4 ms·

The question is wrong

50 points by peter123 18y ago

deleted 18y ago
[deleted]
Eliezer 18y ago
In the correct version of this story, the mathematician says "I have two children", and you ask, "Is at least one a boy?", and she answers "Yes". Then the probability is 1/3 that they are both boys. See here for why I think this problem confuses people: http://www.overcomingbias.com/2008/03/mind-probabilit.html http://www.overcomingbias.com/2008/03/mind-probabilit.html
- MisterMerkin 18y ago
  If I told someone that in real life and they replied by asking, "Is at least one a boy?", I would probably back away very, very slowly. The main point is that given the way Atwood asked the question the correct answer is 50% which wasn't the answer he intended to discuss and indicates he himself didn't fully understand the subject he was writing about. It wouldn't be the first time either.
  - lg 18y ago
    Atwood doesn't seem wrong or particularly unclear to me. If you "met someone who told you they had two children, and one of them is a girl", then presumably we should imagine this person saying: "I have two children and one of them is a girl," and not "I have two children, X and Y, and X is a girl." Obviously in the first case, we don't know if it's X or Y that's the girl, so the set of possible worlds is [(X-G & Y-G), (X-G & Y-B), (X-B & Y-G)], so we get the 2/3 answer. But maybe other people don't share my linguistic intuitions...
    - tspiteri 18y ago
      If I met someone who told me "I have two children, and one of them is a girl," I would be pretty sure they have one girl and one boy. If they had two girls, they would have said "I have two children, and both of them are girls." So the English statement has some implicit information. On the other hand, if I were given a riddle with that statement, I would know the implicit information may be intentionally misleading. In this case, as Paul Buchheit says, the statement does not have enough information.
  - ntoshev 18y ago
    Writing about something is a way to understand it better. Why people write about things they fully comprehend is beyond me ;)
- paul 18y ago
  Yes, the question must be very carefully worded (which is kind of my point, if I have one). In order to get the "unintuitive" outcome, there must be some element of selection (much like the Monty Hall problem has). In your formulation of the question, the GG possibility has already been eliminated because the mathematician answers "yes".
- paul 18y ago
  I just realized, even your "correct" version, although more precise, is still underspecified and can't be answered. :) Can you spot the problem?
  - mattmcknight 18y ago
    What do you think of this story? I get a frantic call from my friend at the airport. He's been captured by the TSA for bringing a nose hair trimmer onto the plane. He was taking his brothers two kids (whom I know nothing of) back home, but now needs me to pick them up at the airport. The line goes dead. I show up at the airport and there are 4 pairs of children waiting behind the glass to be claimed, conveniently arranged in pairs of GG, GB, BG, and BB. Each pair is equally likely to be the one I am supposed to pick up. If my friend had told me "their names are Sarah and--" before the line went dead, I would mentally eliminate BB. Sarah could be any of the four girls. The three remaining options are not equally likely to contain Sarah, because the first one has two girls, and either one could be Sarah. So, the probability of Sarah being the girl in BG is 1/4 and of being the girl in GB is 1/4. The probability of the other child being a girl is also 1/2. It's not so much that BG or GB is eliminated in this case, but that the probability of her being in the GG group is better. But wait a second, I am not trying to find Sarah, I am trying to find the pair of kids that is his pair of kids. Is the probability that a group contains Sarah different from the probability that a group of kids is his? Maybe that's my bad assumption...maybe "at least one is a girl" is some precise formulation I don't understand that means the GG group isn't twice as likely to contain that girl. I'll get back to that. (Possibly her name is "at least once"?). But what if he instead told me something over the phone that would eliminate the possibility of it being two boys only? What if when he had looked out over the groups of children waiting to be picked up before they blindfolded him and panicked and said "it's not the two boys--"? Is knowing that one of the children is a girl different from knowing they both aren't boys? I think it is, because the latter is a statement about the set of events, and the former is a statement about a single event. When we talk about the pairs of children, the set of events, we are not conveying direct information about the individual events. In this second case he is not conveying information about the individual events and the three remaining choices remain equally likely, 1/3 each. Looking back to "at least one is a boy"- if we interpret that as a statement about the sets of probabilities, I would more precisely restate Eliezer's question as "Does the set of two children contain at least one boy?". This is why the important part of the story is the word "mathematician". The mathematician is talking about eliminating sets of events when he says "Yes". Ordinary people would just talk about the gender of a child.
