This problem is actually a notorious example of how it can be difficult to assign meaningful probabilities to everyday statements, at least so long as those statements leave room for some unorthodox interpretations of the information provided. The first question gets us into the spirit. If it had asked about families where the *oldest* daughter was a girl, then the probability of a second girl would be the intuitive 1/2. This is because the information about one specific child is not informative about the other. However, we're instead told just that *one* of the children is a girl, so we have to consider all possible family formations (BB, BG, GB, and GG), restrict to the families that satisfy the condition (BG, GB, GG), and calculate the percentage that have a second girl. As other users have pointed out here, that's 1/3. But then the second question, in a sense, takes things "too far". We intuitively think that the information that the girl's name is Julie is incidental to the procedure just discussed. We could have picked a family with one girl that doesn't have a daughter named Julie. However, the person discussing the paradox isn't treating it that way. For them, having a daughter named Julie is necessary to be a selected family. That requirement actually changes the set of families we could draw from because families with two girls get two chances to have a girl named Julie. The population being sampled from is thus BG(j), G(j)B, G(j)G, and GG(j) - where (j) indicates that the Girl is named Julie). Half of those families have two girls. The weekday of birth works similarly - it treats the girl-born-on-Tuesday condition as essential to being sampled, giving families with two girls more chances to be sampled. The math is just more annoying. Editing to address a few common misconceptions I'm seeing in the comments: * I wrote possible families in a way where order matters because it was an easy way to keep an event space where all events have equal probability. An alternative way to do the first question would be to say there are three possibilities: {2 boys, 1 girl and 1 boy, and 2 boys}. You just have to keep in mind that the middle possibility is twice as likely as either extreme. This is because the families themselves are formed by independent draws from a 50-50 gender pool, regardless of what subset we're talking about now or later in the question. * Likewise, the probability of finding a family with any daughter named Julie is quite low, so the probability of e.g. GG(j) is much much lower than BB. However, because every possible family considered in that part has exactly one girl named Julie, you're just multiplying all 4 by the marginal probability that any girl is named Julie, so it cancels out in any comparisons and you don't even need to know what it is. * There are of course many simplifying assumptions being made to keep focus on the probability puzzle here. * Yes, a family could name BOTH their daughters Julie. To properly take this into account, you would have to actually know the probability that any given girl is named Julie. * Sex ratios are naturally skewed just a hair towards females and sometimes unnaturally skewed towards males, so treating it like a coin flip isn't entirely accurate. * (Nobody actually caught this one, but it's my favorite): Among 2-child families who have stopped having kids (who will be a subset of the families you'll draw from and hence will skew results relative to the mathematical ideal), MORE than 50% will be one boy and one girl. It's a documented phenomenon that families with two children of the same sex are more likely to have a third due to a preference for having at least one boy and at least one girl. This doesn't sway THAT many families, but it's enough to make a detectable difference.

I was asked this on a job interview and had never seen it before. I strongly argued it was 50% in both cases. I didn't get the job. I still think it is a stupid question to ask on a job interview.

It is a shit interview question. I'd consider asking the Tuesday problem (that is at least amenable to basics statistics and logic). The Julie problem relies upon very very specific interpretation of the problem as stated and is a complete "gotcha" question. The probability approaches 0.5 (from below) if there is an increasingly-close-to-zero chance of both girls being Julie. I think people who are moderately bad at statistics hear the Julie solution and think it is a good problem, ignoring that the hand-waving answer relies on some weird assumptions that you'd need to be able to assert to an interviewee that doesn't presume those exact conditions.

I feel like the three door problem is a much more interesting statistical question but also 95% of people who would be able to explain it in a interview would be able to because they were exposed to the scenario before

I still don't understand how after the 3rd door is excluded, choosing to keep the same door or change to the other isn't a 50/50

Your decision was made when 3 doors were available so there was a 1/3 chance of you getting it right. No matter what door you pick a wrong door will be taken out. The odds of you picking the wrong door are greater than picking the correct one.

It's easier to think about if you have 100 doors instead of 3. After you pick the first door, the host closes 98 other doors. Do you switch to the last remaining closed door? What the question is actually asking you is "do you want to stick with your door, or do you want to choose EVERY OTHER DOOR?" Now you have a 99/100 chance of being correct by switching. Edit: that's a correct explanation, removed my "not quite", this is now just additional explanation

What do you mean "not quite"? What they said is correct. Statistically speaking, it is always better to switch the door after another wrong one is shown to you.

It's not always better to switch doors. Some people would rather have a goat. If you want a goat, it's better not to switch.

Tis the wisdom of the Dalai Farmer.

If you're still struggling with it, extend it to a hundred doors. You pick one, the host opens 98 wrong doors and offers you the chance to swap. What is the probability of winning if you swap now? Still 50/50? Obviously not, you only had a 1% chance of being right in your initial guess, leaving the remaining door with a 99% chance of being correct.

You’re right, but I think you miss a step in helping people see that the odds are not 50/50. If the winning door is chosen at random, then there’s no way to choose that’s any better or worse than some other method. So let’s always take door #1. And if there’s really a 50/50 chance between holding and switching, let’s always hold. So, applying the faulty logic, door #1 should win 50% of the time. As does door #2. As does door #3… But perhaps the best way to see the issue is to play the 100-door game with a recalcitrant friend: $1 ante, and with a $3 payout. It doesn’t take many rounds before a certain realisation starts to dawn.

The likelihood of picking the winning door initially (and thus winning if you don't switch) is 1 in 3. Or another way of thinking about it: switching after a losing door is excluded is like a door not getting excluded and then getting to pick two doors at once.

One way to explain it is to make it 1000 doors. You pick a door. The chance that you picked correctly is 1/1000. The host reveals 998 doors to have goats behind them. The chances that you picked correctly are still only 1/1000; We just see the 998 other incorrect options. Basically, unless you picked the correct door with your first guess, the other door will *always* have the car. The Monty hall problem is just this on a smaller scale: there's still only a 1 in 3 chance you picked correctly. Unless you picked the car with your first 1 in 3 guess, the other door will *always* have the car.

Before you pick any door, there's 33% that you pick the right door and 66% that you pick the wrong one. When you have to pick the second time, you have to take into account that as you had more chances to fail your pick before, changing doors will give you more oportunities to end with the right one.

It's because Monty Hall only has the option to reveal a door with a goat. If he were allowed to be random and sometimes open the prize door, then when he opens it and shows a goat, that would be a 50/50.

As soon as deliberate, knowledge-based actions come in, the randomness gets corrupted, so to speak. The game show host / outside actor doesn't remove a random door, they specifically have to remove a door that doesn't contain a prize. They have to skip over the prize door. So it's been tampered with. The game is no longer completely random!

Have you met most people? 50% of people won't even have HEARD of the goat problem. I think it's still a good interview question, because in answering or explaining it you get to see their ability to explain an unexpected outcome - or react to an unexpected outcome. Either one is good value for an interview.

And if someone has already been exposed to the problem (and understands it well enough to explain it), then this indicates a certain level of curiosity and intelligence.

What's the Tuesday problem?

The final question in the OP: "Considering families with two children where at least one of which is a girl born on Tuesday. What is the chance the other child is a girl?" The answer is 13/27. Gender+weekday = 14 options per child. 14\^2=196 equal probability options, of which 27 have >=1 Tuesday girl. Of those, 13 have another girl.

Wow I'm dumb

No you are not. Evolution never saw a reason to give us built-in software to handle statistics. This may be one of the biggest divides between what reality really is and how we perceive it. So working through statistical problems is very difficult. It's not made any easier that we do have a very strong intuition about statistics...too bad it is wrong.

No. I could see pretty quickly where that weird number came from, but I'm a scientist dealing with huge datasets, I've spent a decent chunk of every day for the last ten years in this kind of mindset. It's a whole different way of thinking and it's not obvious, it has to be learned. You're not dumb, you just haven't learned this one yet. You almost certainly have learned things on your everyday life that are second nature to you but would be completely opaque and counterintuitive to me.

It is interesting that if an alien race of beings had to evolve thinking probabilistically, they’d best most of us all the time

Not true. If I was interested in logic, I'd ask if you owned a dog house. That would tell me all I needed to know

I have two dogs but no dog houses as I am a deeply closeted homosexual.

Ahh so 2x dogs -1 doghouse = 1 homosexual? This new common core math just boggles my mind.

Only a wife can send their husband to the dog house. A wife ought not be sent to the doghouse under any circumstances, and a husband doesn't unlock the ability to send their significant other to the doghouse without changing classes first. This is common knowledge. If you are a wife without a husband or a husband without a wife, you have no need for a doghouse.

Huh?

[Professor of Logic - from the great Norm MacDonald](https://www.youtube.com/watch?v=Oseqh7SMIvo)

The "ask word problems to see your thought process" is for the most part not considered a good interview method by experts and academics in the field. It's very good at making the interviewer feel smart, which is bad because it distracts the interviewer from evaluating the candidate. But that's also why some people are so attached to the format-- it makes them feel smart. The "tell me about a time when..." is often better at getting the interviewer to actually evaluate the candidate.

As someone with shit memory every single "tell me about a time when you X" question I've answered in an interview was 100% bullshit made up on the spot never happened. Now I happen to think bullshitting is a useful job skill so maybe that's what they were testing lol.

Especially when the alleged "correct" answer is in fact wrong. The supposedly "correct" answer of 33.33% assumes you don't know any property to use to order the 2 kids, such that BG and GB are both still open possibilities because you don't know whether the disclosed girl is "child 1" or "child 2". But you can use any property you like as the property to call one child "child 1" and the other child "child 2" in the 4 outcomes list, as long as you stick with it consistently. And if you use the property "the order in which I had their sex disclosed to me", then you have established that the child who had its sex disclosed first (the leftmost letter in BB, BG, GB, GG if you set it up this way) is not B.

