Why not? I don’t see what’s wrong with the analysis provided, which also tallies with Pleonast’s.
Alas, this situation is different. See my earliest explanation.
Your description is picking an individual first, then looking at the sibling. While there’s twice as many mixed gender pairs as boy-only pairs, there’s twice as many boys in a boy-only pair as a mixed-gender pair. Which gives a 50% chance that the unseen sibling is also a boy.
This is probably the crux of the matter. Many people have difficult understanding that how the information is found out is vital to determining the probabilities.
That’s why I adjusted to:
You pick the pair by picking the individual.
Which is a different question than the one in the OP. It’s extremely important to make that distinction.
Very interesting point. I hadn’t considered that.
Let’s say - At least one child is a girl, but we don’t know that.
We have 3 possible cases - BG, GB, GG.
One of the children at random answers the door.
So we have a 2-in-6 chance that a boy will answer the door, 4 in-6 that a girl answers. Yes, the odds then are 50-50 the other child is a girl.
The catch here is again, you have additional information for asking odds, which then changes the odds. You are restricting the question to “there is at least one girl AND a girl answers the door”. So who answers the door is not random now.
So perhaps my wording needs to be clarified. We weren’t picking on girls, or making them answer the door. Who answers the door is random.
The original question really is “We know nothing about the genders of the children. Gender A answers the door, what are the odds of the other child being A or being B?” We don’t eliminate one gender ahead of time. If a boy answered the door, the question would be the same - what are the odds the other child is a girl? A boy? Instead of eliminating the BB case, we’ve eliminated the GG case.
Yes, because he’s making the minimal assumption that the way that he learns that at least one child is a boy is the way that people almost always learn that.
I don’t see a difference. You can count the Monty Hall problem too, there’d be 18 possible outcomes, based on you selecting one of three doors and Monty Hall selecting one of the two remaining (3 * 3 * 2). The key is that Monty selects a door randomly even if it has the prize, but the door he actually reveals must always be a goat and Monty revealing a different door than the one he picked doesn’t increase (ETA: or decrease) the sample space as it’s deterministic. So under 12 of the 18 possible outcomes, your door hides a goat and the other door hides the prize, compared to 6 out of 18 where you picked the right door to begin with.
~Max
I’m not sure what you are not seeing a difference to.
Monty Hall is well understood and doesn’t impact the question on the cited web site.
Monty Hall Problem, counted
"Spoiler (click to show/hide)
There are three doors. Behind one door is a prize, but there are goats behind the other two. You pick one of the doors at random. Monty Hall selects one of the remaining goat-hiding doors at random: if it hides a goat, he reveals the door he picked; however, if it hides the prize, he reveals the other door.\text{Probability that a given door hides a prize: } \frac{6}{18} = \frac{1}{3}
\text{Probability that the chosen door hides a goat: } \frac{12}{18} = \frac{2}{3}
\text{Probability that the unrevealed door hides the prize: } \frac{12}{18} = \frac{2}{3}
Door 1 (prize), Door 2 (goat), Door 3 (goat), you pick Door 1 (prize), Monty Hall reveals the goat behind Door 2, leaving Door 3 (goat)
Door 1 (prize), Door 2 (goat), Door 3 (goat), you pick Door 1 (prize), Monty Hall reveals the goat behind Door 3, leaving Door 2 (goat)
Door 1 (prize), Door 2 (goat), Door 3 (goat), you pick Door 2 (goat), Monty Hall picks Door 1, but knows it contains the prize, so he reveals the goat behind Door 3, leaving Door 1 (prize).
Door 1 (prize), Door 2 (goat), Door 3 (goat), you pick Door 2 (goat), Monty Hall reveals the goat behind Door 3, leaving Door 1 (prize)
Door 1 (prize), Door 2 (goat), Door 3 (goat), you pick Door 3 (goat), Monty Hall picks Door 1, but knows it contains the prize, so he reveals the goat behind Door 2, leaving Door 1 (prize).
Door 1 (prize), Door 2 (goat), Door 3 (goat), you pick Door 3 (goat), Monty Hall reveals the goat behind Door 2, leaving Door 1 (prize)
Door 1 (goat), Door 2 (prize), Door 3 (goat), you pick Door 1 (goat), Monty Hall picks Door 2, but knows it contains the prize, so he reveals the goat behind Door 3, leaving Door 2 (prize).
