The sisters paradox is madenning example of counter-intuitive probability. The resolution is straightforward, but it's really easy to get tied up in knots.
"This might seem abstract, but I've seen variations of this problem pop up in business and I've had difficult conversations with non-technical people as a result."
Does anyone have some real life examples? i cannot think of any off hand but would like to be able to cite a couple if someone says "So, what is this good for?".
"We noticed that our app reviews on TV, Watch, and VR suck, why could that be? Are the OSes on those platforms terrible? Are user expectations just super high? Can we ever make money on those platforms?"
"When you're on TV, you need to use a DPad to laboriously type in a review. Same with the watch and its tiny screen. When you're on VR, you're using a laser pointer. That makes the barrier to leaving a review very high. As a result, people only bother to leave reviews if they have particularly strong feelings. The strongest motivating emotion is anger. Ergo, we are getting a disproportionate number of reviews from people who are angry, and everyone that is happy is not going to bother leaving a review."
Same goes for many kinds of user feedback situations. The feedback you're measuring is P(user holds that opinion | user bothers to tell you that opinion). The denominator needs to be understood in order to evaluate that probability, but by definition, you don't have any data on people who don't bother to engage with you, unless you can measure their interaction in some other way (eg. usage data). This is why online reviews of many offline businesses skew negative; why businesses will often nag you to leave a review if you're satisfied; why corporate executives often overestimate employee satisfaction, particularly in prestigious businesses (anyone who is dissatisfied will leave and get a job elsewhere), and why Congress has such a low approval rating (by definition, you can't escape the laws enacted by Congress, so there is no "exit" option, and you get an accurate appraisal of what people really think).
Perhaps people would be more likely to give the correct answer if "at least one of them is a girl" is rephrased as the equivalent "the youngest is a girl or the oldest is a girl".
It's not even worth mentioning this problem unless you talk about how the result depends on the data generating process. If you take it to be something like "you randomly sample from families with two children, discarding any without at least one girl", you get the 1/3 result, but there are various other ways to read a sampling process from the problem statement which lead to other results.
Right, you have to very clearly define how you do the sampling. Here's some possible cases:
Assume all houses have 2 children, and each child has equal probability of being either a boy or a girl. It will help to treat children as distinguishable, e.g. by eldest vs youngest.
* You randomly choose a house, and a random child answers the phone. You question the person who answers the phone about his/her gender and that of his/her sibling. You repeat this experiment. Given that the person who answered the phone was a girl, the conditional probably she has a sister is 1/2. Note that this is not bertrand’s box, as the 1 girl house can occur as either eldest vs youngest girl (unlike Bertrand box where only 1 box has exactly 1 gold) so they cancel out and so a girl you spoke with is equally likely to have been from either a 2 girl or 1 girl house.
* You categorize houses into 2 girl, 1 girl, and 0 girl houses. You randomly pick a category, then pick a house from that category, then phone them. As before a random child answers, and you question them. given that the person who answered the phone was a girl, the conditional probably she has a sister is 2/3. This is bertrand’s box, you’re more likely to have spoken with a girl from from a 2 girl house than a 1 girl house. Explicitly grouping by # of girls first before sampling breaks the previous symmetry.
* You randomly choose a house, and ask for the eldest child. You question him/her. You repeat this experiment. Given that the person whom you spoke with was a girl, the probability she has a sister is 1/2. Nothing new here, as seen in case (1) you were already equally likely to speak with a girl from a 2 girl vs 1 girl house, so asking for the eldest person (which by symmetry is equally likely to be a boy or a girl) doesn’t change anything here.
* You categorize houses into 2 girl, 1 girl, and 0 girl houses. You randomly pick a category, then pick a house from that category, then phone them and ask for the eldest child. Given that the person who answered the phone was a girl, the conditional probably she has a sister is 1/2. Explicitly choosing the eldest disrupts the asymmetry in bertrand’s box: since every house has only 1 eldest which is the one you speak with, being from a two girl house no longer makes that girl more likely to have spoken with the caller.