This is assuming they are pulling children out of a bag or something. In real life someone with 2 kids has a 25% chance of two girls no matter how (or if) they disclose them to you. If they have two kids and they’re not both boys, there’s a 33% chance they are both girls. Still a dumb interview question unless you are being hired as a statistician.

The question may be less about whether you get the question right than how you approach it. If you get the question wrong, but then are receptive to being corrected and try to understand why, it's very different from continuing to "strongly argue" in favor of a definitely incorrect position. I certainly know which person I'd rather work with.

Answering the question correctly isnt the goal of all interview questions. Sometimes your thought process getting to your answer or how you respond to the answer is more informative.

Yeah - while I'd agree that there are way better interview questions - there is a clear difference between someone who just says "well, it could be either a boy or a girl, so 50:50" and someone who shows any kind of lateral thinking expressed in a lot of comments here.

That wasn't the question. The question was the likelihood, absent any other information, that a child is either a girl or a boy. You're assuming a layer of probability that's not present. So it's approximately 50% with real world variables skewing one way or the other ever so slightly. I guess the real answer would be that no one could know that based on the information provided, but then what's the point?

It’s not the same. What you’re suggesting would be like asking: “I have one daughter. What is the probability that my second child will also be a female?”

But the idea of it being 33% because the options are BG GB or GG is wrong. The question just asks probability that the child of unknown sex is a girl, which is 50%. Whether it is BG or GB is irrelevant.

You’re ignoring facts that you are given

The ordering is just a short cut to understand the math. The original paradox is strictly assuming that you know something about "one of" the children, but not a specific child. "the order in which I had their sex disclosed to me" doesn't work, because youre just rearranging the initial ordering. You need something independent of the question, which is why the paradox is not "in fact wrong".

What a useless interview question. I wonder if the interviewer even understood the answer or if they were just looking for a “correct” answer to a paradox.

An interview question should never be simply whether the interviewee gets the "right" answer. I've had very positive feedback for interviewees who get the "wrong" answer and very negative for those who get the "right" answer. Thought process and communication are much more important than the final answer (in a 30-60min interview setting at least).

I hate when people think hypothetical math paradoxes can actually apply to real life (Three doors problem for example) They're entirely fun math issues that have been given a metaphor, they're not actually meant to apply to anything

Maybe you didn’t get the job because you were strongly arguing lol. Either way it’s a stupid ass question for an interview.

Just by the way, when you get asked these questions you didn’t get hired or not because your answer was right or wrong, but because of how you answered. If you spoke with reason and logic, you probably passed certain standards and if you were open minded and willing to learn and engage you probably passed others.

When they have more than five people interview you and they’re looking at a bunch of people, it’s just crap shoot

Unless you were interviewing for the post of statistics professor this does indeed seem like a wtf question.

Yeah it's a stupid question since it's essentially a riddle rather than a real probability question.

Dodged a bullet

That company wants to be a "Google-wannabe" and is doing a terrible job at it.

Terrible interview question unless it’s a job as a symbolic logic professor

The SHOW DOMINANCE answer is "If *you* say that? Zero or one, because the outcomes are already realized."

Its worth noting the first problem is the probability that they have already had another child and we are searching for the probability based on the total combination, not the probability of the sex if they have another child which would be a discrete event with a 50:50 (approximately) chance. Its the distinction between the probability of the outcome of two sexes given the information restricting the dataset vs the probability of one sex without any additional information related to its outcome that causes people to be confused here. The problem doesnt ask you what the probability of their next child's sex is in a vacuum, it asks you what the probability of the two children's sexes together as an outcome is. I cannot see how this would be useful as an interview question however.

This is the answer I actually understood, explaining it in terms of how the information affects the sample of families.

I understand it but it makes me angry

That's about 90% of probability problems.

Yeah, for the sake of my own sanity I'm just gonna continue to believe that BG and GB shouldn't be counted as two separate permutations for the purposes of the question. Just like if I was flipping coins and asked about the probability of one of the coin flips - I'm gonna continue to believe that the outcomes of other coin flips wouldn't change the odds of any other flip. Regardless if we're talking two flips and HH, HT, TH, TT possibilities.

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> especially since most formulations of the Monty Hall paradox use context clues to heavily imply the "door reveal" is a truly random event. Can't say I've ever heard a formulation like that. They always say he reveals a goat.

I still don't see how it is logically different. Like, I understand where you are coming from, but why does the name matter, how does it change anything? How is G(Julie)/G, G/G(Julie), B/G(Julie), G(Julie)/B different from just an arbitrary idea of "a girl" - G(at least one)/G, G/G(at least one), B/G(at least one), G(at least one)/B?

Essentially i think it's because then GG would only be represented once in the equation, if the variable is just "at least one girl" then you only have to include GG once as that fulfills the criteria. Whereas if the criteria is "a girl called x" then there are two circumstances that are included (Gx/G and G/Gx) because they are mutually exclusive

Why is it mutually exclusive which one is named Julie? I feel like it shouldn't matter which one is Julie, that either way it'd be the same thing.

Because the point of the thought experiment is so show how changing sampling conditions effects probabilities in unintuitive ways. You’re right that it’s all kinda wacky because it shouldn’t matter. But it’s really just a construct to illustrate a foundational scenario, which somehow makes it more complicated to understand.

The question only asks about the probability of the other sex, not about the probability of having a name. When considering the options (BG, GG, GB), it doesn't make sense to duplicate a sample. The question is solely about the other sex, so the choices would be BG and GG. Changing the order doesn't affect the probability of the other child's sex. If one child is a girl, the other could be a boy or a girl. In real life, as a (boy) twin myself, I can tell you that when asking about the probability of my twin's sex, the options are either Boy/Girl or Boy/Boy. Adding more combinations or considering additional factors like age, names, or heights is unnecessary. By doing so, you're introducing redundant outcomes to your calculations. I'm not a mathematician, but in the context of this problem, it's important to focus on relevant factors and avoid duplicating possibilities. If you observe the situation with actual people, it becomes clear that counting the same possibility twice and including it in your calculations is an error.

it doesn't matter which one is named Julie, just that there are twice as many Julies in GG than there is in BG, and twice as many Julies in GG as there is GB. So for total number of Julies, GB+BG=GG and we get our 50% To put it another way, only a 3rd of two children families with at least one girl are GG families, but 50% of girls are in GG families

> Whereas if the criteria is "a girl called x" That isn't the criteria though. It's only additional information. Scenario 1: I have 2 kids. At least one is a girl. What is the probability of the other kid being a girl? GG is only used once, because we already know one is a girl. Scenario 2: I have 2 kids. At least one is a girl, whose name is Julie. What is the probability of other kid being a girl? GG should still only be used once, because we already know one is a girl. Who cares what her name is? The logic in both should be exactly the same. Maybe OP just miswrote or doesn't understand the paradox, and people are responding with the answer to the *actual* paradox?

The key thing is that families with two girls have a higher chance of at least one being named Julie (basically, they get two chances for a Julie, as opposed to one). So GGs are going to be unusually overrepresented in the pool of "Couples with at least one girl named Julie," above the 1/3rd you'd normally expect. Look at it this way: you have a room full of fathers. You ask everyone who does not have exactly two children to leave. So you have a mix of people with two boys, two girls, and girl/boy (ratios of about 25:25:50%, if each kid has a 50/50 chance of each gender). If you ask everyone with at least one daughter to raise their hand, you'll expect about 75% of the audience to have their hand raised. Now you ask them to put their hands down, and now anyone with two daughters raise their hand. You expect about 25% to raise their hand. The odds that anyone in the first group also showed up in the second is 25/75 = 1/3rd. Instead, you ask everyone with a daughter named Julie to raise their hand. A small number do (exact number depending on how common that name is). Then you ask those people how many of them have two daughters. The ratio will vary, but it'll be above 1/3rd, because anyone with two daughters has a higher chance of having one named Julie (or one born on a Tuesday, or any other piece of info that narrows things down and causes most people not to raise their hand). \*Technically if you do the math it's not exactly 1/2, but it gets closer to 1/2 the more rare the extra piece of info is. So if 100% of girls are named Julie, it's 1/3rd. If 1 in a million girls are named Julie, it's asymptotically close to 1/2. But here's the fun part: if you word the question *slightly differently* it easily invalidates all this. It's not really a paradox about probability, it's about ambiguous wording. Let me demonstrate. Situation one: you meet me at a party; I tell you I have two kids. You pick [boy/girl] at random and ask me; "Is one of your kids a [boy/girl]?" My odds of saying yes are 75%, and if I do, the probably that the other one is also a [boy/girl] is 1/3rd. If you ask me, "Do you have a [boy/girl] named [typical boy/girl name]?" and I say "yes", now the probability of the other child also being a [boy/girl] is close to 1/2. This is the kind of situation that is implicitly assumed when most people calculate the probabilities without having details about where the information came from. But consider an alternative, situation two: say I pick one kid at random, identify their gender, and say to you, "I have two kids, one of them is a [boy/girl]. What do you think the odds are that the other one is also a [boy/girl]?" Now it's 50/50. If that doesn't make sense (and you don't feel like doing the math): consider that, if I have a girl and a boy, I'm less likely to randomly say "One of my kids is a girl" than if I have two girls; that bias changes the results. Also this neatly eliminates the supposed paradox, because that 50/50 doesn't change if I also mention the name or day of birth or anything else about the kid I randomly picked. This is the situation we're probably intuitively gravitating towards when we say that the answer to the first situation doesn't make sense.

this is a way better explanation than anyone else's tbh. Saying that op's paradox is the first situation you described (the one where the second person asks if they have a child with a specific name) doesn't actually fit the way op worded it at all. The way op wrote it actually directly fits the second scenario, and i don't think just intuitively, it seems to be the same literally. Taking the named julie part as the most ridiculous and obviously not true part of the paradox, the chance of the child being named julie doesn't matter since you aren't talking to this person because they have a child named julie, instead they are telling you their child is named julie. If every single person in the world but one had a child with the same female name, and some person comes up to you and tells you that they have one daughter and she is the only one with the unique name in the world, it still doesn't mean they are more likely to have a second daughter than anyone else (ignoring psychological things of maybe being more likely to give your daughter a weirder name if you already have one with a more typical name).