Door 1 (goat), Door 2 (prize), Door 3 (goat), you pick Door 1 (goat), Monty Hall reveals the goat behind Door 3, leaving Door 2 (prize)
Door 1 (goat), Door 2 (prize), Door 3 (goat), you pick Door 2 (prize), Monty Hall reveals the goat behind Door 1, leaving Door 3 (goat)
Door 1 (goat), Door 2 (prize), Door 3 (goat), you pick Door 2 (prize), Monty Hall reveals the goat behind Door 3, leaving Door 1 (goat)
Door 1 (goat), Door 2 (prize), Door 3 (goat), you pick Door 3 (goat), Monty Hall reveals the goat behind Door 1, leaving Door 2 (prize)
Door 1 (goat), Door 2 (prize), Door 3 (goat), you pick Door 3 (goat), Monty Hall picks Door 2, but knows it contains the prize, so he reveals the goat behind Door 1, leaving Door 2 (prize).
Door 1 (goat), Door 2 (goat), Door 3 (prize), you pick Door 1 (goat), Monty Hall reveals the goat behind Door 2, leaving Door 3 (prize)
Door 1 (goat), Door 2 (goat), Door 3 (prize), you pick Door 1 (goat), Monty Hall picks Door 3, but knows it contains the prize, so he reveals the goat behind Door 2, leaving Door 3 (prize).
Door 1 (goat), Door 2 (goat), Door 3 (prize), you pick Door 2 (goat), Monty Hall reveals the goat behind Door 1, leaving Door 3 (prize)
Door 1 (goat), Door 2 (goat), Door 3 (prize), you pick Door 2 (goat), Monty Hall picks Door 3, but knows it contains the prize, so he reveals the goat behind Door 1, leaving Door 3 (prize).
Door 1 (goat), Door 2 (goat), Door 3 (prize), you pick Door 3 (prize), Monty Hall reveals the goat behind Door 1, leaving Door 2 (goat)
Door 1 (goat), Door 2 (goat), Door 3 (prize), you pick Door 3 (prize), Monty Hall reveals the goat behind Door 2, leaving Door 1 (goat)
The two-kids problem in the link from post #1 (not the OP’s reformulation, the citation itself: “Suppose a family has two children and we know that one of them is a boy. What is the probability that the family has two boys?”) has a sample space of 4 possibilities, as it explains in the solutions section. The Monty Hall problem, as answered on the same page in the same terms, has 18 possibilities, but they simplify it to 9 by not differentiating when Monty Hall picks between two goat doors.
The two problems described by the text of post #1 have a sample space of 8 and solutions of 1/2.
A family with three binary gendered kids could be arranged with equiprobability in 2^3 (two genders, three kids) ways,
BBB (3 boys, 0 girls)
BBG (2 boys, 1 girl)
BGB (2 boys, 1 girl)
BGG (1 boy, 2 girls)
GBB (2 boys, 1 girl)
GBG (1 boy, 2 girls)
GGB (1 boy, 2 girls)
GGG (3 girls)
And there are other ways to count, we did tree diagrams in school. Either way with simple examples counting is the easy way to get the probability distribution:
\text{Probability of there being three boys: } \frac{1}{8}
\text{Probability of there being two boys and one girl: } \frac{3}{8}
\text{Probability of there being one boy and two girls: } \frac{3}{8}
\text{Probability of there being three girls: } \frac{1}{8}
Otherwise you have to calculate each possibility, e.g.
\text{Probability of there being exactly two girls:}
\text{ } = \frac{\text{Combinations of two girls among three binary gendered kids}}{\text{Sample space}}
\text{ } = \frac{3!}{2! (3 - 2)!} \div \text{Sample space}
\text{ } = \frac{3 \times \not 2 \times \not 1}{\not 2 \times \not 1 \times \not 1} \div 2^3
\text{ } = 3 \div 8
\text{ } = \frac{3}{8}
If you were to know the oldest one was a boy, for example, you could eliminate the latter four possibilities from the above sample space and adjust the probabilities accordingly (now 50% probability of there being two boys and one girl). You can do this based on order because relative age among triplets is a trinary value, but it doesn’t have to be age. You can use the order for any single trinary value.
If we knew the kids were named Jo, Teri, and Alex, name is a trinary value and we can superimpose a name on each column of letters. If we know Alex is a girl, we can eliminate the four possibilities where the rightmost column is “B”.