* You randomly choose a house, and a random child answers the phone. You question the person who answers the phone. You repeat this experiment. Given that EITHER the child you spoke with OR their sibling is a girl, the probability you spoke with someone from a 2 girl house is 1/3. It might seem counterintuitive at first that loosening the criteria _reduces_ the probability of speaking with someone from a 2 girl house. But this makes sense, since there’s still only 2 ways you can speak with someone from a 2 girl house (either the eldest or youngest sister), but now 4 ways you can speak with someone from a 1 girl house, since you’re allowed to speak with the boys of that house as well.
* You randomly choose a house, and ask for the eldest child, and question them. Given that EITHER the child you spoke with OR their sibling is a girl, the probability you spoke with someone from a 2 girl house is again 1/3. Explicitly speaking with the eldest doesn’t make a difference here because we’re already conditioning on either the eldest or youngest being a girl.
Peter Winkler shares some great variations of this: "Boy Born on Tuesday" (p. xix) and "Men with Sisters" (p. xxii) in "Mathematical Puzzles".
"Mrs. Chance has two children of different ages. At least one of them is a boy born on Tuesday. What is the probability that both of them are boys?"
(note: it is a puzzle, not a biology or data demography problem. so there are 50/50 independence assumptions on gender and uniform day of week assumptions prior to adding the conditioning.)
One of the challenges with puzzles like this it that it gives you "at least one of them is a girl" as a mathematical assertion where you're not supposed to further introspect the context of how/why you're being given that fact.
But that's unrealistic. In real life, the context for how and why there would be a speaker telling you such a thing in the first place can be relevant and affect the probability!
How is this possible? Suppose among all the math-riddle-loving parents of two children who would ask such a puzzle in the first place there are an equal number of parents of B-B, B-G, G-B, G-G, and that each is equally likely to ask you such a riddle when you meet them.
Suppose when asking such a riddle the B-B parents tell you "at least one of them is a boy" (they don't have any girls, so that's the only way they can ask this kind of riddle), the G-G parents tell you "at least one of them is a girl" (same thing but in reverse), while the B-G and G-B parents say one of "at least one of them is a boy" and "at least one of them is a girl" equally at random.
Then, conditioned on being told that "at least one of them is a girl", the probability of another girl is actually 1/2, not 1/3 like the paradox answer claims. To see this, imagine 40 examples of the above puzzle asking taking place. You get 10 B-B parents saying "at least one of them is a boy", 10 G-G parents saying "at least one of them is a girl", and among the 20 (B-G and G-B) parents since they choose randomly, you have 10 saying "at least one of them is a boy" and saying "at least one of them is a girl".
So out of the 20 times where "at least one of them is a girl" is said, there are 10 cases where it's a G-G family and 10 cases where it's a B-G or G-B family, therefore conditioned on being told "at least one of them is a girl", the probability of two girls is actually 1/2.
If there were some gender bias in how the B-G and G-B families might ask the question, or other differences that affect how likely different of these people would be posing the puzzle to you, then the probability could be yet different than either of 1/3 or 1/2.
So there's a difference in being present something as a flat mathematical assertion that you're supposed to take at face value and not supposed to question further (where the probability is 1/3, as the article claims). Versus being told something in real life, where you always need to take into account the context and situation of the speaker, and the probability could be different.
There are real life implications of this too - the big classic one being publication bias / newsworthiness bias. As most people intuitively know by now, it is also often wrong to take the statistical analysis or claims of a particular research study or paper entirely at face value, because there is a bias in the fact that "positive" and "exciting" results are more likely to be reported in the first place, and so statistical outliers that aren't actually replicable are disproportionately likely to be reported (see also https://xkcd.com/882/). And publication bias still occurs with respect to the reporting of results, amplification or not in the media etc, even when the the authors themselves are trustworthy and have done their analysis within the paper in a statistically proper way. So conditioned on you hearing about the result in the first place, it is often less likely to be true (and less likely to replicate in the future, etc) than you would think if you just took the statistical analysis in the paper at face value, even when that analysis was done correctly. The situation in the "sisters paradox" of computing a probability taking a statement entirely at logical face value is rare in real life.