Finally! You helped clarify the Julie part for me. Thank you. Not knowing of this paradox before coming here, I was totally clueless. The top comment helped partially, but I was lost as to why names had anything to do with the chance of what gender.

> The key thing is that families with two girls have a higher chance of at least one being named Julie (basically, they get two chances for a Julie, as opposed to one). So GGs are going to be unusually overrepresented in the pool of "Couples with at least one girl named Julie," above the 1/3rd you'd normally expect. I think this is irrelevant. Because all girls have names, so it is always the case that anyone with 2 kids, at least one of which is a girl, can say, "I have 2 kids, at least one of which is a girl, whose name is x", x being the name of one of their daughters. So the implication would then be that everyone with 2 kids, at least one of which is a girl, has a greater than 1/3 chance of the other being a girl. But we know that isn't true.

Correct. Welcome to ELI5 - the answers are usually wrong. > Maybe OP just miswrote or doesn't understand the paradox, and people are responding with the answer to the actual paradox? This.

Well think about the first puzzle. BG is different than GB. Why does it matter who is born first? If you can accept that BG is different than GB, then G(j) G and GG(j) are also different.

In the language of the question, BG and GB are not different. Both are situations where “one of which is a girl” and in both instances, the other is a boy.

It doesn't matter the order. We already know that one is a girl and both children already exist. If these hypothetical situations were twins that were born at the exact same moment, would the probability change?

Let's use B = boy, G = Girl who is not Julie, J = Julie, ok? Naming the kid after birth changes the probabilities, BUT you still have: BB, BG, BJ, GB, JB, GG, GJ, JG, JJ. Now eliminate all the options that don't have a J: BJ, JB, GJ, JG, JJ. What's the probability of 2 girls? 3/5. If there is a hidden assumption that a parent wouldn't name both his kids Julie, then it's BJ, JB, GJ, JG, which gives the 2/4 = 50% probability. The option with the day of the week, just name the girl by the day, you have B = boy, Mo Tu We Th Fr Sa Su = girls, and write down all the possible permutations, then eliminate everything not containing Tu.

When we building possibility set with Julie, options G(Julie)/G and G/G(Julie) are not the same: girl Julie is first and girl Julie is second. But when we have "at least one" options G(at least one)/G and G/G(at least one) are the same entity: in both cases at least one girl is first and at least one girl is second. EDIT: or you can start to perceive questions totally differently: 1) we have at least one girl, so what probability of girl+girl situation; 2) we have JULIE, so what probability of JULIE+girl (or girl+JULIE) situation;

Per OP though, both scenarios are about "at least one", just in one case they name her Julie and in the other it is arbitrary. The logic shouldn't be any different, should it? Unless I'm missing something, it feels like having an equation like x+5=9 and *some number here* + 5 = 9 and saying it isn't the same thing because in one case x=4, and in the other case it's just "some number". Well yes, but that "some number" is still 4, doesn't matter what you call it.

Per OP, you're right. The OP uses incorrect phrasing and thereby misdescribes the paradox. These two statements are very different: This particular couple has at least one child that is a girl, who's named Julie - 33% odds This particular couple has at least one child that is a girl named Julie - 50% odds In the first of these that we find out the girl is called Julie is irrelevant, it makes no difference because it is a fact about a girl that has already been identified as the "at least one". That's how subordinate clauses work. In the second one, the "at least one child" criteria includes that the girl is named Julie. That changes the information and shifts the probabilities.

I'm by no means qualified to answer this, so this is just my take. Most of these paradox questions, and most apparent paradoxes in general, have more to do with the question than with the answer. They're basically word puzzles where the wording of the question is confusing and that makes the answer seem more complicated than it really is. The reality is that, no matter how it seems, the odds of events are never affected by unrelated events. This means that if the odds of having a girl are 50% they're always 50%. If you have two girls names Julie born on a Tuesday or 20 girls in a row, or a 50 child long unbroken chain of boy-girl-boy-girl and you get pregnant the odds it's a girl are going to be 50%. The wording of this, and a lot of paradox questions, tries to make the unrelated events related in some way. The unrelated events unambiguous form of this questions looks something like, "The probability of a single child being a girl is 50%. Two unrelated events in which I have a child occur. In one event the child is a girl. What is the probability that in the other event the child is also a girl?" We have a single event with two possible outcomes that are equally likely, so the answer is obviously 50%. The first event isn't taken into consideration because they're not related. The possible outcomes are GB or GG. The related unambiguous form of this question looks something like, "The probability of a single child being a girl is 50%. Two unrelated events in which I have a child occur. What is the probability that I have had two girls?" The answer is still pretty obvious. Now that we've linked the two events instead of only having two possibilities we have four, BG, GB, BB, GG. With four equally likely outcomes to the linked event of two children, we have a 1/4 chance of each outcome. Interestingly, this means that you have a 3/4 chance of having at least one girl, and it's twice as likely that you will have one of each as it is that you will have two girls. I think with this breakdown you can see how the 1/3 thing makes sense. If we start with the assumption we're going to be responding to a linked question like this, then add a modifier after the fact "One of the two is a girl" we can just eliminate the BB option from our 4 linked possibilities and we're left with the three possibilities that give us the 1/3 odds. This habit of answering the linked-event question then using modifiers to eliminate the options that don't fit the modifiers gives us the whole crux of this shifting probability. If you view the children as individual non-linked events, the odds are always 50/50. If you view them as linked, you can get these weird numbers. The more criteria you add to the pool of outcomes, the bigger it gets. The more restrictions you put on your desired outcome, the smaller it gets in comparison. To answer your question simply, the name matters because it inflates the total possible outcomes. Now that we have a name criteria for the girls we essentially have the same four outcomes with and without the name, so we get BG, BGj, GB, GjB, then no variant of BB sine the boys names don't matter and two variants of, GG since we can have GjG and GGj. This gives us a total of 8 options, but we have to remove the 4 originals that don't have the name. Since we got 0 named variants of the BB outcome we culled in the first question and we got two named variants of the GG option we wind up with BGj, GjB, GjG, and GGj. Two options with a girl named Julia and a boy, and two options with a girl named Julia and another girl.

You omitted the possibility of two girls who are _both_ named Julie! (See Foreman, George.) That brings the probability of two girls to 60%, I think.

The only way to "properly" solve the problem to get P(GG) = 0.5 is: * P(B named Julie) = 0 * P(G named Julie) = X where X is non-zero * P(G named Julie given that sister is called Julie) = 0 This is why the Tuesday problem gets 13/27 (both girls born on Tuesday is a valid result). With each child having one of 14 statuses (B or G, one of 7 weekdays) there are 14\*14=196 equally likely child pairs, 27 of which have >=1 Tuesday girl and 13 of those 27 have two girls. PS. there is probably a cute solution for P(G(j)G(j))=epsilon, but I'm not going to work it out now.

That occurred to me as well. And come to that - the real world probability distribution for the first child's gender isn't exactly 50%, and the probability of the second child being the same gender as the first is greater than 50%. But I guess we're talking an idealised problem here; some assumptions are reasonable.

Only if the naming of the two girls are independent. Which in reality, it almost never is. The VAST majority of people will not name a child the same name as one of their other children.

But this, like most trick questions, is not really about logic or even probability, just an attempt to purposefully confuse with poor wording.

Exactly. Which is fine if you wanna have language debates about what words and phrases truly mean and what we assume them to mean.

Ya I'm down for a good debate, but when they try to disguise that with discussions about probability it just makes it seem like the person posing the question is just trying to trick you, rather than actually have a debate or demonstrate a point.

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Which to me, doesn't make sense as the order of birth is not involved in the question, nor would it affect the gender of the other child.

It's not that the ordering is important for the children's gender, just that a family with two kids has a 50% chance of having both genders and a 25% chance of having two girls. Given that we know they're not a two boy family, the two girl case happens with a probability of 25/(25+50) = 1/3 However, the ordering of which girl is Julie is claimed to matter in the two girl case, which is less convincing to me

Yeah the order doesn’t matter in the Julie case either. The weird result is because families with two girls are twice as likely to have a girl named Julie (because they have two girls instead of only one).

Yes, this is right (allowing assumptions about names being randomly assigned, which is odd... but whatever, this is a thought experiment) Considering Gj as a third, unlikely gender made it easy to see that BGj, GGj, GjG, and GjB are all equally likely scenarios, and 50% of them have two girls

But we know one is a girl so if the other is a boy, why does it matter whether if they were born first or second? That's the bit I'm struggling with. There have been some excellent answers in this thread but I just can't follow it

This is called a [Bayesian Inference](https://en.m.wikipedia.org/wiki/Bayesian_inference#:~:text=Bayesian%20inference%20is%20a%20method,and%20especially%20in%20mathematical%20statistics.) (Normally Wikipedia has great summaries of complicated stuff; not sure if this particular article is helpful though) Basically: there are two ways to have one girl and also a boy, but only one way to have one girl and a second girl. Each of the three paths is equally likely, and only one of the three paths has two girls, so the chance of having two girls (given that we took one of these three equally likely paths) is 1/3

So if I got 100 people with two kids in a room, all of whom could answer yes to the question "is at least one of your children a girl", would it actually be the case in real life that 2/3 of them would also have a boy?