mpk 18y ago
The author apparently doesn't understand where the confusion in the original question (or variations) comes from. One way of phrasing it is asking, "If you have an unrelated man and woman and they both have two children (one of which is a boy), where the oldest son of the man is a boy - what are the odds that they both have two boys?" (There are any number of variations on this - what are the odds of the man having two boys? what are the odds of the woman having two boys what are the odds of the man but not the woman having two boys .. all the same problem, but stated differently). This 'loads' the question by implying (superficially incorrect) that there might be a difference between the chances of a man and a woman having two boys. Next up is a bit of probability theory. In the case of the woman, no order is stated, so the chances of her two children have no connection - the events are unrelated. The man, however, has as a first child a boy (which eliminates the possibility of this being a girl). This is another variation on the Monty Hall problem, http://en.wikipedia.org/wiki/Monty_hall_problem http://en.wikipedia.org/wiki/Monty_hall_problem Read up on some historical background on this here, http://www.marilynvossavant.com/articles/gameshow.html http://www.marilynvossavant.com/articles/gameshow.html And as for the overall birthrate of men vs women or the possibilities of having twins/triplets/etc (and their male/female ratio) ... well, that's really out of the scope of a fairly trivial question statement such as this. [edit for non-trivial detail]
- trominos 18y ago
  No, you're wrong. The thing is that there are two kinds of confusion that arise from Atwood's problem. The first kind of confusion comes from not understanding how probabilities work, which you discuss. The second kind -- which is what Paul Buchheit is talking about -- comes from noticing that the statement "I have two children, and one of them is a girl" can be parsed in two different ways. It can be parsed as, "I have two children, and at least one of them is a girl," or as "I have two children, and the gender of one of them is #{my_first_child.gender}." Despite what many commenters in this thread are saying, these are not the same. That's the real problem here: the English language is inexact. The words that Atwood used to describe the scenario actually describe at least two mathematically distinct scenarios. The Monty Hall problem suffers from this fact, too, but not as badly -- because both interpretations yield the same conclusion, namely, that you should switch doors. It's just that under one interpretation, you get a car 1/2 of the time, and under the other you get it 2/3 of the time. Also, under the interpretation that yields a car 1/2 the time, it's logically implied that the host is willing to open a door and reveal a car -- which most people use to rule out that interpretation, if only subconsciously.
BvS 18y ago
If Jeff Atwood argument would be right, couldn't it than be applied to roulette as well? If you play two games for example, there are 4 different orders of black/red possible (neglecting 0/00): b,b b,r r,b r,r Each has the same possibility of 1/4. If you place 1$ in the first round on black, your chance to win is 1/2. If it is black (or red), only 3 of the original 4 option are left and two will lead to red...
- randallsquared 18y ago
  No, you've reintroduced ordering.
  - mpk 18y ago
    randallsquared is correct. You are misunderstanding sequences of related events. Running the roulette wheel twice is a sequence of unrelated events. The (ideal) roulette wheel is totally random. The results of run #1 are unrelated to the results of run #2. However, in this case of the man and woman both having two children, one of which is a boy, but the man having as an oldest child a boy we are talking about two different scenarios. The events are suddenly related. If you want it spelled out, the matrices are, The childbirth options are this, BB BG GB GG However, for the man it's been stated that his oldest child is a boy, so it becomes, BB BG XX XX For the woman (who has as a condition 'at least one boy, but not necessarily the eldest') it is, BB BG GB XX (Where 'B' is Boy, 'G' is Girl and 'X' is a non-option).
- jfarmer 18y ago