> Following classical probability arguments, we consider a large urn containing two children.
I like how they modified a classic from probability texts, drawing items from an urn, and made sure it would be big enough in this example to accommodate two kids.
The "paradox" problem is in the setup. It's easy to mistake it as "a couple has one girl, what is the probability that their next child will be a girl," in which case the answer is 50%.
There is no sisters paradox. The trick is how the question is weirdly framed and has to be interpreted. What people think about when they hear the question would effectively lead to a probability of 0.5: if you see a family in the street with a girl and know they have two kids, the probability of the other kid being a girl is indeed 0.5.
The trick of the so-called "paradox" is turning the question into the Monty Hall but with an ambitious enough formulation that you might be confused it’s not.
The other interpretation that leads to 1/3 probably is also pretty intuitive. That's the fun part of this question is that it leaves crucial information unspecified.
I think this is a reasonable interpretation:
You meet a family at a party. They say "We have two children". You ask "Do you have any girls"? They say "yes!"
This will give you 1/3 probability that the other child is also a girl.
I think this interpretation is more intuitive because it doesn't make any assumptions about how you get your information. Usually in probability questions you assume any information you have is given to you from on high. For example, you just "know" that the family has two children, you don't somehow deduce it. Therefore I assume the same for "one child is a girl" information.
This reminds me of the Monty Hall. Which doesn't explicitly tell that host never reveals the car. Which for lot of people would be sensible option in game show. Just screw them up instantly. Well at least the winner gets some mutton.
Q1: "A family has two children. You're told that at least one of them is a girl. What's the probability both are girls?"
Q2: "A family has two children. You're told that at least one of them is a boy. What's the probability both are boys?"
Note that these are symmetric problems, and must have the same answer.
Q3: "A family has two children. You're told that a gender, that applies to at least one, is written inside a sealed envelope. What's the probability both have that gender?"
In Q3, we have no information. So the answer is the proportion of two-child families that are single gendered. That is, 1/2.
But if we open the envelope, and read what is written inside, the problem becomes either Q1 or Q2. Which have the same answer. So we don't have to open it; whatever the answer to Q1 and Q2 is, opening the envelope in Q3 make its answer the same. If that answer is 1/3, we have a paradox. The answer has to be 1/2 of we don't look.
This is what is known as "Bertrand's Box Paradox." Well, if we add a fourth box to his problem, with one gold and one silver coin. I realize that in modern times the problem itself is called the paradox, but what Bertrand actually wrote (edited to this problem) was "How can it be that opening the envelope suffices to change the probability from 1/2 to 1/3?"
The resolution is that probability must be based on the full set of possibilities, not the possibilities that _could_ result from the full set of _states._ These are the possibilities for this problem:
1) BB and you are told that there is at least one boy.
2A) BG and you are told that there is at least one boy.
2B) BG and you are told that there is at least one girl.
3A) GB and you are told that there is at least one boy.
3B) GB and you are told that there is at least one girl.
4) GG and you are told that there is at least one girl.
Each numbered case has a prior probability of 1/4. Let's say the "A" subcases have a probability of Q/4, so the "B" subcases have a probability of (1-Q)/4.
The answer to the first problem is the probability of case 1, which is 1/4, divided by the total probability of cases 1, 2A, which is (1+2Q)/4. That's 1/(1+2Q).
The answer to the second problem is the probability of case 4, divided by the total probability of cases 4, 2B, and 3B. Which is (3-2Q)/4.
Bertrand's paradox, stated another way, is that these must be equal, but can only be equal if Q=1/2 and both answers are 1/2.
I think the explanation is wrong. It is based on the probability of having combinations of boys and girls and then counting the combinations that have at least one girl, but this is not the situation: there is no probability in question for one of the kids, it is a confirmed, past event and the only other probability is the sex of the other kid.
Otherwise you can derive any probability as a branch of a probability tree that contains it and calculate the probabilities of the tree and then the one of the branch. This makes no sense.