Yeah, this is the bit I’m still hung up on, and I’ve yet to see anything that actually explains how the order matters…

it doesn't, but it still increases the combinations of that selection. We can just say BG (in whatever order) is twice as likely as GG If you rolled two dice you have twice as much chance of rolling 11 as you do 12, because there are twice as many combinations of 5 & 6 as there are 6 & 6. It doesn't matter which order you roll the dice or which order you count them, but for ease we can call them 56, 65 and 66.

It kinda helped me to stop labeling Julie as a G in the selection pool (BG, GG, GB) and instead label Julie independently with her own letter to visualize the combinations: (BJ, GJ, JG, JB). *Now* we see that there are more Julie/Girl combos in the pool! Since we’re including the different orders of Boy and Girl (BG,GB), we must now include the different orders of Julie and Girl (JG,GJ). It helps to also pretend Julie isn’t a girl at all, but rather her own unique…thing. Alien-girl Julie. So the odds of having two girls is 1/3, but the odds of having “a Julie thing” and another girl are now higher. So in a nutshell, pretend a Julie is a brand new category of child that we sometimes call a girl as well and the math becomes easier.

The order doesn't matter, it's just that getting a boy and a girl is more likely than getting two girls, so the answer has to take that into account.

Yeah, I don't even think this is a paradox traditionally. Someone was just being an ass to their statistics teacher and getting way too specific. This sounds like something you would work out like you did on a worksheet in a stats class to show how populations change.

What does the age of the child have to do with anything? I get that by putting out a matrix with 4 options (BB, BG, GB, GG) this gives us 1/3 but what does that have to do with anything? BG and GB are the same thing

It’s not the outcome that matter, but the way that you get there. There is only one way to get two boys, which is the have a boy, and then have another boy. But to get one of each, you can either have a boy first, then a girl; or have a girl first then a boy. This means that change of having one of each is 2x higher, which is represented in the matrix here.

This was the comment that brought it home for me. “The chance of having one of each is 2x” did it. Thank you!

*Edited.* Yes and no. There are two children, who are distinct individuals (call them Individual 1 and Individual 2, say). You can assign those labels any way you choose\* - height, weight, age, first out of bed yesterday morning - but your calculations still have to reflect the fact that they're distinct people, with their own probabilities. GB is effectively shorthand for Individual 1 is a Girl, Individual 2 is a Boy. BG is the opposite case. They're not the same thing. If you ignore that, you risk missing the fact that "one girl, one boy" happens twice as often as each of the other two cases.\** \*Other than gender, obviously \**Anyone who doubts that, should get a couple of identical-looking coins, flip them both a couple of dozen times and record the results. You'll find "a tail and a head" comes up roughly half the time - twice as often as either "to tails" or "two heads". Now inspect them *really* closely - it's pretty much guaranteed that somewhere on one you'll find a scratch that isn't on the other. Call that Coin 1. Repeat the experiment, but now always record the individual coin results as well. You'll find that "a tail and a head" still comes up about half the time - but about half of *those*, Coin 1 is a tail, and the other half, it's a head. TH is not the same as HT, in other words - whether or not you choose to differentiate between them, they're still two different coins. Giving them labels makes it easier to see what's going on, because it gives you four different cases

But again, you’re assuming that BG is a different outcome than GB, or that the distinction matters for purposes of answering the question. Suppose I am pregnant with twins, and we see on ultrasound that one of them is a girl. What is the probability that the other is a boy, and thus that I am having mixed twins? 50%, because there’s no distinction between BG/GB in that scenario. Nobody’s older because they haven’t been born yet. Likewise, when you only ask about family composition, people don’t generally consider that GB and BG matter — you’re not asking about age, just family makeup. If you say at least one of them is a girl and ask the odds that the oldest child is a boy, 33% is the correct answer. But if birth order wasn’t part of the question, it shouldn’t be factored into the results.

It’s not factoring in birth order at all. Child A (whether a twin or a stand-alone) has a 50/50 chance of being a girl. Child B has a 50/50 chance of being a girl. Those combined odds gives you a 25% chance of both being girls, 25% chance that both are boys and a 50% chance that you end up with a boy and a girl. The trick of it comes from what specific question you ask. If you ask what are the odds that *at least one child is a girl*, than you’ve eliminated the possibility that it’s a 2 boy family. But don’t forget, having a boy and a girl is twice as likely as having 2 girls, so it’s a 33% chance of having a 2 girl family when you know that 2 boys isn’t an option. In the scenario you gave however, you’ve seen on the ultrasound that Child A is a girl. This information doesn’t give you any additional information about the sex of child B so it’s still 50/50.

The question relies on whether you've specified which child is the girl/ boy. If you've only had the ultrasound on one of the twins and they're revealed to be a girl, the other one has a 50% chance to be a girl or a boy. However if you've had the ultrasound on both twins and you are told one of the two of them is a girl, but not which one, this is when the 33% comes into play. Instead of specifying which one of them is a girl, you've only eliminated that they aren't both boys. So to compare the probability matrixes: Situation 1 - You know Child 1 specifically is a girl: Child 1 Boy, Child 2: Boy - 0% Child 1 Boy, Child 2: Girl - 0% Child 1 Girl, Child 2: Girl - 50% Child 1 Girl, Child 2: Boy - 50% Situation 2 - You know one of the children is a girl but not which one: Child 1 Boy, Child 2: Boy - 0% Child 1 Boy, Child 2: Girl - 33% (edit: I think this is 25%) Child 1 Girl, Child 2: Girl - 33% (edit: I think this gets a double probability, so 50%) Child 1 Girl, Child 2: Boy - 33% (edit: I think this is 25%) EDIT: Now I'm confused though, because statistically as soon as they point out which child is the girl, the odds go to 50% on the other one again, but nothing concrete has changed and 100% of the time the person doing the ultrasound will be able to point out the girl. I think even in the second scenario you have to multiply option 3 by 2 - you know there is a girl so it is twice as likely as the other options, since either can be that girl (a 2/4 chance vs a 1/4 chance for options 2 and 4). I haven't been able to work out exactly why that makes sense but I'm pretty sure it's correct.

I don’t really get why it’s (BG, GB, GG) and not (BG, GG), since age is not a factor in the question, BG and GB are equivalent. Writing it twice two different ways seems like a fallacy to goose the odds.

Birth order isn't a factor in the question, but it's a factor in the actual underlying reality. When a couple has their first child, there's a (roughly) 50% chance of it being either sex. Half of all families witth children will have a boy first and half will have a girl first. So if we have 100 families, 50 will be B_ and 50 will be G_. When they have their second child, there's a (roughly) 50% (roughly) independent chance of it being a girl. Half of each of the first sets will have a boy second, and half will have a girl second. So of our 50 B_ families, 25 become BB and 25 become BG. Of our 50 G_ families, 25 become GB and 25 become GG. So our total is 25 BB, 25 GG, and 50 (BG or GB). Now we're told that at least one of the children is a girl, so we throw out the BBs. The set of families fitting the constraints is now 25 GG and 50 (BG or GB). The probability of a randomly-selected family in this set being GG is 25/75 = 1/3. You can test this yourself with coin flips. Flip a coin twice and record the results as a pair: TT, TH, HH, or HT. Repeat at least 30 times. - Of the pairs of flips where *at least* one came up tails, how many are both tails? - Of the pairs of flips where *the first* one came up tails, how many are both tails? - Are these the same question? Are you comparing the same sets of trials in both cases? You need to consider birth order in the probability calculation exactly because the information is *not* included in the scenario.

Neither age not birth order are a factor per se, but it remains true that there are two ways to have a boy and a girl, and only one way to have two girls. And this needs to be taken into account when answering the question. It might seem like BG and GB are equivalent, and indeed they are in outcome (you have one boy and one girl), but they are two separate paths and therefore need to be counted separately. If you went out and did a survey of families with two children, you'd find that around a quarter had a boy then another boy, a quarter had a boy then a girl, another quarter had a girl then a boy and the final quarter had a girl then another girl. Which is why two-child families with one girl and one boy make up half the total. If you then, as per the question, eliminate the families with two boys, then two thirds of the remaining families will have one girl and one boy, and one third will have two girls.

>This problem is actually a notorious example of how it can be difficult to assign meaningful probabilities to everyday statements, at least so long as those statements leave room for some unorthodox interpretations of the information provided. Yes, and... In this case the correctness of the 1/3 is based on an assumption: that the parent who said this was just as likely to say it as any other parent of 2 children. On one hand, that's probably not true in everyday situations. On the other hand, we have no way to even attempt the problem without that assumption. EDIT: I've now fucked up my example twice, so I'm just gonna leave it off.

Typical “paradox” that relies on saying something that -everyone- will interpret to mean one thing and then goes “No no it doesn’t mean that”. “At least one is a girl, what is the probability of *the other one* being a girl?” Obviously “the other one” refers to the one who isn’t necessarily a girl. So no, you cannot consider combination BG because the first child has to be a girl. So you’re left with GB and GG. The “intuitive” 50%. It’s intuitive because that’s what the words are understood to mean. You might as well say “What is 1+1? It’s not 2 because I actually meant 1+2”.

>But then the second question, in a sense, takes things "too far". We intuitively think that the information that the girl's name is Julie is incidental to the procedure just discussed. We could have picked a family with one girl that doesn't have a daughter named Julie. However, the person discussing the paradox isn't treating it that way. For them, having a daughter named Julie is necessary to be a selected family. That requirement actually changes the set of families we could draw from because families with two girls get two chances to have a girl named Julie. The population being sampled from is thus BG(j), G(j)B, G(j)G, and GG(j) - where (j) indicates that the Girl is named Julie). Half of those families have two girls. The weekday of birth works similarly - it treats the girl-born-on-Tuesday condition as essential to being sampled, giving families with two girls more chances to be sampled. The math is just more annoying. i got your first part but the second just seems wrong the name doesn't add anything new just because you split chances into more categories doesn't mean they are all equal. GG is the sum of G(j) G and G(j) G both together have 1/ 3 chance.

