10 ms·

The sisters “paradox” – counter-intuitive probability

88 points by Vermin2000 1y ago

Vermin2000 1y ago
The sisters paradox is madenning example of counter-intuitive probability. The resolution is straightforward, but it's really easy to get tied up in knots.
- bell-cot 1y ago
  > maddening example of counter-intuitive probability. Not how I'd describe it. The setup is mundane enough for people to just assume that their intuition will work fine. The difference between the naive and correct answers is too small to spot in a small-n dataset. And ~0% of the population is actually familiar with analyzing such situations, for their "intuition" to be applicable. It's a bit like Gell-Mann amnesia - people are too quick to apply an easy cognitive strategy, when (in theory) they know enough to rule that strategy out.
  - spadros 1y ago
    Yes, I found this one easy. Was surprised my data management intuition came back after all these years since school. There’s really only three options: - boy - boy - boy - girl - girl - girl So it must be 1/3 chance. If you’re looking at permutations in order, that’s a different question.
    - AIPedant 1y ago
      This intuition is wrong even if turned out to get the right answer. The three unordered options do not have equal probabilities, boy+girl is twice as likely to occur as boy+boy and girl+girl. To get the right answer you must be careful about conditional probabilities (or draw out the sample space explicitly). The crux of the issue is that you are told extra information, which changes your estimate of the probability. (This question as written is very easy to misinterpret. The Monty Hall problem, which illustrates the same thing, is better since the sample selection is much more carefully explained.)
      - taeric 1y ago
        Oddly, this is a part I'm sticking with on this problem. Specifically, if you know that one is a girl, then the unordered options seem like they are back on equal footing? That is, it isn't twice as likely if you know that one ordering can't happen? (Or, stated differently, you don't know which version of two girls you are looking at.) So, for this one, you know that either the youngest is a girl (so, girl-boy is not possible) or that the oldest is a girl (so boy-girl is out). That puts you back to the rest of the possibilities. Boy-boy is out, sure, as you have a girl. But every other path remains? So, you have one of (boy-girl(known), girl-girl(known), girl(known)-boy, girl(known)-girl). Which drops you back to 50/50?
        AIPedant 1y ago
        Like others in the thread have said, the question could have been phrased more precisely. Technically you are misreading it but in an annoying and trivial way. What the problem is really saying is this: 1) You have a large collection of families with two kids of varying genders. 2) You draw one of them at random. At this point, your only estimate of P(2 girls) is 0.25. 3) Someone tells you that the family you drew has at least one girl. 4) This extra information changes your probability estimate because the possibility of two boys has been ruled out; the naive 1/4 estimate is refined to 1/3. The way you are interpreting it is this: 1) You have a large collection of families with two kids, at least one of whom is a girl. 2) Then the probability that the other child is a girl is clearly 50%. As a reminder this is how the original post phrased the question: Here's the problem: a family has two children. You're told that at least one of them is a girl. What's the probability both are girls? This is just too vague and admits both interpretations, they needed to be more specific about where the family "came from." That's why Monty Hall is a better illustration: it starts with you explicitly choosing a door at random. Here the family has been chosen at random from the pool of families with two children, but that's totally unclear.
        kgwgk 1y ago
        > 4) This extra information changes your probability estimate because the possibility of two boys has been ruled out; the naive 1/4 estimate is refined to 1/3. That’s not correct in general. It’s only correct if you assume that “3) Someone tells you that the family you drew has at least one girl.” was equally likely to happen whether or not there were two girls. That’s a quite strong assumption. One can make different assumptions and get answers different from 1/3. For example, 1/2.
- EMM_386 1y ago
  I don't understand at all how this is maddening or counter-intuitive. When you have a child, the odds are ~50% ... so the chance the next child is a boy or girl is almost equal. Is it because of the way it's framed that makes people think harder than they need to be? This is like when I (very rarely) play something like "pick six". I play 1, 2, 3, 4, 5, and 6. People think I'm crazy. They don't realize I have the same odds as any ticket they purchase.
  - n4r9 1y ago
    If you have to share the prize money with other people that guessed the same numbers, then it seems sensible to avoid obvious patterns.
    - EMM_386 1y ago
      Good point, others have said the same. In the rare scenario where I buy one of these tickets, I'd be more than happy to share the winnings! One billion dollars goes a long way ... even after taxes!
tocs3 1y ago
"This might seem abstract, but I've seen variations of this problem pop up in business and I've had difficult conversations with non-technical people as a result." Does anyone have some real life examples? i cannot think of any off hand but would like to be able to cite a couple if someone says "So, what is this good for?".
- deleted 1y ago
  [deleted]
- JHonaker 1y ago
  Any time you start conditioning on something, i.e. selecting subsets of data to analyze. You can fool yourself quite often if you do something seemingly innocuous like select "everyone with at least one X" and compare expectations to what's true unconditionally (meaning not conditioning on anything, not "in all cases") with conditional computations.
- nostrademons 1y ago
  "We noticed that our app reviews on TV, Watch, and VR suck, why could that be? Are the OSes on those platforms terrible? Are user expectations just super high? Can we ever make money on those platforms?" "When you're on TV, you need to use a DPad to laboriously type in a review. Same with the watch and its tiny screen. When you're on VR, you're using a laser pointer. That makes the barrier to leaving a review very high. As a result, people only bother to leave reviews if they have particularly strong feelings. The strongest motivating emotion is anger. Ergo, we are getting a disproportionate number of reviews from people who are angry, and everyone that is happy is not going to bother leaving a review." Same goes for many kinds of user feedback situations. The feedback you're measuring is P(user holds that opinion | user bothers to tell you that opinion). The denominator needs to be understood in order to evaluate that probability, but by definition, you don't have any data on people who don't bother to engage with you, unless you can measure their interaction in some other way (eg. usage data). This is why online reviews of many offline businesses skew negative; why businesses will often nag you to leave a review if you're satisfied; why corporate executives often overestimate employee satisfaction, particularly in prestigious businesses (anyone who is dissatisfied will leave and get a job elsewhere), and why Congress has such a low approval rating (by definition, you can't escape the laws enacted by Congress, so there is no "exit" option, and you get an accurate appraisal of what people really think).
tromp 1y ago
Perhaps people would be more likely to give the correct answer if "at least one of them is a girl" is rephrased as the equivalent "the youngest is a girl or the oldest is a girl".
- justonceokay 1y ago
  Well constructing a different question to remove the “trick” of the problem is one approach. Kind of the “no child left behind” approach to riddles.
- renewiltord 1y ago
  Another way is to say: the probability is 1/3, what is the probability? In this way, more people can answer it correctly, though perhaps not all.
- kgwgk 1y ago
  Maybe people would be more likely to give the correct answer if the problem had _one_ correct answer. :-)
sixo 1y ago
It's not even worth mentioning this problem unless you talk about how the result depends on the data generating process. If you take it to be something like "you randomly sample from families with two children, discarding any without at least one girl", you get the 1/3 result, but there are various other ways to read a sampling process from the problem statement which lead to other results.
- ndr 1y ago
  I took this to mean exactly that: > Assume the family is selected at random because they have at least one girl. And then again, if they sampled all families with 2 children the posterior would not change, would it? Still assuming boy vs girls are completely iid and equally probable
- two_handfuls 1y ago
  That's how I read it. What other ways were you thinking about?
  - bloak 1y ago
    Well, one way of getting families with two children, at least one of which is a girl, would be to go to a girls' school and ask the children to raise their hand if they have exactly one sibling.
    - aidenn0 1y ago
      I would expect that would yield a 50% chance of the other being a girl, right?
- the_gipsy 1y ago
  Why can you not frame it as: "a random family has been sampled, the sample family has two childs, one of them is a girl"? I.e. without "discarding", just giving some additional, but not complete, information on the random sample. Is adding information about the picked sample the same as discarding all contrarian samples? Why is this relevant?
pmg101 1y ago
GG / BG GB GG is 1 / 3. What's the paradox?
- teekert 1y ago
  Paradoxes don't exist in reality (they do in hypothetical situations), so there is indeed no paradox as you correctly observe. Instead, most people answer this wrongly, for some reason. And for some reason we call situations where this happens "a paradox". Though I agree that we shouldn't. Edit, ok, there are things like "This statement is false.", but we should perhaps stick to "self-referential problems" with those. I think paradoxes just exist in our theories, languages, and formal systems when we make flawed assumptions or create inconsistent frameworks. But physical reality itself just is what it is - no contradictions, just phenomena we sometimes struggle to describe accurately. If contradictions (paradoxes) can exist, then anything becomes possible through the principle of "explosion in logic". From a contradiction, any statement can be "proven" true. The whole foundation of rational thought would be undermined. Right?