For example, a family has a kid and the kid is a girl. The family wants another kid; what is the probability to be a girl? Is it 1/4 because having 2 girls is 1/4? No, it is 1/2 as it is for any new kid.
So, my problem with how this is modeled is it assumes order doesn't matter in one aspect, but that it does in another.
Simply stated, if you allow the possibility space of "boy-girl" and "girl-boy", you have to also have two "girl-girl" states. Since you don't know which of the kids is known. Why is that not correct?
State it with coins, if I know that you flipped a quarter and a dime and one turned up heads, what are the odds that both are heads?
When I first heard of the Monty Hall problem, I assumed the naive answer was true, then I thought more and drew up the decision tree and thought the “correct” answer was right. Now I think it is an underspecified problem and literally any probability can be assigned to the result. There are no bad answers because it is a bad(ly stated) problem.
“Here's the problem: a family has two children. You're told that at least one of them is a girl. What's the probability both are girls?”
— This is the complete statement of the problem! Everything else is an assumption that may or may not be correct. And is certainly not necessarily a complete set of underlying assumptions relevant to the problem statement.
“Assume that the probability of having a girl or boy is 50% and that the birth order has no effect on the probability. Assume the family is selected at random because they have at least one girl.”
— This is not a part of the statement of the problem! These are a subset of assumptions that can choose to accept, or not. As a modeler or decision analyst you have to make that distinction. Eh, let’s accept them, for the time being. We’ll even assume the narrator is honest, which isn’t a stated assumption.
But let’s add to that list of assumptions. The narrator telling you that one of them is a girl gets all winnings from bets on the outcome of the unknown gender of the “other child” and wants those winnings. The narrator knows that a probability tree analysis of the problem, with perhaps unwarranted assumptions of independence and prior probabilities, will lead to an assignment of 1/3 probability for the other sibling being a girl, and knows you know that result and believes you want to win. [A valid credible interpretation, not misinterpretation, of the original problem statement.]
“What do you think the probability is that both children are girls?”
— Let’s make this question more actionable. “Should you take the even odds bet on both children being girls made by the narrator?. $100 - if they are both girls, the narrator wins $100 and you lose $100; if they aren’t, you win $100 and the narrator loses $100. The narrator and you want to win the money.”
The answer to this question, which seemingly follows from the question of probabilities to be “yes”, is, in fact, “no” - under the additional valid, and quite credible, assumptions made. Because you will only be presented sets of two-girl pairs by the narrator. Let’s assume the “assumptions” are actually correct, and the families will be indeed selected at random, and in the general population there is a 50-50 mix of boys and girls. There is nothing, even in the “assumptions”, that precludes the narrator from preselecting and only presenting two-girl pairs to you, thus always winning when you believe and follow the 1/3 two-girl result.
The statement of the problem, and only the statement of the problem, underspecified as it is, leads to a whole suite of possibly correct answers. The problem is the territory, the problem statement and assumptions are the map.
None of these maps are the territory, necessarily. The probability tree answer is just as sloppy, from a decision analyst perspective, as the naive answer.
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[ 4.3 ms ] story [ 55.6 ms ] threadDoes anyone have some real life examples? i cannot think of any off hand but would like to be able to cite a couple if someone says "So, what is this good for?".
"When you're on TV, you need to use a DPad to laboriously type in a review. Same with the watch and its tiny screen. When you're on VR, you're using a laser pointer. That makes the barrier to leaving a review very high. As a result, people only bother to leave reviews if they have particularly strong feelings. The strongest motivating emotion is anger. Ergo, we are getting a disproportionate number of reviews from people who are angry, and everyone that is happy is not going to bother leaving a review."
Same goes for many kinds of user feedback situations. The feedback you're measuring is P(user holds that opinion | user bothers to tell you that opinion). The denominator needs to be understood in order to evaluate that probability, but by definition, you don't have any data on people who don't bother to engage with you, unless you can measure their interaction in some other way (eg. usage data). This is why online reviews of many offline businesses skew negative; why businesses will often nag you to leave a review if you're satisfied; why corporate executives often overestimate employee satisfaction, particularly in prestigious businesses (anyone who is dissatisfied will leave and get a job elsewhere), and why Congress has such a low approval rating (by definition, you can't escape the laws enacted by Congress, so there is no "exit" option, and you get an accurate appraisal of what people really think).