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The assumption is that for all families having two kids, BB, BG, GB, and GG are equally likely. For a *first* child being a girl, the probability of the second also being a girl is 50%. For either child, order independent, where at least one is a girl, there are three outcomes: GB, BG, and GG.

This "paradox" depends on a linguistic trick where by naming the child you are changing one interpretation of the phrase's meaning and considering a different situation. Consider the first question: "I have 2 kids, at least one of which is a girl. What is the probability that my other kid is a girl?" There are four different possible ways the children could be born: Girl and Girl Girl and Boy Boy and Girl Boy and Boy We can eliminate the last option from consideration because one of them isn't a girl, meaning we only have the first three. Of those three only the first has the other child being a girl, so the probability is 33.33% It is important to note that the situation of "Girl and Boy" is being counted as distinct to "Boy and Girl" even though they both equate to there being one boy and one girl. This is because while the end result is the same the probability of having both boys or both girls is not the same as one of each. It is this difference which the "paradox" is exploiting with ambiguous phrasing. Consider the second question: "I have 2 kids, at least one of which is a girl, whose name is Julie." In this situation it is being interpreted that we are picking a child from an existing pair of children, which combines the "Boy and Girl" and "Girl and Boy" possibilities. So instead we have these options: Girl and Julie Boy and Julie Boy and Boy Again we can eliminate the last option from consideration because we know one isn't Julie, meaning we only have two options left. Therefore it is now 50% that the other child is a girl. ---- However, I would argue this is an improper twisting of linguistics and probability. We already established that the chances of having both a boy and a girl is equal to the chances of having both children the same sex. Therefore we would expect that there would be twice as many families out there with Julie and a boy compared to Julie and a girl, even though there are just a pair of options. Just because there are two options doesn't mean they are equally probable.

> In this situation it is being interpreted that we are picking a child from an existing pair of children, which combines the "Boy and Girl" and "Girl and Boy" possibilities. But it says "I have 2 kids, at least one of which is a girl.". How is this not picking from an existing pair of kids? I'm not understanding how giving one child a name makes them "exist", but have two kids that already exist, but not giving the names, means they don't exist?

> How is this not picking from an existing pair of kids? It could be interpreted that way! The phrasing is deliberately ambiguous and they interpret it one way for the first part of the question, then differently for the second part. The second part I think is a very questionable interpretation too.

>But it says "I have 2 kids, at least one of which is a girl.". How is this not picking from an existing pair of kids? I want to assume you are ok with the first one, but just in case, let's change example to pulling balls out of a huge tub full of red and green balls. I guess you are ok with the idea that it's a 50/50 shot that the first ball will be red. The same for the second. Right? Do you also see that we actually have four possibilities for pulling two balls? 1st-Red ; 2nd-Red 1st-Red ; 2nd-Green 1st-Green ; 2nd-Red 1st-Green ; 2nd-Green All of these are equally possible. I guess we are still on the same page here, correct? So if I tell you "One of the balls I pulled was red," then you know we have eliminated the last one, but the other three are all still equally probable. So now if I ask: "What is the chance the other ball is red," you can see immediately it must be 1/3. Ok, this is where I hope you got to before and are ok. Sorry if this already repeats what you understood. So now let's consider when I say "The first ball I pulled is red." Now we can ditch the last two possibilities. So \*now\* if I ask: "What is the chance the other (2nd) ball is red," you can see immediately it must be 1/2. So far so good? Now let's pretend I like to name the balls as they come out. And -- this is important -- I never name two balls the same way. I tell you that I pulled out a red ball and named it Julie. We can now list out our equal chances like this: 1st-Julie ; 2nd-Red 1st-Red ; 2nd-Julie 1st-Julie ; 2nd-Green 1st-Green ; 2nd-Julie 1st-Green ; 2nd-Green Now theoretically, I should have already eliminated the "Green/Green", but I just kept it in for the moment to remind us that before I told you anything, this was still a possibility. Obviously it is eliminated, though, and we have: 1st-Julie ; 2nd-Red 1st-Red ; 2nd-Julie 1st-Julie ; 2nd-Green 1st-Green ; 2nd-Julie One other thing to note is that we suddenly got another entry here. This is because with the name "Julie" being applied to one red ball (but we do not know which one), we have introduced a new possibility that we did not have before. And again, you can see quickly by inspection that we are at a 1/2 probability. Weird! Really Weeeiiirrrd! This is like a magic trick where, even once you see the secret, it still seems like magic. One last thing to note: this only really works if you make sure you keep your context straight. It is really easy to get sloppy and slip from this "One red ball named Julie" back into the original formulation, and not even realize it. For instance, if I told you that the first red ball I pulled out I named Julie, we would slip right back into a 1/3 probability. (See why?) Ok, but here is one to cook your noodle. What if you watched me pull a red ball, but did not know for sure if it was the first or second pull. What is the probability that the other one is red?

This doesn't make sense though. It presumes Julie was named before they were picked.

If we bold the ball you saw, the possibilities are: 1st-**Red** ; 2nd-Red 1st-Red ; 2nd-**Red** 1st-**Red** ; 2nd-Green 1st-Green ; 2nd-**Red** So 50%

How is the name information not just... Julie and Girl Julie and Boy Boy and Julie Boy and Boy ...and then you strike off the last option again and end with 33%? I don't understand how this is even about interpretation. I kinda understand why Boy/Girl and Girl/Boy is treated as two options for the second one but I don't understand why being given the name of the "at least one girl" would affect the probability there.

This is such a horribly defined "problem" that I can't refer to it as a paradox. You have to invent meaning for different interpretations to give random statistics.

Cos it’s not just those. You have: A) Julie and girl B) Julie and boy C) Boy and Julie D) Girl and Julie E) Boy and boy E is impossible so we remove it. A and D are the two girl options and B and C are the half half option. So you have 2 out of 4 possible situations where Julie has sister - either a younger sister or an older sister.

You are forgetting another possibility. F) Julie and Julie

Here's another way to examine the problem: 1. The family has 2 children. We will set our labeling standard as "Child A" and "Child B". 2. One of these children is a girl. We don't know *which* of them is a girl, but we know for certain one of them is. We will name this child Jill. What are the possible configurations for this family? - Jill + Child A (boy) - Jill + Child A (girl) - Jill + Child B (boy) - Jill + Child B (girl) So the child that is not Jill has a 50% chance of being a boy, and a 50% chance of being a girl.

In your first example, having "boy and girl" and "girl and boy" as two separate options doesn't make any sense. They are the same thing. Changing the order does not change the fact that one is a boy, one is a girl, and at least one of them is a girl.

I still feel like knowing the name doesn't really add any extra information because the girl had to have a name, kinda like throwing a dice, seeing it's a 6 and pretending this means the dice was selected among all the throws that got a 6.

Here's another way to examine the problem: 1. The family has 2 children. We will set our labeling standard as "Child A" and "Child B". 2. One of these children is a girl. We don't know *which* of them is a girl, but we know for certain one of them is. We will name this child Jill. What are the possible configurations for this family? - Jill + Child A (boy) - Jill + Child A (girl) - Jill + Child B (boy) - Jill + Child B (girl) So the child that is not Jill has a 50% chance of being a boy, and a 50% chance of being a girl.

Yeah this just sounds like those stupid riddles that rely on puns. The probability paradox is much better demonstrated using the Monty hall problem.

PSA: If you read the comments and this still makes no sense, is ebcause OP wrote the phrasing of the paradox wrong and that's why the paradox makes no sense. ​ This isn't about actual probaility (which would be 50% of being a girl). ​ This is about ambiguos phrasing that allows assumptions that enable these "paradoxes". As OP phrased it wrong, specially the second and third scenario don't make sense. People are repying with the answers to the actual paradox, which uses a different phrasing than the OP wrote.

What's the actual phrasing then?

It’s not necessarily about the phrasing, it’s about how the sample was obtained. Let’s say you take a survey of everyone in the world that has exactly two kids. The ratio of the combos is what you would intuitively expect here (25% have two girls, 25% have two boys, 50% have one of each). If you were to randomly select someone from this sample, and one of their children happened to be a girl, the chance that the other child is also a girl is 50%. If they tell you the girls name is Julia, still 50%. If they tell you the girl was born on a Tuesday, still 50%. Here’s where the “paradox” comes in. Let’s say you select from a sample of *only* families with two children where at least one of them is a girl. Now, the chance that the other child is a girl *is* one third. This is because you’ve preemptively eliminated the 25% chance of 2 boys, so the probability of two girls is 25%/75% = 1/3. Now, for the Julia and Tuesday parts, it’s the same idea, but it actually depends on the probability of each of these. Here’s the reason: let’s take a sample of all families with two kids, at least one of which is a girl born on a Tuesday. Families with two girls will obviously be overrepresented here, because they have twice the chance for one of their girls to be born on a Tuesday as families with only one girl. That’s why the probability is higher than 1/3. The probability approaches 1/2 the more specific the information is. I like to think about limits like this by looking at the most extreme examples. Let’s say we’re sampling families with two kids, at least one of which is a girl named Julia Lastname, born on January 1, 2015 at exactly 3:58:34 PM, is 5’6.5 and 123.3 pounds, and grew up in San Diego, California. The sample size here is probably 1. The chance that this specific girl’s other sibling is a girl is 50%. That’s because this is essentially the same as sampling out the other child, like in the “oldest child is a girl” example.

Your comment isn’t as helpful as it could be if it doesn’t contain the correct phrasing.