  - luxcem 1y ago
    The Medical Test Paradox or what's that called do exist in the sense that when a test is positive for a rare disease we always run a second one.
    - teekert 1y ago
      To me that is not a paradox, just logical, with very low incidence you just find much more false positives than real positives. What is paradoxical about that? It's why we don't screen for just any condition in the general population. I.e. we just do it for 65+ y/o's, 3 packs/day smokers because there we may actually find it worth the cost of the program. There's no contradiction anywhere in this scenario, just people's incorrect intuitions meeting (mathematical) reality.
- hammock 1y ago
  Confusing permutations and combinations
fkyoureadthedoc 1y ago
> a family has two children. You're told that at least one of them is a girl. What's the probability both are girls? +-----------+-----------+-----------+-----------+ | Boy | | Girl | | | (0.5) | | (0.5) | | +-----------+-----------+-----------+-----------+ | Boy-Boy | Boy-Girl | Girl-Boy | Girl-Girl | | (0.25) | (0.25) | (0.25) | (0.25) | +-----------+-----------+-----------+-----------+ Does this mean the probability that they are both boys is also 25% lol
- randallsquared 1y ago
  That one is ruled out by the "at least one is a girl".
  - fkyoureadthedoc 1y ago
    Well brother I'm looking at the table and it clearly says 25%. You may have been lied to about the at least one girl it seems.
    - furyofantares 1y ago
      The text directly above that table: > A simpler question > Let's image you're asked a simpler question. > A family has two children. What's the probability both are girls?
      - fkyoureadthedoc 1y ago
        Under the 1 child policy, probably slim U+1F614
hammock 1y ago
This is Monty hall problem, is it not?
- luxcem 1y ago
  The 'sample space' reduction method is indeed also used to solve the Monty Hall problem.
- deleted 1y ago
  [deleted]
- teekert 1y ago
  Is it? Would it change your intuition when I tell you "A couple has 100 kids, at least 99 are girls, what is the probability 100 are girls?" I'm a bit at a loss I have to admit.
  - meatmanek 1y ago
    That would change my intuition quite a bit. Even assuming 100 kids were feasible in a human lifespan, the odds of 99 girls or 100 girls happening is (1 [arrangement where all the kids are girls] + 100 [arrangements where there are 99 girls and 1 boy]) / 2^100, or about 8e-29. Practically zero. If we assume that each child really did have a 50/50 chance of being boy or girl, then the result would be that there's a 1/101 chance that it's 100 girls. Given what I know about the world and genetics and such, I think it's much more likely that there's some predisposition by the couple to have girls. If we think it's, say, 90/10, then the prior probability of the 100-girls case would be 0.9 ^ 100, and the prior probability of each of the 99-girls cases would be 0.9^990.1 -- i.e. the all-girls case is 9x more likely. 0.9^100 / (100 0.9^990.1 + 0.9^100) = 0.9 / (1000.1 + 0.9) = 0.9/10.9 = about 8% probability of having 100 girls, 92% probability of having 99 girls and 1 boy. If we think the couple has 99:1 odds of girl:boy on each birth, then it's 0.99^100 / (100 * 0.99^990.01 + 0.99^100) or about 50/50 on whether they have 99 girls or 100 girls. If we think the odds are 999:1, then it's 0.999^100 / (100 0.999^99*0.001 + 0.999^100) = around 90% chance they have 100 girls. Someone else can do the math assuming an uninformative prior on the couple's girl:boy odds and calculating the posterior distribution given that we know there are 99 girls.
jmount 1y ago
Peter Winkler shares some great variations of this: "Boy Born on Tuesday" (p. xix) and "Men with Sisters" (p. xxii) in "Mathematical Puzzles". "Mrs. Chance has two children of different ages. At least one of them is a boy born on Tuesday. What is the probability that both of them are boys?" (note: it is a puzzle, not a biology or data demography problem. so there are 50/50 independence assumptions on gender and uniform day of week assumptions prior to adding the conditioning.)
- Rickasaurus 1y ago
  Great book, I highly recommend it too.
- layer8 1y ago
  Here “on Tuesday” is ambiguous, in my opinion. I first thought it meant “on a Tuesday” and that it was just a diversion. But it is likely intended to mean “last Tuesday” or “this Tuesday” (which excludes the boy-then-girl case). Wording it more clearly would likely reduce the ratio of wrong answers. Furthermore, “of different ages” is likely intended to exclude the case of twins. However, even with twins, one is generally nominally older than the other. (Not to mention that it’s possible for two non-twin siblings to be the same age in years, at certain points in time.) Why not just say “that aren’t twins”? I loathe when logic puzzles are obscured by ambiguous language, turning them more into “gotcha” text interpretation riddles than logic puzzles.
  - stronglikedan 1y ago
    I agree with the first one, but age is measured in days at a minimum, so twins are always the same age. (I'm sure there are cases where they are born farther apart due to some issue with the pregnancy, but that is statistically insignificant here.)
    - layer8 1y ago
      Even if you take days, one can be born at 23:58 and the other at 00:03 the next day. (And it could be New Year’s day — in some cultures that would even imply different ages in years.) Regardless of days, it’s not uncommon to talk about who is the older twin. Of course colloquially twins are the same age, but we are talking about a mathematical puzzle about probabilities here, where precision is paramount.
lightvector 1y ago
One of the challenges with puzzles like this it that it gives you "at least one of them is a girl" as a mathematical assertion where you're not supposed to further introspect the context of how/why you're being given that fact. But that's unrealistic. In real life, the context for how and why there would be a speaker telling you such a thing in the first place can be relevant and affect the probability! How is this possible? Suppose among all the math-riddle-loving parents of two children who would ask such a puzzle in the first place there are an equal number of parents of B-B, B-G, G-B, G-G, and that each is equally likely to ask you such a riddle when you meet them. Suppose when asking such a riddle the B-B parents tell you "at least one of them is a boy" (they don't have any girls, so that's the only way they can ask this kind of riddle), the G-G parents tell you "at least one of them is a girl" (same thing but in reverse), while the B-G and G-B parents say one of "at least one of them is a boy" and "at least one of them is a girl" equally at random. Then, conditioned on being told that "at least one of them is a girl", the probability of another girl is actually 1/2, not 1/3 like the paradox answer claims. To see this, imagine 40 examples of the above puzzle asking taking place. You get 10 B-B parents saying "at least one of them is a boy", 10 G-G parents saying "at least one of them is a girl", and among the 20 (B-G and G-B) parents since they choose randomly, you have 10 saying "at least one of them is a boy" and saying "at least one of them is a girl". So out of the 20 times where "at least one of them is a girl" is said, there are 10 cases where it's a G-G family and 10 cases where it's a B-G or G-B family, therefore conditioned on being told "at least one of them is a girl", the probability of two girls is actually 1/2. If there were some gender bias in how the B-G and G-B families might ask the question, or other differences that affect how likely different of these people would be posing the puzzle to you, then the probability could be yet different than either of 1/3 or 1/2. So there's a difference in being present something as a flat mathematical assertion that you're supposed to take at face value and not supposed to question further (where the probability is 1/3, as the article claims). Versus being told something in real life, where you always need to take into account the context and situation of the speaker, and the probability could be different. There are real life implications of this too - the big classic one being publication bias / newsworthiness bias. As most people intuitively know by now, it is also often wrong to take the statistical analysis or claims of a particular research study or paper entirely at face value, because there is a bias in the fact that "positive" and "exciting" results are more likely to be reported in the first place, and so statistical outliers that aren't actually replicable are disproportionately likely to be reported (see also https://xkcd.com/882/ https://xkcd.com/882/). And publication bias still occurs with respect to the reporting of results, amplification or not in the media etc, even when the the authors themselves are trustworthy and have done their analysis within the paper in a statistically proper way. So conditioned on you hearing about the result in the first place, it is often less likely to be true (and less likely to replicate in the future, etc) than you would think if you just took the statistical analysis in the paper at face value, even when that analysis was done correctly. The situation in the "sisters paradox" of computing a probability taking a statement entirely at logical face value is rare in real life.
hdgvhicv 1y ago
> Select only the families that have at least one girl. That’s not what the first question said. The first question was select a family (Bb,Bb,gb,gg) Then that they happen to have a girl.
- tantalor 1y ago
  I fail to see the distinction.
  - hnbad 1y ago