  If the FIRST round is black, only two of the four possibilities are left, one of which leads to red. Similarly, if Atwood's problem stated that the FIRST child is a girl then the probability that the second is a boy is 1/2. The problem, however, states that one of the children is a girl -- it could be the first or the second. In this case the probability that the other child is a boy is 2/3.
randallsquared 18y ago
It's true, as a commenter on this article mentioned, that successive children are (more-or-less) independent events, but that doesn't mean what that commenter (or our author) apparently thinks. If you specify that the older child is a girl, then the chances of the younger child being a girl are 1/2. If you don't specify which child is a girl, only that one is, then it's 1/3. This and the Monty Hall problem are just examples of the same limited-information situation.
- paul 18y ago
  Nope. Announcing the gender of one child does not magically alter the gender of the other child. It's like a print statement in code. If you don't believe me, try writing some code to simulate this. I will bet you an arbitrary amount of money that I'm right :) Here's another way of looking at it: By your logic, if I announce that one of the children is a girl, then the other child only has a 1/3 chance of also being a girl. Likewise, if I announce that one of the children is a boy, then the other child only has a 1/3 chance of being a boy. Therefore, by your logic, the act of my arbitrarily announcing the gender of one of the children increases the probability that the other child is of the opposite gender from 1/2 (what it was before I spoke) to 2/3, regardless of whether I said it was a girl or boy. Hopefully you can see why this is not correct.
  - randallsquared 18y ago
    "Announcing the gender of one child does not magically alter the gender of the other child." Nor is that what I said. :) Assuming you meant "chance of being a boy" where you said "chance of being a girl", I agree that announcing the gender of one of the two children increases the probability of the other child being the opposite gender from 1/2 to 2/3. The reason this is so is that some of the original probability was sunk in a case you've now eliminated: the case where there were two of the gender you didn't announce. [Edit: Jeff already wrote some code, and while I haven't bothered to review it, I assume the lack of outcry about it indicates its correctness. Care to share where he got that wrong?]
    - 18y ago
cduan 18y ago
Isn't the probability still 2/3 even in the proposed alternative? The fact that we "arbitrarily announce the gender of one of the children" provides new information for calculating the probability, which is ignored if you assume that the original 50/50 distribution is unchanged by the announcement. In an extreme example, consider this algorithm, analogous to the one in the article: 1. Choose a random parent that has exactly two children 2. Announce the gender of _both_ children 3. Ask about the odds that the parent has both a boy and a girl In this situation, it would be ridiculous to say that the odds calculated at step 3 are still 50/50. Because of step 2, it is either certainly true (100%) or not true (0%) that the parent had both a boy and a girl. Granted, on average with repeated trials, half the time the step-3 value will be 100% and the other half 0%. But still, the step 2 information changes the value at step 3.
- trominos 18y ago
  No. You've gained no useful information. The person who announces the gender of one child can say one of two things: "this child is a girl" or "this child is a boy." However, these two statements are symmetric in their effect on P(one child is a girl and one is a boy). There's a huge difference between announcing the gender of one child and announcing the gender of both children. If the announcer announces both children's genders, he can say that "both are boys," that "both are girls," or that "one is a boy and one is a girl." But the last statement is not symmetric to the first two in how it affects P(one child is a boy and one is a girl).
peakok 18y ago
You are right about the question ("Announcing the gender of one child does not magically alter the gender of the other child"), but wrong about the statistics, even in your second formulation the probability is 2/3, not 1/2. This problem, the way Jeff has formulated it, has nothing to do with the Monty Hall problem. But still, the end probability is the same : 2/3 (that's the only common point). Tell me if there is a mistake : The question is : What are the odds that the person has a boy and a girl, if we already know that one of the child is a girl ? (If we agree on the question, then we must agree on the following probabilities). Possibilities are : 1/ boy / boy 2/ boy / girl 3/ girl / boy 4/ girl / girl Since one of them is a girl, we must remove possibility number 1. That leaves us with 3 possiblities, and 2 of them have a boy. Probability : 2/3 (edited for correction, we search the probability of having a boy, not a girl :p)
