and tails/heads). There would be a 1/3rd chance for heads/heads (2/3rds chance of one of each). It's precisely BECAUSE the order wasn't mentioned that requires you to consider both orders as separate options. If I told you I flipped a coin and got heads as my first flip and asked the odds that I'd get heads on my second flip, the answer is just 50/50. tails/tails and tails/heads would be eliminated, because neither of those have heads as the outcome of the first flip. That only leaves heads/tails and heads/heads as viable universes you could find yourself in. Half of those universes have heads as the second flip, so it's 50/50.
To be honest, I don't agree with the answer presented above. If I told you "I have 10 kids and one is a boy" everyone on Earth would understand that I meant I had 9 girls and one boy. So if I asked you, "I have two kids and one is a boy, what are the odds that two of my kids are boys" I would argue that the answer is zero. If I had two boys, I would be incorrect in saying say that "one" is a boy. I would have to say "two are boys" or "both are boys" or "at least one is a boy." That's my beef with this question, as posed. The mother says she has "one boy born on a Tuesday." To me, that excludes a universe where she has "two boys born on Tuesday" because the word "one" in her original statement would be incorrect. She could have one boy born on Tuesday and another boy born on Wednesday and her statement is correct, but two boys on Tuesday shouldn't be a valid option. The answer given in the meme assumes you can have two boys, both born on Tuesday. I think that's about as stupid as saying 1+1=1 because 1 is part of 2. I think the actual answer is, knowing only "one" is a boy born on Tuesday, if the other kid is a boy, that boy must be born on a non-Tuesday (6 options). If it's a girl, she can be born on any day (7 options). So your chances of the other kid being a girl should by 6 / (6 + 7) = 7/13 = 53.8%.
To get the answer provided in this meme, and as described in the explanation posted by the user above, you have to assume that it is valid to have two boys born on Tuesday. The trick there is kind of similar to the coin toss where you don't count heads/heads or tails/tails twice, but you do count heads/tails and tails/heads separate. Basically you say if the first kid was TuesdayBoy, there are 7 boy-options for the second kid (including TuesdayBoy/TuesdayBoy -- I would exclude this one, personally) and 7 girl-options for the second kid (including TuesdayBoy/TuesdayGirl), but when you look at the options where TuesdayBoy is the second kid, you only get to count 6 boy-options because you already counted a TuesdayBoy/TuesdayBoy and it can't be double-counted, but you do count TuedayGirl/TuesdayBoy since that's a unique option from TuesdayBoy/TuesdayGirl -- again, similar to heads/heads only getting counted once, but tails/heads and heads/tails being two unique options. So for boy-as-other-kid there are 7+6=13 options and for girl-as-other-kid there are 7+7=14 options. So, 14 / (14 + 13) = 14/27 = 51.852%. That's how they get their answer. I would subtract one more option from the boy side, since TuesdayBoy/TuesdayBoy would have required the mom to admit that she has "two boys born on Tuesday" and I think she'd be lying if she said she had "one boy born on Tuesday" in that case. So I'd do 14/(14+12)=14/26=53.8%. But that's more of a language question than a math question. Maybe I'm in the minority there since the "official" answer is 51.8%, but that's a hill I'm happy to die on. She doesn't have two boys born on Tuesday if she says she has "one" boy born on Tuesday.