    It's explained in the Wikipedia article linked in TFA. If the family was picked at random, a GG family is twice as likely to have resulted in us finding out one of the children is a girl as a BG or GB family (and a BB family would be ruled out by observation) so you end up with the intuitive but in this case also entirely correct 1/2 chance. If the family was picked from only those that have at least one girl, a GG family is equally likely as a BG or GB family (and a BB family would be ruled out again but in this case by definition) you end up with the very unintuitive 1/3 chance the article describes. So the initial filtering is necessary to create this "paradox" (it's actually not a "paradox" but a "problem", as others have mentioned). Without it, the intuitive answer is actually the correct answer.
jihadjihad 1y ago
From the Wikipedia article linked in TFA: > Following classical probability arguments, we consider a large urn containing two children. I like how they modified a classic from probability texts, drawing items from an urn, and made sure it would be big enough in this example to accommodate two kids.
- bitwize 1y ago
  This is better than an "assume a spherical cow" joke in the wild.
D13Fd 1y ago
The "paradox" problem is in the setup. It's easy to mistake it as "a couple has one girl, what is the probability that their next child will be a girl," in which case the answer is 50%.
- tantalor 1y ago
  If you allow misunderstanding the question, then any answer is allowed.
  - in_cahoots 1y ago
    But it's a valid point, the question is not well-posed. If you said, "I looked at both children and saw that at least one was a girl" more people would get the right answer. Many people will assume that the author looked at only one child, not both. And there's nothing in the wording to indicate either way. As others are pointing out, this is just the Monty Hall problem. But the way the question is posed there is much clearer.
    - tantalor 1y ago
      I don't know how this could be made more clear: "You're told that at least one of them is a girl" > Many people will assume that the author looked at only one child There is no mentioning of "looking"
      - in_cahoots 1y ago
        How did you determine at least one is a girl? Presumably you looked in some way. But did you look at one child or both? That's the crux of the ambiguity.
        tantalor 1y ago
        I think you are asking "how did the person who told you there is at least one girl learn that". The answer is: it doesn't matter how because that is an unambiguous statement. It means "you can assume the family does not have two boys". I think people are actually getting hung up on "you are told" as if that could be a lie, or some kind of trick, when it is really just supposed to mean "here is some more information that you can rely on".
flappyeagle 1y ago
This is exactly the Monty hall problem
bitwize 1y ago
Related to the Monty Hall "paradox". Spoiler: You'll get the car if you switch doors with 2/3 probability. https://en.m.wikipedia.org/wiki/Monty_Hall_problem https://en.m.wikipedia.org/wiki/Monty_Hall_problem
- JeffJor 1y ago
  Bertrand's Box Paradox, which I wrote about in my own comment, applies to it. The upshot is that probability is not based on which prize placements _could_ lead the current game state, it is the set of all possible game states. Lets assume that the contestant starts off with door #3. Case 1: The prize is behind door #1, and the host must open door #2. Probability 1/3. Case 2: The prize is behind door #2, and the host must open door #1. Probability 1/3. Case 3: The prize is behind door #3, and the host has a choice. Case 3A: The host opens door #1. Probability Q/3. Case 3B: The host opens door #2. Probability (1-Q)/3. If the host actually opens door #1, the probability that door #2 has the prize is (Case 2)/(Case 2 + Case 3A) = (1/3)/(1/3+Q/3) = 1/(1+Q). If the host actually opens door #2, the probability that door #1 has the prize is (Case 1)/(Case 1 + Case 3B) = (1/3)/(1/3+(1-Q)/3) = 1/(2-Q). My point is that, since you get to see which door is opened, 2/3 is correct only if you assume Q=1/2. We aren't told what Q is, but we must assume it is 1/2 because otherwise the answer is different depending on which door is chosen.
  - zeroonetwothree 1y ago
    Well if we frame the question as “what is the probability of winning by always switching” then this doesn’t play into it and the answer is indeed 2/3. Hence as a question about general strategy the standard answer is correct. You’re right if we are asking about a specific case though.
- wpollock 1y ago
  This always bugged me. It isn't switching your choice that gives the 2/3 probability, it is exercising a choice once one of the doors has been eliminated. The odds are the same as flipping a fair coin to decide on which of the two doors remaining to pick. Am I wrong?
smohare 1y ago
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rml 1y ago
For something this small we can enumerate all the cases (this is a Scheme version of the Mathematica `Tuples` function) > (list-tuples '(B G) 2) ((B B) (B G) (G B) (G G)) 3 cases have at least one girl of those 3 cases, 1/3 are both girls
- florbnit 1y ago
  A coworker approached you and goes “hi, I have two children and one is a boy…” and is promptly vaporized because he doesn’t fit the selection criteria of the problem statement, another approaches and goes “err hi, I have two children and one is a girl…” looks nervously at the vaporizer but is left standing. What is the chance their other child is a girl, who has not been vaporized? If you phrase the question as “someone with two children tell you the gender of a random one, what is the chance the other is the same gender?” Chance is 50/50 because 50% will have BB or GG and the vaporizer isn’t active.
StopDisinfo910 1y ago
There is no sisters paradox. The trick is how the question is weirdly framed and has to be interpreted. What people think about when they hear the question would effectively lead to a probability of 0.5: if you see a family in the street with a girl and know they have two kids, the probability of the other kid being a girl is indeed 0.5. The trick of the so-called "paradox" is turning the question into the Monty Hall but with an ambitious enough formulation that you might be confused it’s not.
- guy2345 1y ago
  [dead]
- zahlman 1y ago
  >ambitious ambiguous?
  - deleted 1y ago
    [deleted]
- 6gvONxR4sf7o 1y ago
  The way to see this is bayes rule. p(answer | data) = p(data | answer) * p(answer) / (sum_{all possible answers'} p(data | answer') * p(answer')). So for this question, that's expands to: p(both are girls | you're told at least one is a girl) = p(you're told at least one is a girl | both are girls) * p(both are girls) / ( p(you're told at least one is a girl | both are girls) * p(both are girls) + p(you're told at least one is a girl | they aren't both girls) * p(they aren't both girls) ) The problem is that we don't know p(you're told at least one is a girl | they aren't both girls). Clearly if both are boys, then you won't be told at least one is a girl (or at least it's implied that you're told the truth). But that still leaves us p(you're told at least one is a girl | one boy and one girl). This is the crux of the thing. Different readings of the setup imply different answers to p(what you're told | the unknowns). It's also a great case of where bayes rule shorthands can be slippery. You'll usually abbreviate it out (hell, it was tedious to write this way even with copy-paste). But if you abbreviate "you're told there's at least one girl" to "there's at least one girl", then you've stopped modeling a crucial part of the setup. p(there's at least one girl | they aren't both girls) has an unambiguous answer.
Ekaros 1y ago
This reminds me of the Monty Hall. Which doesn't explicitly tell that host never reveals the car. Which for lot of people would be sensible option in game show. Just screw them up instantly. Well at least the winner gets some mutton.
puppycodes 1y ago
And then theres that intersex kid that makes all the numbers wrong
JeffJor 1y ago
Q1: "A family has two children. You're told that at least one of them is a girl. What's the probability both are girls?" Q2: "A family has two children. You're told that at least one of them is a boy. What's the probability both are boys?" Note that these are symmetric problems, and must have the same answer. Q3: "A family has two children. You're told that a gender, that applies to at least one, is written inside a sealed envelope. What's the probability both have that gender?" In Q3, we have no information. So the answer is the proportion of two-child families that are single gendered. That is, 1/2. But if we open the envelope, and read what is written inside, the problem becomes either Q1 or Q2. Which have the same answer. So we don't have to open it; whatever the answer to Q1 and Q2 is, opening the envelope in Q3 make its answer the same. If that answer is 1/3, we have a paradox. The answer has to be 1/2 of we don't look. This is what is known as "Bertrand's Box Paradox." Well, if we add a fourth box to his problem, with one gold and one silver coin. I realize that in modern times the problem itself is called the paradox, but what Bertrand actually wrote (edited to this problem) was "How can it be that opening the envelope suffices to change the probability from 1/2 to 1/3?" The resolution is that probability must be based on the full set of possibilities, not the possibilities that _could_ result from the full set of _states._ These are the possibilities for this problem: 1) BB and you are told that there is at least one boy. 2A) BG and you are told that there is at least one boy. 2B) BG and you are told that there is at least one girl. 3A) GB and you are told that there is at least one boy. 3B) GB and you are told that there is at least one girl. 4) GG and you are told that there is at least one girl. Each numbered case has a prior probability of 1/4. Let's say the "A" subcases have a probability of Q/4, so the "B" subcases have a probability of (1-Q)/4. The answer to the first problem is the probability of case 1, which is 1/4, divided by the total probability of cases 1, 2A, which is (1+2Q)/4. That's 1/(1+2Q). The answer to the second problem is the probability of case 4, divided by the total probability of cases 4, 2B, and 3B. Which is (3-2Q)/4. Bertrand's paradox, stated another way, is that these must be equal, but can only be equal if Q=1/2 and both answers are 1/2.