Assume all houses have 2 children, and each child has equal probability of being either a boy or a girl. It will help to treat children as distinguishable, e.g. by eldest vs youngest.
* You randomly choose a house, and a random child answers the phone. You question the person who answers the phone about his/her gender and that of his/her sibling. You repeat this experiment. Given that the person who answered the phone was a girl, the conditional probably she has a sister is 1/2. Note that this is not bertrand’s box, as the 1 girl house can occur as either eldest vs youngest girl (unlike Bertrand box where only 1 box has exactly 1 gold) so they cancel out and so a girl you spoke with is equally likely to have been from either a 2 girl or 1 girl house.
* You categorize houses into 2 girl, 1 girl, and 0 girl houses. You randomly pick a category, then pick a house from that category, then phone them. As before a random child answers, and you question them. given that the person who answered the phone was a girl, the conditional probably she has a sister is 2/3. This is bertrand’s box, you’re more likely to have spoken with a girl from from a 2 girl house than a 1 girl house. Explicitly grouping by # of girls first before sampling breaks the previous symmetry.
* You randomly choose a house, and ask for the eldest child. You question him/her. You repeat this experiment. Given that the person whom you spoke with was a girl, the probability she has a sister is 1/2. Nothing new here, as seen in case (1) you were already equally likely to speak with a girl from a 2 girl vs 1 girl house, so asking for the eldest person (which by symmetry is equally likely to be a boy or a girl) doesn’t change anything here.
* You categorize houses into 2 girl, 1 girl, and 0 girl houses. You randomly pick a category, then pick a house from that category, then phone them and ask for the eldest child. Given that the person who answered the phone was a girl, the conditional probably she has a sister is 1/2. Explicitly choosing the eldest disrupts the asymmetry in bertrand’s box: since every house has only 1 eldest which is the one you speak with, being from a two girl house no longer makes that girl more likely to have spoken with the caller.
* You randomly choose a house, and a random child answers the phone. You question the person who answers the phone. You repeat this experiment. Given that EITHER the child you spoke with OR their sibling is a girl, the probability you spoke with someone from a 2 girl house is 1/3. It might seem counterintuitive at first that loosening the criteria _reduces_ the probability of speaking with someone from a 2 girl house. But this makes sense, since there’s still only 2 ways you can speak with someone from a 2 girl house (either the eldest or youngest sister), but now 4 ways you can speak with someone from a 1 girl house, since you’re allowed to speak with the boys of that house as well.
* You randomly choose a house, and ask for the eldest child, and question them. Given that EITHER the child you spoke with OR their sibling is a girl, the probability you spoke with someone from a 2 girl house is again 1/3. Explicitly speaking with the eldest doesn’t make a difference here because we’re already conditioning on either the eldest or youngest being a girl.
What's the paradox?
"Mrs. Chance has two children of different ages. At least one of them is a boy born on Tuesday. What is the probability that both of them are boys?"
(note: it is a puzzle, not a biology or data demography problem. so there are 50/50 independence assumptions on gender and uniform day of week assumptions prior to adding the conditioning.)
But that's unrealistic. In real life, the context for how and why there would be a speaker telling you such a thing in the first place can be relevant and affect the probability!
How is this possible? Suppose among all the math-riddle-loving parents of two children who would ask such a puzzle in the first place there are an equal number of parents of B-B, B-G, G-B, G-G, and that each is equally likely to ask you such a riddle when you meet them.
Suppose when asking such a riddle the B-B parents tell you "at least one of them is a boy" (they don't have any girls, so that's the only way they can ask this kind of riddle), the G-G parents tell you "at least one of them is a girl" (same thing but in reverse), while the B-G and G-B parents say one of "at least one of them is a boy" and "at least one of them is a girl" equally at random.