- timr 18y ago
  Ordering is irrelevant here. Options 2 and 3 are identical.
  - randallsquared 18y ago
    Options 2 and 3 each still have the same probability weight as each of options 1 and 4.
    - deleted 18y ago
      [deleted]
    - timr 18y ago
      No, they don't. Again, ordering is irrelevant in this problem. We want to know only the probability that there will be a boy/girl pair, not the probability that the boy/girl pair was born in a particular order. But the same result can be obtained when taking ordering into account -- the key observation is that if you subdivide options 2 and 3 to account for sibling ordering, then you also must subdivide options 1 and 4 to account for sibling order, resulting in 6 total possibilities, of which 2 are M/F sibling pairs, 2 are M/M pairs, and 2 are F/F pairs: 1) M/m 2) m/M 3) M/F 4) F/M 5) F/f 6) f/F Given the knowledge that one sibling is female, you then exclude 2 of the 6 possibilities (the m/M and M/m pairs), to obtain 2/4 = 50% probability that the pair of siblings is of mixed sex. The mistake you're making is that you're including ordering on the mixed-sex pairs, but not including ordering on the same-sex pairs.
mixmax 18y ago
In Jeff's post the question was posed to show how bad humans are at probabilities, and it got the point across very well. I know that I fell right into the trap... This post takes what was meant to be an enlightening and refreshing take on how bad we as humans are at probabilities and turns it into a numbers-duel. No the original questions doesn't clearly state the definition of who is asked, it doesn't take into account that slightly more girls than boys are born, and it doesn't answer the question about whether this all happened in a cult where all boys are killed at birth. It was a simple metaphor. And it worked... Here is a TED talk about the subject of how we are fooled by probabilities: http://www.ted.com/index.php/talks/peter_donnelly_shows_how_stats_fool_juries.html http://www.ted.com/index.php/talks/peter_donnelly_shows_how_...
- jerf 18y ago
  Really, a (not the, but a) problem is that almost everybody is right. We are certainly fooled by probability. But also, if someone volunteers that one out of their two children is a girl, then it's effectively-100% that the other is a boy, in real life. Slightly amending a quote in a similar context from "Principles of Economics, Explained" (http://www.youtube.com/watch?v=VVp8UGjECt4 http://www.youtube.com/watch?v=VVp8UGjECt4), nobody says "I have a girl. I have another girl. I have another girl." One way of interpreting this perennial debate is as proof of how just how fuzzy English can be. People aren't just arguing about the answer, they argue about the question too.
  - randallsquared 18y ago
    "But also, if someone volunteers that one out of their two children is a girl, then it's effectively-100% that the other is a boy, in real life." Well, yes, because saying one is implies that the other isn't, in speech. But we could talk about sets of two of anything that has a 50% probability.
  - mixmax 18y ago
    Really, a problem is that almost everybody is right. Yes - this is why I pointed out that in Jeff's post this riddle was used as a metaphor. Otherwise we could spend all our time bickering about the correct spelling of "colour" or other trivialities instead of addressing the interesting questions. The interesting question in Jeff's post was the poor human understanding of probabilities.
cedsav 18y ago
Manolis comments on Paul's post summed it up pretty well for me. What feels counter intuitive is that announcing the gender of one child seems to increase the chances of the other child being of a different gender, but it's actually the opposite. "Family with 2 children" -> chance of having at least one boy: 75% "but I have at least one girl" -> chance of having at least one boy: 66% "but the first one is a girl" -> chance of having at least one boy: 50% "but they're both girls" -> chance of having at least one boy: 0%
Andi 18y ago
It's all a question about the correct precodition definition. I think people who heard about conditional probability know that the correct answer is "2/3". Most people will just imagine the first precondition "they have two children" and forget the second one. What about going in the opposite direction and simply ask "What is the probability that a mother has a boy and a girl?" I think many people would answer "1/2", too, simply adding the precodition "she has 2 children" ;) The origin for this is - I think - an implicit context people are thinking of without real awareness of it. You think quicker with implicit context.