- meatmanek 1y ago
  In Q3, you've got 8 possibilities, expressed as (gender of 1st child, gender of 2nd child, which child's gender is written inside the sealed envelope?), each with presumably equal probability: 1. B B 1 2. B B 2 3. B G 1 4. B G 2 5. G B 1 6. G B 2 7. G G 1 8. G G 2 in which case 4 of 8 possibilities satisfy the condition (the first two and the last two). Once you open the sealed envelope and it says "girl", it does not become Q1, it becomes a different question: Q4: "A family has two children. I randomly sampled one of the children and it was a girl. What's the probability both are girls?" In which case, we're looking at possibilities 4, 5, 7, and 8, and in only 2 of those 4 possibilities are both children girls. In Q1, you're actually told "A family has two children. I looked at both children and can tell you that at least one of them is a girl. What's the probability that both are girls?". In which case, possibilities 3, 4, 5, 6, 7, 8 are all valid. Only in 2 of those 6 possibilities are both children girls. So as in_cahoots said in https://news.ycombinator.com/item?id=45053187 https://news.ycombinator.com/item?id=45053187, it matters whether the person asking looked at both children or just a single one.
a3w 1y ago
Becomes intuitive if you ask "nth one is a girl. What is nth+1 one likely to be?", for any arbitrary number
AdrianB1 1y ago
I think the explanation is wrong. It is based on the probability of having combinations of boys and girls and then counting the combinations that have at least one girl, but this is not the situation: there is no probability in question for one of the kids, it is a confirmed, past event and the only other probability is the sex of the other kid. Otherwise you can derive any probability as a branch of a probability tree that contains it and calculate the probabilities of the tree and then the one of the branch. This makes no sense. For example, a family has a kid and the kid is a girl. The family wants another kid; what is the probability to be a girl? Is it 1/4 because having 2 girls is 1/4? No, it is 1/2 as it is for any new kid.
- simonh 1y ago
  Your example is different in a critical way. These two questions are not equivalent. Q1: I looked at only one of a pair of two randomly selected children and it was a girl. What is the probability the other I didn’t see is a girl? Q2: I looked at both of two randomly selected children and at least one of the pair of children is a girl. What is the probability the other is also a girl? The question in the article is the second question, not the first. The fact that the observer looked at both children and not just one of them is crucial. As is often the case in these puzzles the exact information available is the critical issue.
taeric 1y ago
So, my problem with how this is modeled is it assumes order doesn't matter in one aspect, but that it does in another. Simply stated, if you allow the possibility space of "boy-girl" and "girl-boy", you have to also have two "girl-girl" states. Since you don't know which of the kids is known. Why is that not correct? State it with coins, if I know that you flipped a quarter and a dime and one turned up heads, what are the odds that both are heads?
- zeroonetwothree 1y ago
  The answer is the same in your coin version. There are four possible outcomes: (Q, D) = HH, HT, TH, TT. Given that one turned up heads that eliminates TT so we see that HH has 1/3 probability. As you can see there aren’t two HH states just as there aren’t two GG states in the original question.
  - taeric 1y ago
    Well, sorta? You were either in the world where dime was heads, in which case the space has TH, HH, or the Quarter was Heads, in which case you have HT, or again HH. Both with 50% probability, no? You were never in a scenario where HT and TH have the same probability as each other. (Well, again, sorta. Point being that you know one of them is possible.) Edit: I want to be clear that my initial thinking was exactly what you said. I was trying to "steelman" the bad intuition and I think I've trapped myself. :D
- pinoy420 1y ago
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- sokoloff 1y ago
  There aren't two "girl-girl" states, because of the stated assumption in the problem: > Assume the family is selected at random because they have at least one girl. Given that plus "a family has two children" and "Assume that the probability of having a girl or boy is 50%" That means you're starting from the set of all two child families: BB, BG, GB, and GG, being told that you do not have the BB case, leaving 3 ways in which the family could be composed and being asked about "the one which is not a G". That's different from the dime and quarter case, and would also be different if you were told "the oldest child is a girl", because being told "the oldest child is a girl" eliminates both BB and BG. Being told "[at least] one of the coins is heads" or "[at least] one of the kids is a girl" only eliminates one of the four cases, while being told "the quarter is heads"/"the oldest is a girl" eliminates two cases.
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onecommentman 1y ago
When I first heard of the Monty Hall problem, I assumed the naive answer was true, then I thought more and drew up the decision tree and thought the “correct” answer was right. Now I think it is an underspecified problem and literally any probability can be assigned to the result. There are no bad answers because it is a bad(ly stated) problem. “Here's the problem: a family has two children. You're told that at least one of them is a girl. What's the probability both are girls?” — This is the complete statement of the problem! Everything else is an assumption that may or may not be correct. And is certainly not necessarily a complete set of underlying assumptions relevant to the problem statement. “Assume that the probability of having a girl or boy is 50% and that the birth order has no effect on the probability. Assume the family is selected at random because they have at least one girl.” — This is not a part of the statement of the problem! These are a subset of assumptions that can choose to accept, or not. As a modeler or decision analyst you have to make that distinction. Eh, let’s accept them, for the time being. We’ll even assume the narrator is honest, which isn’t a stated assumption. But let’s add to that list of assumptions. The narrator telling you that one of them is a girl gets all winnings from bets on the outcome of the unknown gender of the “other child” and wants those winnings. The narrator knows that a probability tree analysis of the problem, with perhaps unwarranted assumptions of independence and prior probabilities, will lead to an assignment of 1/3 probability for the other sibling being a girl, and knows you know that result and believes you want to win. [A valid credible interpretation, not misinterpretation, of the original problem statement.] “What do you think the probability is that both children are girls?” — Let’s make this question more actionable. “Should you take the even odds bet on both children being girls made by the narrator?. $100 - if they are both girls, the narrator wins $100 and you lose $100; if they aren’t, you win $100 and the narrator loses $100. The narrator and you want to win the money.” The answer to this question, which seemingly follows from the question of probabilities to be “yes”, is, in fact, “no” - under the additional valid, and quite credible, assumptions made. Because you will only be presented sets of two-girl pairs by the narrator. Let’s assume the “assumptions” are actually correct, and the families will be indeed selected at random, and in the general population there is a 50-50 mix of boys and girls. There is nothing, even in the “assumptions”, that precludes the narrator from preselecting and only presenting two-girl pairs to you, thus always winning when you believe and follow the 1/3 two-girl result. The statement of the problem, and only the statement of the problem, underspecified as it is, leads to a whole suite of possibly correct answers. The problem is the territory, the problem statement and assumptions are the map. None of these maps are the territory, necessarily. The probability tree answer is just as sloppy, from a decision analyst perspective, as the naive answer.
nicknow 1y ago
Why do we treat boy-girl and girl-boy as different outcomes since they are, per the question, equal outcomes thus representing one possible outcome. We don't care about which came first or second, only what gender each child is. Thus the answer, to the question given the information you have, is 50%. The only possible outcomes are girl-girl or girl-boy (where order is irrelevant.) And this is absolutely NOT the Monty Hall Problem. I don't know why some people are making that reference. The Monty Hall Problem contains three possible choices and one is eliminated by the host, this is what makes the statistical math interesting in that problem. None of that is happening here. Lets look a the exact wording of the question: > a family has two children. You're told that at least one of them is a girl. What's the > probability both are girls? We have a family with two children. Assume we don't know their gender. We'll represent them as XX. We are told one of them is a female. So now they are represented as GX (remember GX = XG, since order doesn't matter.) You are left with the question what is the probability that X is female? Well there are only two choices, F and M, and we are told elsewhere that the probability of having a girl is 50/50. > Assume that the probability of having a girl or boy is 50% and that the birth order has > no effect on the probability. So the chance of X being female is 50%. Thus the answer is 50%. You can't say birth order doesn't matter and then use birth order to say the FM and MF are different results. The only possible results are FM and FF (since birth order is irrelevant.)
- lIl-IIIl 1y ago
  It's not the birth order that is relevant. It's the fact that there are twice as many boy-girl families as there are girl-girl families in the world.
hkkwritesgmail 1y ago
Its all in the wording. Such probability questions have deflated and dumbfounded generations of students and general public trying to learn the subject.