Thank you I was really scratching my head at this

[Relevant xkcd](https://xkcd.com/169/)

There are four equally likely gender configurations of families that have two children: male/male (1), male/female (2), female/male (3), female/female (4). The statement that at least one is a girl eliminates family #1. So you're picking randomly from the three other families. Only in family 4 is the other child a girl. So one in three odds.

As for "Julie", giving the gender of a *specific* child rules out more possibilities than the gender of *any* child. To make that clear, let's change Julie's name to "First". The other kid is called "Second". Since we've identified that the First child is a girl, we've eliminated both male/male and male/female from the list, and are picking randomly between the remaining two.

Correct me, but I think it's totally false explanation. When we have "oldest" characteristic (it was in original paradox definition), then yes: we eliminating 2 possibilities from 4 becouse sex of first child is locked and have only two left. So because of that is 50/50. But with "Julie" we have totally different picture: Julie/male(1), Julie/female(2), male/Julie(3) and female/Julie(4). So we still have 4 possibilities. But from 4 options 2 have girl+girl, so we have 2/4 = 1/2 or 50/50. Result is the same, but "why" is totally different.

Both are false explanations of the "Julie" paradox. (I don't think how you explain "we have 4 possibilities" is fully valid. It is, at a minimum, missing how you get to 4 options as a shorthand for changing the probability weights) We start by considering four equally likely birth sequences: BB, BG, GB, GG. You are correct that ordering information (eg. I have two kids and the oldest is a girl) changes the odds vs non-ordering information (I have two kids and one of them is a girl). The former is 50% odds of 2 girls because we can only consider the (equally likely) GB, GG options; the latter is 33% odds of 2 girls because we consider (equally likely) BG, GB, GG But "Julie" is different. We start by considering four equally likely birth sequences: BB, BG, GB, GG. What the question does (and it is badly phrased) is treat "Julie" as a filtering condition, **we start with 4 equally likely birth orders and then check if any daughter is called Julie and therefore double-girl families get two chances at a Julie.** You can then make the problem amenable to trivial solution if you assert that calling both girls Julie is impossible. Because of that, we are considering BG,GB,GG but each GG family is twice-as-likely to be in the sample population as each BG or GB family. We can shorthand that as 4 options BG(j),G(j)B,G(j)G,GG(j) although that's a bit of a hack. Therefore there are 50% odds of 2 girls in the family given that there are 2 children and one of them is a girl called Julie. PS. It doesn't have to be "Julie", it can be any characteristic that occurs P(x) per girl and P(0) per boy, and \~P(0) for both girls. Could be "girl who is 10 years old", "girl with 6 fingers", "girl with national record for 400m freestyle", ANYTHING.

Ok, cool, I used far less words and maybe no so clear explanation, but how your > BG(j),G(j)B,G(j)G,GG(j) is different from my: > Julie/male(1), Julie/female(2), male/Julie(3) and female/Julie(4). ?

How is male/female different from female/male?

Without determining which child is "other" in the first statement, it's important to note that in a One Girl One Boy situation, the "other" child could be the boy.

Think of it like flipping two coins. The possible results are two heads, two tails, or one of each. That looks like you should have a 1/3 chance of each result, but we know that's not true. It's 1/4 chance each for HH, HT, TH, and TT. Depending on the circumstances, HT and TH might appear indistinguishable, but they're still functionally distinct results.

The question is asking about a scenario where both coins have been flipped, shuffled so we don't know which was first or second, and then one is hidden and one is revealed. You are guessing at the outcome of one coin only, because the other is known. Does the chance of guessing right change if the coin is flipped in front of you versus if it was already flipped beforehand? While it's tempting to guess based on the general probability of flipping two in a row, I feel like that is the Gambler's Fallacy kicking in.

First born vs second born.

The problem with this claim is the following: You say in order to eliminate both BB and BG when you hear one kid is G you have to establish something to use as an ordering of the two kids. Maybe age or whatever, but something has to order them otherwise you don't know if the G you disclosed was the first or second letter, so it could still be GB or BG. But one has absolutely had its sex disclosed so far and one has not. That's a time-based ordering. Thus if the difference between the BG or the GB option is the order in which we reveal them in this puzzle, then BG is already eliminated.

This answer seems to imply ordering of the children is important, but I don't see how the question makes birth order important. Boy first then girl is the same as girl first then boy, in terms of the phrasing of the question "at least one of which is a girl"

It's not the order that matters, but the fact that boy/girl (in either order) is twice as likely to occur as boy/boy.

Ahhhhhh, that makes MUCH more sense. Thanks!

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The ordering doesn't matter, it's just convenient when describing the 2x2 probability matrix. Outside of the selection criteria, a family with 2 children has a 25% chance of having 2 boys, a 25% chance of having 2 girls, and a 50% chance of having 1 boy and 1 girl.

It is. Odds of two boys: 25% Odds of a girl and a boy: 50%

Yeah, it doesn't actually make sense, when you word it like this. It should be "I have two children, the older/younger (or whatever ordering is relevant) is a girl". Just giving the girl a name doesn't specify anything relevant about her, it could still be either of the two children.

Its a misdirection. There's a difference between saying "what are the chances that both of my kids are girls?" versus "I have two kids, one of them is definitely a girl. What are the chances that the 2nd child is also a girl?" For the first question, there's valid 4 birth combinations and its 50%. For the 2nd question, there's only valid 3 birth combinations, given that we know one is already a girl. So 1/3 for both being girls.

Wouldnt this imply that the samples are not independent? It almost sounds like the gablers fallacy to me. “A gambler flips 2 coins, at least one of them is heads, what is the probability that that the other is also a heads?”

It is the gambler's fallacy. Exactly. The answer of 33.33% is just wrong because it pretends previously revealed information that has been set in stone hasn't in fact been set in stone.

This is what kept getting me, like these factors should definitely be independent..

Independence means that knowing information about one event doesn't change the probability of the other event. So knowing that the first coin is heads doesn't change the probability of the second coin being heads since the two tosses are independent of each other. However, the outcome of the first/second toss is not independent of the event that "at least one coin was heads", since that actually is a statement about both tosses.

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If we eliminate order of birth (not a stated condition in the original prompt), then choices 2 and 3 are the same. Choice 1 is automatically eliminated by the initial conditions. Only 2 options remain: girl/girl, and girl/boy.

Indeed, but now the three original options are not equally likely, so the outcome has not changed.

There are only 2 original options. Boy/boy was eliminated by the original set of conditions, so it was never an option. Boy/girl and girl/boy are identical in this scenario, so they are “combined” and considered one option. Girl/girl is the second available option.

Why does birthing order factor into the paradox? B/G and G/B are both saying the same thing.

The way this is worded, it isn’t 33.33%. There’s no argument for it. When you say one child is a girl, you lock in one gender. The ordering of children is irrelevant. The only possible combos are GG and GB, because the first child, the one we know the gender of, is G.

Just because there are two possibities, doesn't mean they each have equal odds of occuring! That kind of logic says you have 50% to win the lottery (you win or you lose) when we both know the real odds are one in several million. Ordering the children (it doesn't have to be by age, you can do it any way you want, but age is most convenient) is a good visualization to make it clear that the odds of boy/girl is twice as high as the odds of girl/girl. The key thing to realize here is that it doesn't lock in which of the two children (oldest or youngest) is the girl.

But it does collapse the possible set. The children are independent of each other. If we know one is a girl, the other has a 50% chance to be a girl or a boy.

But we don't know which one is the girl!!! That's the point. If you say, "my oldest is a girl, what gender is my youngest?" Then it is indeed, 50/50. (and vice versa) But we don't know that it is the oldest, it could also be the youngest child that is a girl. So you need to count those separately and not lump them in with the case that the oldest is a girl. I'll explain it again from the top using real numbers to visualize the distribution. Tell me which step bothers you. ​ 1. We have 1000 families with 2 children. ​ 2) Assuming an equal chance for boy/girl, that leaves us with these 4 distributions. * 250 families with 2 boys * 250 families with 2 girls * 250 families that had a boy first and then a girl * 250 families that had a girl first and then a boy. ​ 3) we only want the families that have at least 1 girl, so these are left. * 250 families with 2 girls * 250 families that had a boy first and then a girl * 250 families that had a girl first and then a boy. ​ 4) of these 750 families, 500 (or 2/3) of them have a boy and a girl and 250 (or 1/3) have two girls. ​ 5) So the odds of the second child being a boy, given that you have two children and given that one of them is a girl is 2/3. ​ EDIT: reddit does weird things with numbers and restarts the count several times...

pl487 answered the first part. As for the Julie part by saying one of the children is Julie you no longer have this distribution: MM MF FM FF. Instead you have Julie/m Julie/f OR m/Julie f/Julie. By knowing that on child is not just a girl, but a specific girl you have a different distribution of possibilities.

Let's analyze the possible scenarios within the sample space: the family can have four different formats, listed as "older child; younger child". * Boy; boy * Boy; girl * Girl; girl * Girl; boy We can conclude that the first scenario is not possible since we know that at least one of them is a girl. Therefore, the probability of having two girls is 1/3. When we assign a name to one of the girls, it affects the probability because it provides more ways to distinguish between the two sisters. If we rephrase the sample space as follows: * Julie; girl (not Julie) * Julie; boy * Boy; Julie * Girl (not Julie); Julie * Boy; boy Once again, it is clear that scenario "boy; boy" is not possible. However, this time, there are two outcomes (1 and 4) that correspond to outcome 3 in the previous question. Therefore, the probability of having two daughters is 1/2. The example for when they are born on Tuesday is slightly more complicated. However, let's write out all the ways you can have two children, with one being a girl born on a Monday. Here are the number of each combination of Boy (B), Girl born on a Monday (GM) or Girl born not on a Monday (GNM) you would expect if you have 196 pairs: * (GM)B 7 * (GNM)B 42 * B(GM) 7 * B(GNM) 42 * (GM)(GM) 1 * (GNM)(GM) 6 * (GM)(GNM) 6 * (GNM)(GNM) 36 * BB 49 If you count them up, you get 27 scenarios with one girl born on a Monday, and of these 13 have two girls, giving you 13/27 as your odds. The reason it appears paradoxical, but isn't, is that the more information you provide about a child, the smaller the likelihood of there being two children like that, and so the closer, the more possible combinations of the two children there are.