50 points by peter123 18y ago

deleted 18y ago

[deleted]

In the correct version of this story, the mathematician says "I have two children", and you ask, "Is at least one a boy?", and she answers "Yes". Then the probability is 1/3 that they are both boys. See here for why I think this problem confuses people: http://www.overcomingbias.com/2008/03/mind-probabilit.html http://www.overcomingbias.com/2008/03/mind-probabilit.html

MisterMerkin 18y ago
If I told someone that in real life and they replied by asking, "Is at least one a boy?", I would probably back away very, very slowly. The main point is that given the way Atwood asked the question the correct answer is 50% which wasn't the answer he intended to discuss and indicates he himself didn't fully understand the subject he was writing about. It wouldn't be the first time either.
- lg 18y ago
  Atwood doesn't seem wrong or particularly unclear to me. If you "met someone who told you they had two children, and one of them is a girl", then presumably we should imagine this person saying: "I have two children and one of them is a girl," and not "I have two children, X and Y, and X is a girl." Obviously in the first case, we don't know if it's X or Y that's the girl, so the set of possible worlds is [(X-G & Y-G), (X-G & Y-B), (X-B & Y-G)], so we get the 2/3 answer. But maybe other people don't share my linguistic intuitions...
  - tspiteri 18y ago
    If I met someone who told me "I have two children, and one of them is a girl," I would be pretty sure they have one girl and one boy. If they had two girls, they would have said "I have two children, and both of them are girls." So the English statement has some implicit information. On the other hand, if I were given a riddle with that statement, I would know the implicit information may be intentionally misleading. In this case, as Paul Buchheit says, the statement does not have enough information.
- ntoshev 18y ago
  Writing about something is a way to understand it better. Why people write about things they fully comprehend is beyond me ;)
paul 18y ago
Yes, the question must be very carefully worded (which is kind of my point, if I have one). In order to get the "unintuitive" outcome, there must be some element of selection (much like the Monty Hall problem has). In your formulation of the question, the GG possibility has already been eliminated because the mathematician answers "yes".
paul 18y ago
I just realized, even your "correct" version, although more precise, is still underspecified and can't be answered. :) Can you spot the problem?
- mattmcknight 18y ago
  What do you think of this story? I get a frantic call from my friend at the airport. He's been captured by the TSA for bringing a nose hair trimmer onto the plane. He was taking his brothers two kids (whom I know nothing of) back home, but now needs me to pick them up at the airport. The line goes dead. I show up at the airport and there are 4 pairs of children waiting behind the glass to be claimed, conveniently arranged in pairs of GG, GB, BG, and BB. Each pair is equally likely to be the one I am supposed to pick up. If my friend had told me "their names are Sarah and--" before the line went dead, I would mentally eliminate BB. Sarah could be any of the four girls. The three remaining options are not equally likely to contain Sarah, because the first one has two girls, and either one could be Sarah. So, the probability of Sarah being the girl in BG is 1/4 and of being the girl in GB is 1/4. The probability of the other child being a girl is also 1/2. It's not so much that BG or GB is eliminated in this case, but that the probability of her being in the GG group is better. But wait a second, I am not trying to find Sarah, I am trying to find the pair of kids that is his pair of kids. Is the probability that a group contains Sarah different from the probability that a group of kids is his? Maybe that's my bad assumption...maybe "at least one is a girl" is some precise formulation I don't understand that means the GG group isn't twice as likely to contain that girl. I'll get back to that. (Possibly her name is "at least once"?). But what if he instead told me something over the phone that would eliminate the possibility of it being two boys only? What if when he had looked out over the groups of children waiting to be picked up before they blindfolded him and panicked and said "it's not the two boys--"? Is knowing that one of the children is a girl different from knowing they both aren't boys? I think it is, because the latter is a statement about the set of events, and the former is a statement about a single event. When we talk about the pairs of children, the set of events, we are not conveying direct information about the individual events. In this second case he is not conveying information about the individual events and the three remaining choices remain equally likely, 1/3 each. Looking back to "at least one is a boy"- if we interpret that as a statement about the sets of probabilities, I would more precisely restate Eliezer's question as "Does the set of two children contain at least one boy?". This is why the important part of the story is the word "mathematician". The mathematician is talking about eliminating sets of events when he says "Yes". Ordinary people would just talk about the gender of a child.