88 points by Vermin2000 1y ago

Vermin2000 1y ago

The sisters paradox is madenning example of counter-intuitive probability. The resolution is straightforward, but it's really easy to get tied up in knots.

bell-cot 1y ago
> maddening example of counter-intuitive probability. Not how I'd describe it. The setup is mundane enough for people to just assume that their intuition will work fine. The difference between the naive and correct answers is too small to spot in a small-n dataset. And ~0% of the population is actually familiar with analyzing such situations, for their "intuition" to be applicable. It's a bit like Gell-Mann amnesia - people are too quick to apply an easy cognitive strategy, when (in theory) they know enough to rule that strategy out.
- spadros 1y ago
  Yes, I found this one easy. Was surprised my data management intuition came back after all these years since school. There’s really only three options: - boy - boy - boy - girl - girl - girl So it must be 1/3 chance. If you’re looking at permutations in order, that’s a different question.
  - AIPedant 1y ago
    This intuition is wrong even if turned out to get the right answer. The three unordered options do not have equal probabilities, boy+girl is twice as likely to occur as boy+boy and girl+girl. To get the right answer you must be careful about conditional probabilities (or draw out the sample space explicitly). The crux of the issue is that you are told extra information, which changes your estimate of the probability. (This question as written is very easy to misinterpret. The Monty Hall problem, which illustrates the same thing, is better since the sample selection is much more carefully explained.)
    - taeric 1y ago
      Oddly, this is a part I'm sticking with on this problem. Specifically, if you know that one is a girl, then the unordered options seem like they are back on equal footing? That is, it isn't twice as likely if you know that one ordering can't happen? (Or, stated differently, you don't know which version of two girls you are looking at.) So, for this one, you know that either the youngest is a girl (so, girl-boy is not possible) or that the oldest is a girl (so boy-girl is out). That puts you back to the rest of the possibilities. Boy-boy is out, sure, as you have a girl. But every other path remains? So, you have one of (boy-girl(known), girl-girl(known), girl(known)-boy, girl(known)-girl). Which drops you back to 50/50?
      - AIPedant 1y ago
        Like others in the thread have said, the question could have been phrased more precisely. Technically you are misreading it but in an annoying and trivial way. What the problem is really saying is this: 1) You have a large collection of families with two kids of varying genders. 2) You draw one of them at random. At this point, your only estimate of P(2 girls) is 0.25. 3) Someone tells you that the family you drew has at least one girl. 4) This extra information changes your probability estimate because the possibility of two boys has been ruled out; the naive 1/4 estimate is refined to 1/3. The way you are interpreting it is this: 1) You have a large collection of families with two kids, at least one of whom is a girl. 2) Then the probability that the other child is a girl is clearly 50%. As a reminder this is how the original post phrased the question: Here's the problem: a family has two children. You're told that at least one of them is a girl. What's the probability both are girls? This is just too vague and admits both interpretations, they needed to be more specific about where the family "came from." That's why Monty Hall is a better illustration: it starts with you explicitly choosing a door at random. Here the family has been chosen at random from the pool of families with two children, but that's totally unclear.
        kgwgk 1y ago
        > 4) This extra information changes your probability estimate because the possibility of two boys has been ruled out; the naive 1/4 estimate is refined to 1/3. That’s not correct in general. It’s only correct if you assume that “3) Someone tells you that the family you drew has at least one girl.” was equally likely to happen whether or not there were two girls. That’s a quite strong assumption. One can make different assumptions and get answers different from 1/3. For example, 1/2.
EMM_386 1y ago
I don't understand at all how this is maddening or counter-intuitive. When you have a child, the odds are ~50% ... so the chance the next child is a boy or girl is almost equal. Is it because of the way it's framed that makes people think harder than they need to be? This is like when I (very rarely) play something like "pick six". I play 1, 2, 3, 4, 5, and 6. People think I'm crazy. They don't realize I have the same odds as any ticket they purchase.
- n4r9 1y ago
  If you have to share the prize money with other people that guessed the same numbers, then it seems sensible to avoid obvious patterns.
  - EMM_386 1y ago
    Good point, others have said the same. In the rare scenario where I buy one of these tickets, I'd be more than happy to share the winnings! One billion dollars goes a long way ... even after taxes!

tocs3 1y ago

"This might seem abstract, but I've seen variations of this problem pop up in business and I've had difficult conversations with non-technical people as a result." Does anyone have some real life examples? i cannot think of any off hand but would like to be able to cite a couple if someone says "So, what is this good for?".

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JHonaker 1y ago
Any time you start conditioning on something, i.e. selecting subsets of data to analyze. You can fool yourself quite often if you do something seemingly innocuous like select "everyone with at least one X" and compare expectations to what's true unconditionally (meaning not conditioning on anything, not "in all cases") with conditional computations.
nostrademons 1y ago
"We noticed that our app reviews on TV, Watch, and VR suck, why could that be? Are the OSes on those platforms terrible? Are user expectations just super high? Can we ever make money on those platforms?" "When you're on TV, you need to use a DPad to laboriously type in a review. Same with the watch and its tiny screen. When you're on VR, you're using a laser pointer. That makes the barrier to leaving a review very high. As a result, people only bother to leave reviews if they have particularly strong feelings. The strongest motivating emotion is anger. Ergo, we are getting a disproportionate number of reviews from people who are angry, and everyone that is happy is not going to bother leaving a review." Same goes for many kinds of user feedback situations. The feedback you're measuring is P(user holds that opinion | user bothers to tell you that opinion). The denominator needs to be understood in order to evaluate that probability, but by definition, you don't have any data on people who don't bother to engage with you, unless you can measure their interaction in some other way (eg. usage data). This is why online reviews of many offline businesses skew negative; why businesses will often nag you to leave a review if you're satisfied; why corporate executives often overestimate employee satisfaction, particularly in prestigious businesses (anyone who is dissatisfied will leave and get a job elsewhere), and why Congress has such a low approval rating (by definition, you can't escape the laws enacted by Congress, so there is no "exit" option, and you get an accurate appraisal of what people really think).

tromp 1y ago

Perhaps people would be more likely to give the correct answer if "at least one of them is a girl" is rephrased as the equivalent "the youngest is a girl or the oldest is a girl".

justonceokay 1y ago
Well constructing a different question to remove the “trick” of the problem is one approach. Kind of the “no child left behind” approach to riddles.
renewiltord 1y ago
Another way is to say: the probability is 1/3, what is the probability? In this way, more people can answer it correctly, though perhaps not all.
kgwgk 1y ago
Maybe people would be more likely to give the correct answer if the problem had _one_ correct answer. :-)

sixo 1y ago

It's not even worth mentioning this problem unless you talk about how the result depends on the data generating process. If you take it to be something like "you randomly sample from families with two children, discarding any without at least one girl", you get the 1/3 result, but there are various other ways to read a sampling process from the problem statement which lead to other results.

ndr 1y ago
I took this to mean exactly that: > Assume the family is selected at random because they have at least one girl. And then again, if they sampled all families with 2 children the posterior would not change, would it? Still assuming boy vs girls are completely iid and equally probable
two_handfuls 1y ago
That's how I read it. What other ways were you thinking about?
- bloak 1y ago
  Well, one way of getting families with two children, at least one of which is a girl, would be to go to a girls' school and ask the children to raise their hand if they have exactly one sibling.
  - aidenn0 1y ago
    I would expect that would yield a 50% chance of the other being a girl, right?
the_gipsy 1y ago
Why can you not frame it as: "a random family has been sampled, the sample family has two childs, one of them is a girl"? I.e. without "discarding", just giving some additional, but not complete, information on the random sample. Is adding information about the picked sample the same as discarding all contrarian samples? Why is this relevant?

pmg101 1y ago

GG / BG GB GG is 1 / 3. What's the paradox?