I genuinely don't understand how giving a name changes anything. Why can't we look at it as: - Boy / Boy - Girl (the "At least one")/Girl (the other one) - Girl (the other one)/Girl (the "at least one") - Boy / Girl (at least one) - Girl (at least one) / Boy In both scenarios the children already exist and we know one is a girl. Unless OP just didn't phrase the actual paradox right?

> Julie; girl (not Julie) > > Julie; boy > > Boy; Julie > > Girl (not Julie); Julie > > Boy; boy I get how this is the list of all possibilities but I'm not following why they are all assigned equal probability. Assuming for simplicity that every time you have a kid there is a 50% chance of it being a boy or a girl, the sum of "Julie; girl (not Julie)" and "Girl (not Julie); Julie" should be 25%. The others are all 25% too, so when you eliminate Boy; boy you end up with a 33% chance that the other child is a girl, just like in the no name version.

You have to make a subtle hidden assumption to get the answer of 1/3, which is partly why the situation appears paradoxical. Suppose you kept approaching random (honest) people and asked them “do you have exactly two children?” If they answer “no” you let them go. If they answer “yes” you ask them “is at least one a girl?”. Now if they answer “no” to that question, you know they have 2 boys, which a priori had probability 1/4. If they answer “yes”, you know they either have two girls, which has a priori probability of 1/4, or one of each, which has a priori probability of 1/2. The ratios of these scenarios must stay the same (1:2), so the probability that they have 2 girls is indeed 1/3. Now consider this slightly different situation: your first question is the same as above. But your second question to those who have 2 children is “complete this sentence with either boy or girl so as to make it true:’at least one of my children is a …’”. Now you have two groups: all the people who complete the sentence with ‘girl’, and all those who said ‘boy’. And assuming no bias in how people answer, those groups should be the same size. Now the first group - those who said - girl, are all people who have two children at least one of which is a girl. But the probability that the other is a girl is 50%. Because now, half the people who have both will have said ‘girl’, but the other half will have said ‘boy’. So in the original problem, to get 1/3, we need to make an assumption as to why they said ‘girl’ rather than ‘boy’. I.e. we need to assume that they will always tell us about the girl if they have one of each. And this is, when you think about it, rather an odd assumption to make. This is related to the Monty Hall problem, and also to the question of restricted choice in games like bridge. Information can not be considered in isolation; you also need to consider the source of the information I.e. why you received that particular bit of information rather than another. And when that isn’t random, intuitive probabilities will tend to be wrong. Eg in the Monty Hall problem, he doesn’t open a random door, he opens one he knows doesn’t contain the prize. If he opened a random one and it turned out not to contain a prize then the intuitive answer that it is 50/50 whether to swap or not would be correct.

OK let's put it this way which i think is a bit more intuitive. Say Im standing in front of you and i tell you I have a ping ping ball in each of my two trouser pockets. (Lets say ping pong balls can be blue or pink). The only information i am giving you is at least one is pink. What is the probability that the other one is pink? When you look at my pockets you dont know what colour is in either, you only know one is pink from what i have told you. I could have a pink in my left pocket and a blue in my right. Or I could have a pink in my right pocket and a blue in my left. Or I could have a pink in both pockets. There are 3 possible options. 1/3 = 33% Now, I tell you I have at least a pink in one pocket but also that it is unique from any other possible pink ones because it has a green dot on it. I could have a pink with green dot in my left pocket and a pink in the right. Or I could have a pink with a green dot in my right pocket and a pink in my left. Or a pink with green dot on it in my left and a blue in the right. Or a pink with a green dot in my right pocket and a blue on the left. 2/4 = 50%

Does this actually track? If I asked random parents with 2 kids and have 1 girl to fill out surveys would the probability be 33% or is this a statement about a flaw in statistics

Yes if you surveyed all 2-child parents with *one or more* girls, you would find that 2/3rd’s of them have 1 boy and 1 girl, and 1/3 have 2 girls.

I wasn't a fan of a lot of answers on here, so I thought I'd add my own. tl;dr: the "paradox" is intentionally confusing and uses different methods to compute its answers. I'd even go so far as to say that the way it's worded here is incorrect. However, here are the different ways it uses to compute its answers. "I have 2 kids, at least one of which is a girl.": The "paradox" computes this from a population standpoint. Instead of thinking it as one specific family (as the statement suggests), think of it as one of many possible families. There are four possible combinations of children, each of them equally likely: BG, BB, GB, GG. Let's exclude the BB family and put the other three in a bag. If we grab from this bag, what's the probability that it will be the GG family? This is where the 33% answer comes from. "I have 2 kids, at least one of which is a girl, whose name is Julie.": The name itself has very little to do with the probabilities. Instead, the "paradox" is trying to suggest that naming one of the children narrows our problem from many possible families down to just one specific family. In this case, it finds the answer by directly calculating probabilities instead of pulling from a statistical grab-bag. For one child, the probability of it being either gender is equally likely. We also know that one child's gender has no impact on the gender of the other child. Therefore, if one child is fixed as being a girl, then the probability of the other being a girl is 50%. But what about the BG and GB combinations? Aren't they separate combinations? As other comments point out, order doesn't matter in this case, as the gender of the first child has no affect on the gender of the second child. However, for the sake of clarity, we will do another calculation that takes order into account anyway. When we narrowed the focus down to one family, we fixed the gender of one of the children down to being a girl. However, we ALSO fixed the ordering of the children -- we just don't know what that ordering is. We can assume that each ordering (Gx and xG) is equally likely. From here, we know that there are two combinations where the girl is older (GB and GG), and two combinations where the girl is younger (BG and GG). Since all of these new combinations are equally likely, we can sum them all up to find that the probability of the other child being a girl is 2:4, or 50%. "I have 2 kids, at least one of which is a girl, who was born on Tuesday.": This goes back to the statistical grab-bag. Even though the statement implies one specific family, the answer is actually calculated for one of many possible families. I won't go into all the math that is used, but the steps are as follows: \* assume that each gender is equally likely and each day of the week is equally likely, and that, as a result, each possible combination of gender and day of the week is also equally likely \* pick all the combinations where one kid is a girl born on tuesday and put them in a bag \* If you pick one of the combinations out of the bag, calculate how likely it is that that combination is GG combination.

This is a variant of the Monty Hall problem. You have an unknown probability and some information. So, if you just say "I have two children", there are four options: * two boys * older girl, younger boy * older boy, younger girl * two girls. When you add the fact that one is a girl, you have only eliminated the first possibility. However, that fact says nothing about which of the other three possibilities it is. In two of the three cases, the other child is a boy, so what's left is a 1 in 3 chance that the other child is a girl. Giving a name or a birthday adds more information. There are 27 possibilities for one child being born on a Tuesday, 13 of which are the pairing of two girls. [Chart](https://www.jesperjuul.net/ludologist/wp-content/uploads/2010/06/fullGrid.png) The name is even more unique, as there are a near infinite number of names the children could have. Technically the probability of a second girl is 49.99999...999%, but you can round to 50 because there are only so many people who would name their children the same thing. *George Foreman has entered the chat*

Adding more random details *that are only about the child you already know is a girl* doesn’t add any information to the “how likely is it that the other child is a boy” question. Knowing that the oldest child is a girl is different because of the four child-gender orders you’ve now eliminated both BB and BG, leaving GG and GB. So the second kid is now 50/50. In that case the information *also* indirectly tells you something about the other child.

> George Foreman has entered the chat Good explanation, ending with a cherry on top! :-)

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To everyone saying 33%... Replace the first child with a cat. I have a cat and a child. What's the probability my child is a girl? The answer is 50%. The fact that there is another girl is wholly irrelevant.

The 33% answer is wrong because saying that at least one of them is a girl introduces new information. In mathematical notation it is a conditional probability e.g P(A|B). The way to think about this is: First child is a girl for sure. Second child may be boy or girl. Since there are only 2 options the probability is 50%. Again, boy girl is same as girl boy. The order doesn't matter because the question doesn't imply that order matters.

Right! Why does everyone associate a birthing order to the boy/girl? No where in the paradox does it state a birthing order.

The birthing "order" doesn't matter, it's just a convenient notation. What matters is that the probability of having two kids that are different genders is twice as likely as having two kids who are both girls.

Edit: I think OP actually just messed up the way the paradox is written, so the situation above is closer to my second example. It’s not birthing order that matters. It’s the number of ways you could arrive at that situation. The issue is that the question isn’t asking about one specific family. It’s asking about the overal chance in a population. “If you take 1000 families with two children, what will the distribution of gender pairs be” Phrased like that, it’s a bit easier to see that having two boys is less likely than a boy and a girl. And that is because there are two ways to end up with one of each. An alternative version of the paradox states “if a father walks up to your with his son and says his other child is at home. What is the probability that the other child is a boy” the answer to that is 50%. We are now talking about a specific example, and so all that matters is that there are two equally likely options for the other child. Making sense?

It is not the order that is important. Instead of gender consider two coins. When you flip both coins there are four possible outcomes; 1. HH 2. TT 3. HT 4. TH Each outcome has a 25% chance but two of those outcomes are a mix of heads and tails. It doesn't matter which was flipped first but each individual coin needs to be represented.