But let’s say we want another variable such as whether the kids answer the door. This changes the sample space; each of the three kids belongs to one of two genders, and independently, answers the door or doesn’t. This gives 2^3 * 2^3 = 64 equiprobable permutations.
You could theoretically count all discrete sample spaces and calculate probability as a simple fraction, even when percentages are involved but in more complicated examples it gets out of hand. When I was in school, addition and multiplication rules for probability were taught with consecutive dice rolls, and two independently tossed 6-sided dice have only 6^2 = 36 equiprobable outcomes.
~Max
I’m not sure we are disagreeing.
Chronos the essence of this issue is that exactly how a piece of information is learned changes its value.
Being told that “mr smith doesnt have two boys” is different to knocking on his door and being greeted by a daughter.
Yes, I’m fairly sure that’s been @Chronos’ point this whole thread. It’s certainly been mine.
I don’t quite understand this, unless there is some context that was not written.
If I’m told “Mr. Smith has two children, but not two boys” then I know that Mr. Smith’s kids are either GG, BG, or GB. The BB option is ruled out.
If I’m greeted by a girl who says “I’m one of Mr. Smith’s two kids” then I also know that Mr. Smith’s kids are either GG, BG, or GB. The BB options is ruled out.
If prior to either method of learning there are not two boys, all I know is Mr. Smith has two kids, then in both cases the new information tells me that at least one of Mr. Smith’s kids is a girl. I’m not understanding why it matters how I learn that at least one of the kids is a girl.
Considering a set of two (binary identifying) people “at least one female” and “not two males” are equal statements. My instincts are often poor in these kinds of trick probability questions, so maybe I’m missing something.
Exactly - this puzzle is like a crooked politician - “What did you know, and when did you know it?” determines the odds of a certain result.
No, you’ve also ruled out BG. At least, if we’re listing the children in the order “one who answered the door” and then “one who didn’t”.
If you were told “Mr. smith has two children, but not two boys”, and then you go knock on the door and a girl answers, you’ve added new information - you’ve performed a statistical sample of the household. Prior to the statistical sample, the “either BG or GB” result was the likeliest option (2/3 chance) , compared to GG (1/3 chance) but now since a girl answered the door, the odds shift in the direction of GG (now 1/2).
If you had knocked on the door and a boy answered, you also would have received new information, enough in fact to completely rule out the GG option. Why would a girl answering the door be different?
Then that is part of the context that was left out.
So ordered by speed at which they answer the door, Mr. Smith’s kids are BB, BG, GB, or GG. Then when the door is answered by a girl (assuming both kids are home, etc.), I know that the kids are either GB or GG.
If, again ordered by speed at which they answer the door, I’m told that there are not two boys then BG can still be one of the available choices.
If speed at which they answer the door, and ordering are unrelated, then after the door is answered by a girl the choices are still BG, GB, are GG.
I’m not quite following this. Obviously if a boy answers I rule out GG. If a girl answers (and we’re not sorting by speed of answering the door) then prior to answering the door I new there was at least one girl (not two boys), but after answering I still only know that there is one girl. I don’t think I’ve learned any new information. Unless, as @Chronos says we’re sorting by speed of door answering, and then BG can be ruled out, and I’m left with GB or GG.
Maybe my hangup is that as somebody who does family based research, ordering children by speed at which they answer the door seems absurd. The most obvious and common method of ordering is by age, but proband and control, or randomly are other useful ordering methods.
Pretend you had a bad case of face blindness, and short term memory, so all you have is your notebook to rely on. You’re told “Mr. Smith has two children, but not two boys” and then go to their house, and a girl answers the door. You write “G” in your notebook, and then leave. You try again the next day at a different time, and a girl answers the door (could be same one, could be different), which you write in your notebook and leave. You try again the next day and a girl answers the door one more time.
So you’ve now had three separate visits and a girl has answered the door all three times. You can’t quite rule out the “BG/GB” option, but it’s starting to look less likely for each visit you’ve made, including the first one. It’s clear you’ve added new information with each subsequent visit, but all you can definitely prove is still “there is at least one girl”.
This is just sampling with replacement. If we never see a boy, then at some point we are going to guess that there are two girls, but that assumes that who answers the door is random in relation to gender. If the brother is a lazy couch potato and will never answer the door, then our coin isn’t fair, and our probability calculations will be way off.
(Edit to add: and yeah, there is lots of space between “random in relation to gender” and “boy never answers.” Boy may answer 20% of the time, for example.)
How many visits until I can ask “Is your brother or sister home?”