teekert 1y ago
Paradoxes don't exist in reality (they do in hypothetical situations), so there is indeed no paradox as you correctly observe. Instead, most people answer this wrongly, for some reason. And for some reason we call situations where this happens "a paradox". Though I agree that we shouldn't. Edit, ok, there are things like "This statement is false.", but we should perhaps stick to "self-referential problems" with those. I think paradoxes just exist in our theories, languages, and formal systems when we make flawed assumptions or create inconsistent frameworks. But physical reality itself just is what it is - no contradictions, just phenomena we sometimes struggle to describe accurately. If contradictions (paradoxes) can exist, then anything becomes possible through the principle of "explosion in logic". From a contradiction, any statement can be "proven" true. The whole foundation of rational thought would be undermined. Right?
- luxcem 1y ago
  The Medical Test Paradox or what's that called do exist in the sense that when a test is positive for a rare disease we always run a second one.
  - teekert 1y ago
    To me that is not a paradox, just logical, with very low incidence you just find much more false positives than real positives. What is paradoxical about that? It's why we don't screen for just any condition in the general population. I.e. we just do it for 65+ y/o's, 3 packs/day smokers because there we may actually find it worth the cost of the program. There's no contradiction anywhere in this scenario, just people's incorrect intuitions meeting (mathematical) reality.
hammock 1y ago
Confusing permutations and combinations

fkyoureadthedoc 1y ago

> a family has two children. You're told that at least one of them is a girl. What's the probability both are girls? +-----------+-----------+-----------+-----------+ | Boy | | Girl | | | (0.5) | | (0.5) | | +-----------+-----------+-----------+-----------+ | Boy-Boy | Boy-Girl | Girl-Boy | Girl-Girl | | (0.25) | (0.25) | (0.25) | (0.25) | +-----------+-----------+-----------+-----------+ Does this mean the probability that they are both boys is also 25% lol

randallsquared 1y ago
That one is ruled out by the "at least one is a girl".
- fkyoureadthedoc 1y ago
  Well brother I'm looking at the table and it clearly says 25%. You may have been lied to about the at least one girl it seems.
  - furyofantares 1y ago
    The text directly above that table: > A simpler question > Let's image you're asked a simpler question. > A family has two children. What's the probability both are girls?
    - fkyoureadthedoc 1y ago
      Under the 1 child policy, probably slim U+1F614

hammock 1y ago

This is Monty hall problem, is it not?

luxcem 1y ago
The 'sample space' reduction method is indeed also used to solve the Monty Hall problem.
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teekert 1y ago
Is it? Would it change your intuition when I tell you "A couple has 100 kids, at least 99 are girls, what is the probability 100 are girls?" I'm a bit at a loss I have to admit.
- meatmanek 1y ago
  That would change my intuition quite a bit. Even assuming 100 kids were feasible in a human lifespan, the odds of 99 girls or 100 girls happening is (1 [arrangement where all the kids are girls] + 100 [arrangements where there are 99 girls and 1 boy]) / 2^100, or about 8e-29. Practically zero. If we assume that each child really did have a 50/50 chance of being boy or girl, then the result would be that there's a 1/101 chance that it's 100 girls. Given what I know about the world and genetics and such, I think it's much more likely that there's some predisposition by the couple to have girls. If we think it's, say, 90/10, then the prior probability of the 100-girls case would be 0.9 ^ 100, and the prior probability of each of the 99-girls cases would be 0.9^990.1 -- i.e. the all-girls case is 9x more likely. 0.9^100 / (100 0.9^990.1 + 0.9^100) = 0.9 / (1000.1 + 0.9) = 0.9/10.9 = about 8% probability of having 100 girls, 92% probability of having 99 girls and 1 boy. If we think the couple has 99:1 odds of girl:boy on each birth, then it's 0.99^100 / (100 * 0.99^990.01 + 0.99^100) or about 50/50 on whether they have 99 girls or 100 girls. If we think the odds are 999:1, then it's 0.999^100 / (100 0.999^99*0.001 + 0.999^100) = around 90% chance they have 100 girls. Someone else can do the math assuming an uninformative prior on the couple's girl:boy odds and calculating the posterior distribution given that we know there are 99 girls.

jmount 1y ago

Peter Winkler shares some great variations of this: "Boy Born on Tuesday" (p. xix) and "Men with Sisters" (p. xxii) in "Mathematical Puzzles". "Mrs. Chance has two children of different ages. At least one of them is a boy born on Tuesday. What is the probability that both of them are boys?" (note: it is a puzzle, not a biology or data demography problem. so there are 50/50 independence assumptions on gender and uniform day of week assumptions prior to adding the conditioning.)

Rickasaurus 1y ago
Great book, I highly recommend it too.
layer8 1y ago
Here “on Tuesday” is ambiguous, in my opinion. I first thought it meant “on a Tuesday” and that it was just a diversion. But it is likely intended to mean “last Tuesday” or “this Tuesday” (which excludes the boy-then-girl case). Wording it more clearly would likely reduce the ratio of wrong answers. Furthermore, “of different ages” is likely intended to exclude the case of twins. However, even with twins, one is generally nominally older than the other. (Not to mention that it’s possible for two non-twin siblings to be the same age in years, at certain points in time.) Why not just say “that aren’t twins”? I loathe when logic puzzles are obscured by ambiguous language, turning them more into “gotcha” text interpretation riddles than logic puzzles.
- stronglikedan 1y ago
  I agree with the first one, but age is measured in days at a minimum, so twins are always the same age. (I'm sure there are cases where they are born farther apart due to some issue with the pregnancy, but that is statistically insignificant here.)
  - layer8 1y ago
    Even if you take days, one can be born at 23:58 and the other at 00:03 the next day. (And it could be New Year’s day — in some cultures that would even imply different ages in years.) Regardless of days, it’s not uncommon to talk about who is the older twin. Of course colloquially twins are the same age, but we are talking about a mathematical puzzle about probabilities here, where precision is paramount.

lightvector 1y ago

One of the challenges with puzzles like this it that it gives you "at least one of them is a girl" as a mathematical assertion where you're not supposed to further introspect the context of how/why you're being given that fact. But that's unrealistic. In real life, the context for how and why there would be a speaker telling you such a thing in the first place can be relevant and affect the probability! How is this possible? Suppose among all the math-riddle-loving parents of two children who would ask such a puzzle in the first place there are an equal number of parents of B-B, B-G, G-B, G-G, and that each is equally likely to ask you such a riddle when you meet them. Suppose when asking such a riddle the B-B parents tell you "at least one of them is a boy" (they don't have any girls, so that's the only way they can ask this kind of riddle), the G-G parents tell you "at least one of them is a girl" (same thing but in reverse), while the B-G and G-B parents say one of "at least one of them is a boy" and "at least one of them is a girl" equally at random. Then, conditioned on being told that "at least one of them is a girl", the probability of another girl is actually 1/2, not 1/3 like the paradox answer claims. To see this, imagine 40 examples of the above puzzle asking taking place. You get 10 B-B parents saying "at least one of them is a boy", 10 G-G parents saying "at least one of them is a girl", and among the 20 (B-G and G-B) parents since they choose randomly, you have 10 saying "at least one of them is a boy" and saying "at least one of them is a girl". So out of the 20 times where "at least one of them is a girl" is said, there are 10 cases where it's a G-G family and 10 cases where it's a B-G or G-B family, therefore conditioned on being told "at least one of them is a girl", the probability of two girls is actually 1/2. If there were some gender bias in how the B-G and G-B families might ask the question, or other differences that affect how likely different of these people would be posing the puzzle to you, then the probability could be yet different than either of 1/3 or 1/2. So there's a difference in being present something as a flat mathematical assertion that you're supposed to take at face value and not supposed to question further (where the probability is 1/3, as the article claims). Versus being told something in real life, where you always need to take into account the context and situation of the speaker, and the probability could be different. There are real life implications of this too - the big classic one being publication bias / newsworthiness bias. As most people intuitively know by now, it is also often wrong to take the statistical analysis or claims of a particular research study or paper entirely at face value, because there is a bias in the fact that "positive" and "exciting" results are more likely to be reported in the first place, and so statistical outliers that aren't actually replicable are disproportionately likely to be reported (see also https://xkcd.com/882/ https://xkcd.com/882/). And publication bias still occurs with respect to the reporting of results, amplification or not in the media etc, even when the the authors themselves are trustworthy and have done their analysis within the paper in a statistically proper way. So conditioned on you hearing about the result in the first place, it is often less likely to be true (and less likely to replicate in the future, etc) than you would think if you just took the statistical analysis in the paper at face value, even when that analysis was done correctly. The situation in the "sisters paradox" of computing a probability taking a statement entirely at logical face value is rare in real life.

hdgvhicv 1y ago

> Select only the families that have at least one girl. That’s not what the first question said. The first question was select a family (Bb,Bb,gb,gg) Then that they happen to have a girl.

tantalor 1y ago
I fail to see the distinction.
- hnbad 1y ago
  It's explained in the Wikipedia article linked in TFA. If the family was picked at random, a GG family is twice as likely to have resulted in us finding out one of the children is a girl as a BG or GB family (and a BB family would be ruled out by observation) so you end up with the intuitive but in this case also entirely correct 1/2 chance. If the family was picked from only those that have at least one girl, a GG family is equally likely as a BG or GB family (and a BB family would be ruled out again but in this case by definition) you end up with the very unintuitive 1/3 chance the article describes. So the initial filtering is necessary to create this "paradox" (it's actually not a "paradox" but a "problem", as others have mentioned). Without it, the intuitive answer is actually the correct answer.

jihadjihad 1y ago

From the Wikipedia article linked in TFA: > Following classical probability arguments, we consider a large urn containing two children. I like how they modified a classic from probability texts, drawing items from an urn, and made sure it would be big enough in this example to accommodate two kids.

bitwize 1y ago
This is better than an "assume a spherical cow" joke in the wild.