Then, conditioned on being told that "at least one of them is a girl", the probability of another girl is actually 1/2, not 1/3 like the paradox answer claims. To see this, imagine 40 examples of the above puzzle asking taking place. You get 10 B-B parents saying "at least one of them is a boy", 10 G-G parents saying "at least one of them is a girl", and among the 20 (B-G and G-B) parents since they choose randomly, you have 10 saying "at least one of them is a boy" and saying "at least one of them is a girl".
So out of the 20 times where "at least one of them is a girl" is said, there are 10 cases where it's a G-G family and 10 cases where it's a B-G or G-B family, therefore conditioned on being told "at least one of them is a girl", the probability of two girls is actually 1/2.
If there were some gender bias in how the B-G and G-B families might ask the question, or other differences that affect how likely different of these people would be posing the puzzle to you, then the probability could be yet different than either of 1/3 or 1/2.
So there's a difference in being present something as a flat mathematical assertion that you're supposed to take at face value and not supposed to question further (where the probability is 1/3, as the article claims). Versus being told something in real life, where you always need to take into account the context and situation of the speaker, and the probability could be different.
There are real life implications of this too - the big classic one being publication bias / newsworthiness bias. As most people intuitively know by now, it is also often wrong to take the statistical analysis or claims of a particular research study or paper entirely at face value, because there is a bias in the fact that "positive" and "exciting" results are more likely to be reported in the first place, and so statistical outliers that aren't actually replicable are disproportionately likely to be reported (see also https://xkcd.com/882/). And publication bias still occurs with respect to the reporting of results, amplification or not in the media etc, even when the the authors themselves are trustworthy and have done their analysis within the paper in a statistically proper way. So conditioned on you hearing about the result in the first place, it is often less likely to be true (and less likely to replicate in the future, etc) than you would think if you just took the statistical analysis in the paper at face value, even when that analysis was done correctly. The situation in the "sisters paradox" of computing a probability taking a statement entirely at logical face value is rare in real life.
That’s not what the first question said. The first question was select a family (Bb,Bb,gb,gg)
Then that they happen to have a girl.
> Following classical probability arguments, we consider a large urn containing two children.
I like how they modified a classic from probability texts, drawing items from an urn, and made sure it would be big enough in this example to accommodate two kids.
https://en.m.wikipedia.org/wiki/Monty_Hall_problem
of those 3 cases, 1/3 are both girls
The trick of the so-called "paradox" is turning the question into the Monty Hall but with an ambitious enough formulation that you might be confused it’s not.
I think this is a reasonable interpretation:
You meet a family at a party. They say "We have two children". You ask "Do you have any girls"? They say "yes!"
This will give you 1/3 probability that the other child is also a girl.
I think this interpretation is more intuitive because it doesn't make any assumptions about how you get your information. Usually in probability questions you assume any information you have is given to you from on high. For example, you just "know" that the family has two children, you don't somehow deduce it. Therefore I assume the same for "one child is a girl" information.
Q2: "A family has two children. You're told that at least one of them is a boy. What's the probability both are boys?"
Note that these are symmetric problems, and must have the same answer.
Q3: "A family has two children. You're told that a gender, that applies to at least one, is written inside a sealed envelope. What's the probability both have that gender?"
In Q3, we have no information. So the answer is the proportion of two-child families that are single gendered. That is, 1/2.
But if we open the envelope, and read what is written inside, the problem becomes either Q1 or Q2. Which have the same answer. So we don't have to open it; whatever the answer to Q1 and Q2 is, opening the envelope in Q3 make its answer the same. If that answer is 1/3, we have a paradox. The answer has to be 1/2 of we don't look.
This is what is known as "Bertrand's Box Paradox." Well, if we add a fourth box to his problem, with one gold and one silver coin. I realize that in modern times the problem itself is called the paradox, but what Bertrand actually wrote (edited to this problem) was "How can it be that opening the envelope suffices to change the probability from 1/2 to 1/3?"
The resolution is that probability must be based on the full set of possibilities, not the possibilities that _could_ result from the full set of _states._ These are the possibilities for this problem:
1) BB and you are told that there is at least one boy. 2A) BG and you are told that there is at least one boy. 2B) BG and you are told that there is at least one girl. 3A) GB and you are told that there is at least one boy. 3B) GB and you are told that there is at least one girl. 4) GG and you are told that there is at least one girl.