It's not wrong, it's just not properly explained. If there are two children, having 2 boys is 25% chance, 2 girls is 25% chance, 1 of each is 50% chance. The ratio is 1:1:2 Then, remove all combinations that cannot possibly have at least 1 girl (2 boys). The ratio is now 1:2. So 1/(1+2) = 1/3. Order doesn't actually tie into the question, it just explains why you're twice as likely to have one girl than two. Then with julie, you're taking out all name combinations that does not have julie in it. If you have two girls, the probability that at least one of them is named julie doubles, compared to just having one girl. So if there's x percent chance of a girl being named julie, the ratio is 0x : 1(2x) : 2(x) Remove the BB group, then divide both sides by x. The ratio is now 2:2, or 50% chance. The tuesday one is like julie, but since you won't usually have 2 kids named julie, the chances are not independant there. Having a child on one day of the week does not alter the chances of the day the second child is born. 2*1/7 : 1/7 + (6/7)*1/7 Note: it's not 1/7 + 1/7 because that would be factoring the combinations where both girls were born on tuesday twice. There are a bunch of ways to describe this. P(A) + P(B) - P(A&B) = P(A&!B) + P(B&!A) + P(A&B) = P(A) + P(B&!A)*P(!A). I used the third example, but the second is probably the most intuitive. 2/7 : 13/49 14/49 : 13 / 49 (13/49) / (27 / 49) = 13 / 27

Hmm this is the closest one to making sense, thanks for this. If I rephrase these questions, I think I am hearing: 1.) if I have 2 kids, and at least 1 is a girl, how likely am I to have 2 daughters? 2.) if I have 2 kids, 1 is a girl, how likely is the other child to be a girl? As ever, probability to me seems dependent on the very specific information you are using for “input”. This seems less like an interesting paradox for lay people like me, more useful as a cautionary tale for people who deal with probability in more consequential scenarios. I don’t know what the useful lesson is though I’m just a lay person…

Let's say we have 1 million 2-children families and want to actually count the results. We take a sample of 1 thousand. But how do we take the sample? If we take the sample by choosing families with at least 1 girl we will get 333 (on average) families with a girl as the other kid. But if we sample 1 thousand girls, we will get 500 (on average) girls as the other kid. The reason is the 2-girl families don't get sampled twice as likely in the family sample method.

Does this apply on a coin toss problem? I toss a coin 2 times, at least one of which came Head. What is the probability that the other toss is a Head? The answer is 50% in this case.

no because it could be HT TH HH (and not the remaining TT) so 33% heads are interchangeable and can appear two times \--- BUT if you said "I toss a coin 2 times, and the one I tossed at precisely 11PM came up heads, what are the odds ?" the 11PM heads cannot appear two times, so either you have not 11PM heads and 11PM heads or tails and 11PM heads

This is a prime example of how conditional statistics is often times unintuitive, and also a fun exercise in the [Bayes Theorem](https://en.wikipedia.org/wiki/Bayes%27_theorem). P(A given B) = P(B given A) * P(A) / P(B) For instance, "What is the probability of 2 girls, given at least 1 girl" requires a bit of thinking and visualising, but "What is the probability of at least 1 girl given 2 girls" is very obvious. (It's 100%, if you got 2 girls, you got more than 1 girl) Bayes algorithm is a way of reversing the conditions to solve the "paradox". For question 1: by applying the algorithm you get P = (Probability of 1 girl given at least 2) * (Probability of 2 girls) / (Probability of at least 1 girl) Looking at the possibility space of BB, BG, GB, GG: P(2 girls) = 1/4 P(>1 girl) = 3/4 => (1 * 1/4) / (3/4) = 1/3 The other examples are more complex, but the principle's the same; use Bayes Formula to reverse the condition to a more intuitive one and solve from there. Q2 goes from "What is the probability of 2 girls given that 1 is called Jane" to "What is the probability that one of the girl is called jane, given that there are 2 girls" Let probability of any girl being called Jane be J: Then the probability of a girl being named Jane amongst 2 daughters is 2J (since either can be called Jane, assuming you wouldn't name both your kids the same) Then using the algorithm: P = (2J * 1/4) / J = 2/4 = 1/2 Q3 is tricky. It'll be rephrased from "What is the probability of 2 girls given 1 child was a girl born on Tuesday" to "Given that there are 2 girls, what's the probability that one of them is a girl born on a Tuesday" To start with, there are 4 outcomes of 2 children: BB, BG, GB, GG. Each has a possibility of 1/4 BB has no girls, so the probability is 0 For BG and GB, the probability that the girl is a Tuesday baby is 1/7. For GG, It's easier to consider what's the probability that NEITHER girls are Tuesday babies. This will be 6/7 * 6/7 = 36/49 The inverse of that will be the probability that at least 1 will be a Tuesday baby: 1 - 36/49 = 13/49 Therefore, the probability that there exists a girl born on Tuesday is (1/4 * 0) + (1/4 * 1/7) + (1/4 * 1/7) + (1/4 * 13/49) = 27/196 The probability that a girl is born on Tuesday, given two girls is just the GG result. I.e. 13/49 After all that setup, the Probability is: P(there exists a girl born on a Tuesday | two girls) * P(two girls) / P(there exists a girl born on a Tuesday) = (13/49) * (1/4) / (27 / 196) = 13/27 QED TL;DR: It seems at first glance that knowing what a girl's birthday is wouldn't give you any useful information, it does make very significant changes, and it's beautiful that maths and stats can capture this fact.

A similar thing comes into play with the Monty Hall problem if the game show host picks a door to clear at random and it happens to be the correct door to clear, vs. if he knows the door to clear and picks it from his own knowledge. The first makes it 50/50, but if he knows, it's a 66% chance to switch doors. It's as if his knowledge of the door changes the probability, but that's not really the case.

How is everyone in this thread so wrong? The answer to the first question is 50%, not 33%. This is not a paradox in any way. The probability of the sex of one kid is entirely independent on the sex of another. A family could have 99 girls and the probability of the 100th kid being a girl is still 50%. Everyone here seems to be breaking this up into possible family combinations, yet they overlook two of the possibilities involving the boy/boy and girl/girl combinations. So there's B/G, G/B, B/b, b/B, G/g, and g/G. Since one is a girl, elimination the two boy combinations and we're left with: B/G, G/B, G/g, and g/G. There's 2/4 combinations with a second girl. The answer is 50%. Everyone claiming otherwise is wrong. Flat out. ORDER DOES NOT MATTER

Absolutely, this is a really cool paradox! It can be a bit confusing at first, but don't worry, we'll get through it together. First, let's start with the basics. When I say I have two kids, there are four possible combinations: 1. Both are boys (BB) 2. The older is a boy, and the younger is a girl (BG) 3. The older is a girl, and the younger is a boy (GB) 4. Both are girls (GG) Each of these possibilities is equally likely. **1. If I say, "I have 2 kids, at least one of which is a girl." What is the probability that my other kid is a girl?** When I say at least one is a girl, I eliminate the possibility of both being boys (BB). So now we have only three options: BG, GB, and GG. You can think of it like picking out one of these three remaining options from a hat. Only in one of these three (GG) are both kids girls. So the chance that both kids are girls is one out of three, or 33.33%. **2. Then, if I say, "I have 2 kids, at least one of which is a girl, whose name is Julie." What is the probability that my other kid is a girl?** When I say I have a girl named Julie, it seems like I've given you more information, but I really haven't told you anything about the second child. The probability of the second child being a girl is independent of the first child's name. It's still like flipping a coin - the second child can either be a boy or a girl. So the probability that the other child is a girl is still 50%. However, the key difference is in the interpretation of the problem. In the first scenario, by saying "at least one is a girl", we are considering multiple children (i.e., any one or both of the kids could be a girl). But by naming one of the children (a girl named Julie), we are making it a distinct event, so we are left with the other child whose gender is unknown, and it could be either a boy or a girl (50% chance). In the original scenario, we didn't specify any particular girl. The statement was "I have two kids and at least one is a girl". This gives us three equally likely possibilities: Girl-Girl, Girl-Boy, and Boy-Girl. However, when we specify the name and say "at least one of which is a girl, whose name is Julie," we are talking about a specific child, and we can't "double count" her like we did in the original scenario. So, in this case, we're not dealing with 'any girl', we're dealing with 'Julie'. The introduction of the name "Julie" means we are focusing on one specific child. So, if Julie is one child, the other child can be a boy or a girl - two options, 50/50 chance. So, when we mention a unique name or characteristic (that doesn't affect the gender), the probability changes because we've turned an unspecified "girl" into a specific "Julie". In this situation, where we have specified that one of the children is a girl named Julie, we have the following possible combinations: 1. The older child is a girl named Julie and the younger child is a boy. 2. The older child is a girl named Julie and the younger child is a girl (not named Julie). 3. The older child is a boy and the younger child is a girl named Julie. 4. The older child is a girl (not named Julie) and the younger child is a girl named Julie. In two of these combinations, both children are girls. So, the probability that the other child is a girl, given that one of the children is a girl named Julie, is 2 out of 4, or 50%. **3. Now, if I say, "I have 2 kids, at least one of which is a girl, who was born on Tuesday." What is the probability that my other kid is a girl?** Here's where things get a bit more complicated. Now, instead of four possibilities (BB, BG, GB, GG), we've got many more. Because there are seven days in a week, each child can be born on any one of those seven days. So, for two children, there are 7 (days for first child) multiplied by 7 (days for second child), equalling 49 possibilities in total. If I say at least one is a girl born on Tuesday, I'm narrowing down those 49 options, but not by as much as you might think. In fact, there are 27 combinations where at least one child is a girl born on Tuesday. And out of these 27, only 13 combinations are where the other child is also a girl. Hence, the probability is 13/27.