D13Fd 1y ago

The "paradox" problem is in the setup. It's easy to mistake it as "a couple has one girl, what is the probability that their next child will be a girl," in which case the answer is 50%.

tantalor 1y ago
If you allow misunderstanding the question, then any answer is allowed.
- in_cahoots 1y ago
  But it's a valid point, the question is not well-posed. If you said, "I looked at both children and saw that at least one was a girl" more people would get the right answer. Many people will assume that the author looked at only one child, not both. And there's nothing in the wording to indicate either way. As others are pointing out, this is just the Monty Hall problem. But the way the question is posed there is much clearer.
  - tantalor 1y ago
    I don't know how this could be made more clear: "You're told that at least one of them is a girl" > Many people will assume that the author looked at only one child There is no mentioning of "looking"
    - in_cahoots 1y ago
      How did you determine at least one is a girl? Presumably you looked in some way. But did you look at one child or both? That's the crux of the ambiguity.
      - tantalor 1y ago
        I think you are asking "how did the person who told you there is at least one girl learn that". The answer is: it doesn't matter how because that is an unambiguous statement. It means "you can assume the family does not have two boys". I think people are actually getting hung up on "you are told" as if that could be a lie, or some kind of trick, when it is really just supposed to mean "here is some more information that you can rely on".

flappyeagle 1y ago

This is exactly the Monty hall problem

bitwize 1y ago

Related to the Monty Hall "paradox". Spoiler: You'll get the car if you switch doors with 2/3 probability. https://en.m.wikipedia.org/wiki/Monty_Hall_problem https://en.m.wikipedia.org/wiki/Monty_Hall_problem

JeffJor 1y ago
Bertrand's Box Paradox, which I wrote about in my own comment, applies to it. The upshot is that probability is not based on which prize placements _could_ lead the current game state, it is the set of all possible game states. Lets assume that the contestant starts off with door #3. Case 1: The prize is behind door #1, and the host must open door #2. Probability 1/3. Case 2: The prize is behind door #2, and the host must open door #1. Probability 1/3. Case 3: The prize is behind door #3, and the host has a choice. Case 3A: The host opens door #1. Probability Q/3. Case 3B: The host opens door #2. Probability (1-Q)/3. If the host actually opens door #1, the probability that door #2 has the prize is (Case 2)/(Case 2 + Case 3A) = (1/3)/(1/3+Q/3) = 1/(1+Q). If the host actually opens door #2, the probability that door #1 has the prize is (Case 1)/(Case 1 + Case 3B) = (1/3)/(1/3+(1-Q)/3) = 1/(2-Q). My point is that, since you get to see which door is opened, 2/3 is correct only if you assume Q=1/2. We aren't told what Q is, but we must assume it is 1/2 because otherwise the answer is different depending on which door is chosen.
- zeroonetwothree 1y ago
  Well if we frame the question as “what is the probability of winning by always switching” then this doesn’t play into it and the answer is indeed 2/3. Hence as a question about general strategy the standard answer is correct. You’re right if we are asking about a specific case though.
wpollock 1y ago
This always bugged me. It isn't switching your choice that gives the 2/3 probability, it is exercising a choice once one of the doors has been eliminated. The odds are the same as flipping a fair coin to decide on which of the two doors remaining to pick. Am I wrong?

smohare 1y ago

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rml 1y ago

For something this small we can enumerate all the cases (this is a Scheme version of the Mathematica `Tuples` function) > (list-tuples '(B G) 2) ((B B) (B G) (G B) (G G)) 3 cases have at least one girl of those 3 cases, 1/3 are both girls

florbnit 1y ago
A coworker approached you and goes “hi, I have two children and one is a boy…” and is promptly vaporized because he doesn’t fit the selection criteria of the problem statement, another approaches and goes “err hi, I have two children and one is a girl…” looks nervously at the vaporizer but is left standing. What is the chance their other child is a girl, who has not been vaporized? If you phrase the question as “someone with two children tell you the gender of a random one, what is the chance the other is the same gender?” Chance is 50/50 because 50% will have BB or GG and the vaporizer isn’t active.

StopDisinfo910 1y ago

There is no sisters paradox. The trick is how the question is weirdly framed and has to be interpreted. What people think about when they hear the question would effectively lead to a probability of 0.5: if you see a family in the street with a girl and know they have two kids, the probability of the other kid being a girl is indeed 0.5. The trick of the so-called "paradox" is turning the question into the Monty Hall but with an ambitious enough formulation that you might be confused it’s not.

guy2345 1y ago
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zahlman 1y ago
>ambitious ambiguous?
- deleted 1y ago
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6gvONxR4sf7o 1y ago
The way to see this is bayes rule. p(answer | data) = p(data | answer) * p(answer) / (sum_{all possible answers'} p(data | answer') * p(answer')). So for this question, that's expands to: p(both are girls | you're told at least one is a girl) = p(you're told at least one is a girl | both are girls) * p(both are girls) / ( p(you're told at least one is a girl | both are girls) * p(both are girls) + p(you're told at least one is a girl | they aren't both girls) * p(they aren't both girls) ) The problem is that we don't know p(you're told at least one is a girl | they aren't both girls). Clearly if both are boys, then you won't be told at least one is a girl (or at least it's implied that you're told the truth). But that still leaves us p(you're told at least one is a girl | one boy and one girl). This is the crux of the thing. Different readings of the setup imply different answers to p(what you're told | the unknowns). It's also a great case of where bayes rule shorthands can be slippery. You'll usually abbreviate it out (hell, it was tedious to write this way even with copy-paste). But if you abbreviate "you're told there's at least one girl" to "there's at least one girl", then you've stopped modeling a crucial part of the setup. p(there's at least one girl | they aren't both girls) has an unambiguous answer.

Ekaros 1y ago

This reminds me of the Monty Hall. Which doesn't explicitly tell that host never reveals the car. Which for lot of people would be sensible option in game show. Just screw them up instantly. Well at least the winner gets some mutton.

puppycodes 1y ago

And then theres that intersex kid that makes all the numbers wrong

JeffJor 1y ago

Q1: "A family has two children. You're told that at least one of them is a girl. What's the probability both are girls?" Q2: "A family has two children. You're told that at least one of them is a boy. What's the probability both are boys?" Note that these are symmetric problems, and must have the same answer. Q3: "A family has two children. You're told that a gender, that applies to at least one, is written inside a sealed envelope. What's the probability both have that gender?" In Q3, we have no information. So the answer is the proportion of two-child families that are single gendered. That is, 1/2. But if we open the envelope, and read what is written inside, the problem becomes either Q1 or Q2. Which have the same answer. So we don't have to open it; whatever the answer to Q1 and Q2 is, opening the envelope in Q3 make its answer the same. If that answer is 1/3, we have a paradox. The answer has to be 1/2 of we don't look. This is what is known as "Bertrand's Box Paradox." Well, if we add a fourth box to his problem, with one gold and one silver coin. I realize that in modern times the problem itself is called the paradox, but what Bertrand actually wrote (edited to this problem) was "How can it be that opening the envelope suffices to change the probability from 1/2 to 1/3?" The resolution is that probability must be based on the full set of possibilities, not the possibilities that _could_ result from the full set of _states._ These are the possibilities for this problem: 1) BB and you are told that there is at least one boy. 2A) BG and you are told that there is at least one boy. 2B) BG and you are told that there is at least one girl. 3A) GB and you are told that there is at least one boy. 3B) GB and you are told that there is at least one girl. 4) GG and you are told that there is at least one girl. Each numbered case has a prior probability of 1/4. Let's say the "A" subcases have a probability of Q/4, so the "B" subcases have a probability of (1-Q)/4. The answer to the first problem is the probability of case 1, which is 1/4, divided by the total probability of cases 1, 2A, which is (1+2Q)/4. That's 1/(1+2Q). The answer to the second problem is the probability of case 4, divided by the total probability of cases 4, 2B, and 3B. Which is (3-2Q)/4. Bertrand's paradox, stated another way, is that these must be equal, but can only be equal if Q=1/2 and both answers are 1/2.