Each numbered case has a prior probability of 1/4. Let's say the "A" subcases have a probability of Q/4, so the "B" subcases have a probability of (1-Q)/4.
The answer to the first problem is the probability of case 1, which is 1/4, divided by the total probability of cases 1, 2A, which is (1+2Q)/4. That's 1/(1+2Q).
The answer to the second problem is the probability of case 4, divided by the total probability of cases 4, 2B, and 3B. Which is (3-2Q)/4.
Bertrand's paradox, stated another way, is that these must be equal, but can only be equal if Q=1/2 and both answers are 1/2.
Otherwise you can derive any probability as a branch of a probability tree that contains it and calculate the probabilities of the tree and then the one of the branch. This makes no sense.
For example, a family has a kid and the kid is a girl. The family wants another kid; what is the probability to be a girl? Is it 1/4 because having 2 girls is 1/4? No, it is 1/2 as it is for any new kid.
Simply stated, if you allow the possibility space of "boy-girl" and "girl-boy", you have to also have two "girl-girl" states. Since you don't know which of the kids is known. Why is that not correct?
State it with coins, if I know that you flipped a quarter and a dime and one turned up heads, what are the odds that both are heads?
Yeah, I'm with you here. Assuming heads=girl and tails=boy:
* If order does NOT matter (GB=BG), then it means Alice-Albert (GB, Heads-Tails) is the same as Albert-Alice (BG, Tails-Heads).
* If order DOES matter (GB!=BG), then it means Alice-Barbara (GG, Heads-Heads) is different than Barbara-Alice (GG, Heads-Heads). Thereforce, GG!=GG.
Either way, the stated problem seems badly defined.
“Here's the problem: a family has two children. You're told that at least one of them is a girl. What's the probability both are girls?”
— This is the complete statement of the problem! Everything else is an assumption that may or may not be correct. And is certainly not necessarily a complete set of underlying assumptions relevant to the problem statement.
“Assume that the probability of having a girl or boy is 50% and that the birth order has no effect on the probability. Assume the family is selected at random because they have at least one girl.”
— This is not a part of the statement of the problem! These are a subset of assumptions that can choose to accept, or not. As a modeler or decision analyst you have to make that distinction. Eh, let’s accept them, for the time being. We’ll even assume the narrator is honest, which isn’t a stated assumption.
But let’s add to that list of assumptions. The narrator telling you that one of them is a girl gets all winnings from bets on the outcome of the unknown gender of the “other child” and wants those winnings. The narrator knows that a probability tree analysis of the problem, with perhaps unwarranted assumptions of independence and prior probabilities, will lead to an assignment of 1/3 probability for the other sibling being a girl, and knows you know that result and believes you want to win. [A valid credible interpretation, not misinterpretation, of the original problem statement.]
“What do you think the probability is that both children are girls?”
— Let’s make this question more actionable. “Should you take the even odds bet on both children being girls made by the narrator?. $100 - if they are both girls, the narrator wins $100 and you lose $100; if they aren’t, you win $100 and the narrator loses $100. The narrator and you want to win the money.”
The answer to this question, which seemingly follows from the question of probabilities to be “yes”, is, in fact, “no” - under the additional valid, and quite credible, assumptions made. Because you will only be presented sets of two-girl pairs by the narrator. Let’s assume the “assumptions” are actually correct, and the families will be indeed selected at random, and in the general population there is a 50-50 mix of boys and girls. There is nothing, even in the “assumptions”, that precludes the narrator from preselecting and only presenting two-girl pairs to you, thus always winning when you believe and follow the 1/3 two-girl result.
The statement of the problem, and only the statement of the problem, underspecified as it is, leads to a whole suite of possibly correct answers. The problem is the territory, the problem statement and assumptions are the map.
None of these maps are the territory, necessarily. The probability tree answer is just as sloppy, from a decision analyst perspective, as the naive answer.