meatmanek 1y ago
In Q3, you've got 8 possibilities, expressed as (gender of 1st child, gender of 2nd child, which child's gender is written inside the sealed envelope?), each with presumably equal probability: 1. B B 1 2. B B 2 3. B G 1 4. B G 2 5. G B 1 6. G B 2 7. G G 1 8. G G 2 in which case 4 of 8 possibilities satisfy the condition (the first two and the last two). Once you open the sealed envelope and it says "girl", it does not become Q1, it becomes a different question: Q4: "A family has two children. I randomly sampled one of the children and it was a girl. What's the probability both are girls?" In which case, we're looking at possibilities 4, 5, 7, and 8, and in only 2 of those 4 possibilities are both children girls. In Q1, you're actually told "A family has two children. I looked at both children and can tell you that at least one of them is a girl. What's the probability that both are girls?". In which case, possibilities 3, 4, 5, 6, 7, 8 are all valid. Only in 2 of those 6 possibilities are both children girls. So as in_cahoots said in https://news.ycombinator.com/item?id=45053187 https://news.ycombinator.com/item?id=45053187, it matters whether the person asking looked at both children or just a single one.

a3w 1y ago

Becomes intuitive if you ask "nth one is a girl. What is nth+1 one likely to be?", for any arbitrary number

AdrianB1 1y ago

I think the explanation is wrong. It is based on the probability of having combinations of boys and girls and then counting the combinations that have at least one girl, but this is not the situation: there is no probability in question for one of the kids, it is a confirmed, past event and the only other probability is the sex of the other kid. Otherwise you can derive any probability as a branch of a probability tree that contains it and calculate the probabilities of the tree and then the one of the branch. This makes no sense. For example, a family has a kid and the kid is a girl. The family wants another kid; what is the probability to be a girl? Is it 1/4 because having 2 girls is 1/4? No, it is 1/2 as it is for any new kid.

simonh 1y ago
Your example is different in a critical way. These two questions are not equivalent. Q1: I looked at only one of a pair of two randomly selected children and it was a girl. What is the probability the other I didn’t see is a girl? Q2: I looked at both of two randomly selected children and at least one of the pair of children is a girl. What is the probability the other is also a girl? The question in the article is the second question, not the first. The fact that the observer looked at both children and not just one of them is crucial. As is often the case in these puzzles the exact information available is the critical issue.

taeric 1y ago

So, my problem with how this is modeled is it assumes order doesn't matter in one aspect, but that it does in another. Simply stated, if you allow the possibility space of "boy-girl" and "girl-boy", you have to also have two "girl-girl" states. Since you don't know which of the kids is known. Why is that not correct? State it with coins, if I know that you flipped a quarter and a dime and one turned up heads, what are the odds that both are heads?

zeroonetwothree 1y ago
The answer is the same in your coin version. There are four possible outcomes: (Q, D) = HH, HT, TH, TT. Given that one turned up heads that eliminates TT so we see that HH has 1/3 probability. As you can see there aren’t two HH states just as there aren’t two GG states in the original question.
- taeric 1y ago
  Well, sorta? You were either in the world where dime was heads, in which case the space has TH, HH, or the Quarter was Heads, in which case you have HT, or again HH. Both with 50% probability, no? You were never in a scenario where HT and TH have the same probability as each other. (Well, again, sorta. Point being that you know one of them is possible.) Edit: I want to be clear that my initial thinking was exactly what you said. I was trying to "steelman" the bad intuition and I think I've trapped myself. :D
pinoy420 1y ago
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sokoloff 1y ago
There aren't two "girl-girl" states, because of the stated assumption in the problem: > Assume the family is selected at random because they have at least one girl. Given that plus "a family has two children" and "Assume that the probability of having a girl or boy is 50%" That means you're starting from the set of all two child families: BB, BG, GB, and GG, being told that you do not have the BB case, leaving 3 ways in which the family could be composed and being asked about "the one which is not a G". That's different from the dime and quarter case, and would also be different if you were told "the oldest child is a girl", because being told "the oldest child is a girl" eliminates both BB and BG. Being told "[at least] one of the coins is heads" or "[at least] one of the kids is a girl" only eliminates one of the four cases, while being told "the quarter is heads"/"the oldest is a girl" eliminates two cases.

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onecommentman 1y ago

When I first heard of the Monty Hall problem, I assumed the naive answer was true, then I thought more and drew up the decision tree and thought the “correct” answer was right. Now I think it is an underspecified problem and literally any probability can be assigned to the result. There are no bad answers because it is a bad(ly stated) problem. “Here's the problem: a family has two children. You're told that at least one of them is a girl. What's the probability both are girls?” — This is the complete statement of the problem! Everything else is an assumption that may or may not be correct. And is certainly not necessarily a complete set of underlying assumptions relevant to the problem statement. “Assume that the probability of having a girl or boy is 50% and that the birth order has no effect on the probability. Assume the family is selected at random because they have at least one girl.” — This is not a part of the statement of the problem! These are a subset of assumptions that can choose to accept, or not. As a modeler or decision analyst you have to make that distinction. Eh, let’s accept them, for the time being. We’ll even assume the narrator is honest, which isn’t a stated assumption. But let’s add to that list of assumptions. The narrator telling you that one of them is a girl gets all winnings from bets on the outcome of the unknown gender of the “other child” and wants those winnings. The narrator knows that a probability tree analysis of the problem, with perhaps unwarranted assumptions of independence and prior probabilities, will lead to an assignment of 1/3 probability for the other sibling being a girl, and knows you know that result and believes you want to win. [A valid credible interpretation, not misinterpretation, of the original problem statement.] “What do you think the probability is that both children are girls?” — Let’s make this question more actionable. “Should you take the even odds bet on both children being girls made by the narrator?. $100 - if they are both girls, the narrator wins $100 and you lose $100; if they aren’t, you win $100 and the narrator loses $100. The narrator and you want to win the money.” The answer to this question, which seemingly follows from the question of probabilities to be “yes”, is, in fact, “no” - under the additional valid, and quite credible, assumptions made. Because you will only be presented sets of two-girl pairs by the narrator. Let’s assume the “assumptions” are actually correct, and the families will be indeed selected at random, and in the general population there is a 50-50 mix of boys and girls. There is nothing, even in the “assumptions”, that precludes the narrator from preselecting and only presenting two-girl pairs to you, thus always winning when you believe and follow the 1/3 two-girl result. The statement of the problem, and only the statement of the problem, underspecified as it is, leads to a whole suite of possibly correct answers. The problem is the territory, the problem statement and assumptions are the map. None of these maps are the territory, necessarily. The probability tree answer is just as sloppy, from a decision analyst perspective, as the naive answer.

nicknow 1y ago

Why do we treat boy-girl and girl-boy as different outcomes since they are, per the question, equal outcomes thus representing one possible outcome. We don't care about which came first or second, only what gender each child is. Thus the answer, to the question given the information you have, is 50%. The only possible outcomes are girl-girl or girl-boy (where order is irrelevant.) And this is absolutely NOT the Monty Hall Problem. I don't know why some people are making that reference. The Monty Hall Problem contains three possible choices and one is eliminated by the host, this is what makes the statistical math interesting in that problem. None of that is happening here. Lets look a the exact wording of the question: > a family has two children. You're told that at least one of them is a girl. What's the > probability both are girls? We have a family with two children. Assume we don't know their gender. We'll represent them as XX. We are told one of them is a female. So now they are represented as GX (remember GX = XG, since order doesn't matter.) You are left with the question what is the probability that X is female? Well there are only two choices, F and M, and we are told elsewhere that the probability of having a girl is 50/50. > Assume that the probability of having a girl or boy is 50% and that the birth order has > no effect on the probability. So the chance of X being female is 50%. Thus the answer is 50%. You can't say birth order doesn't matter and then use birth order to say the FM and MF are different results. The only possible results are FM and FF (since birth order is irrelevant.)

lIl-IIIl 1y ago
It's not the birth order that is relevant. It's the fact that there are twice as many boy-girl families as there are girl-girl families in the world.

hkkwritesgmail 1y ago

Its all in the wording. Such probability questions have deflated and dumbfounded generations of students and general public trying to learn the subject.