mpk 18y ago

The author apparently doesn't understand where the confusion in the original question (or variations) comes from. One way of phrasing it is asking, "If you have an unrelated man and woman and they both have two children (one of which is a boy), where the oldest son of the man is a boy - what are the odds that they both have two boys?" (There are any number of variations on this - what are the odds of the man having two boys? what are the odds of the woman having two boys what are the odds of the man but not the woman having two boys .. all the same problem, but stated differently). This 'loads' the question by implying (superficially incorrect) that there might be a difference between the chances of a man and a woman having two boys. Next up is a bit of probability theory. In the case of the woman, no order is stated, so the chances of her two children have no connection - the events are unrelated. The man, however, has as a first child a boy (which eliminates the possibility of this being a girl). This is another variation on the Monty Hall problem, http://en.wikipedia.org/wiki/Monty_hall_problem http://en.wikipedia.org/wiki/Monty_hall_problem Read up on some historical background on this here, http://www.marilynvossavant.com/articles/gameshow.html http://www.marilynvossavant.com/articles/gameshow.html And as for the overall birthrate of men vs women or the possibilities of having twins/triplets/etc (and their male/female ratio) ... well, that's really out of the scope of a fairly trivial question statement such as this. [edit for non-trivial detail]

trominos 18y ago
No, you're wrong. The thing is that there are two kinds of confusion that arise from Atwood's problem. The first kind of confusion comes from not understanding how probabilities work, which you discuss. The second kind -- which is what Paul Buchheit is talking about -- comes from noticing that the statement "I have two children, and one of them is a girl" can be parsed in two different ways. It can be parsed as, "I have two children, and at least one of them is a girl," or as "I have two children, and the gender of one of them is #{my_first_child.gender}." Despite what many commenters in this thread are saying, these are not the same. That's the real problem here: the English language is inexact. The words that Atwood used to describe the scenario actually describe at least two mathematically distinct scenarios. The Monty Hall problem suffers from this fact, too, but not as badly -- because both interpretations yield the same conclusion, namely, that you should switch doors. It's just that under one interpretation, you get a car 1/2 of the time, and under the other you get it 2/3 of the time. Also, under the interpretation that yields a car 1/2 the time, it's logically implied that the host is willing to open a door and reveal a car -- which most people use to rule out that interpretation, if only subconsciously.

BvS 18y ago

If Jeff Atwood argument would be right, couldn't it than be applied to roulette as well? If you play two games for example, there are 4 different orders of black/red possible (neglecting 0/00): b,b b,r r,b r,r Each has the same possibility of 1/4. If you place 1$ in the first round on black, your chance to win is 1/2. If it is black (or red), only 3 of the original 4 option are left and two will lead to red...

randallsquared 18y ago
No, you've reintroduced ordering.
- mpk 18y ago
  randallsquared is correct. You are misunderstanding sequences of related events. Running the roulette wheel twice is a sequence of unrelated events. The (ideal) roulette wheel is totally random. The results of run #1 are unrelated to the results of run #2. However, in this case of the man and woman both having two children, one of which is a boy, but the man having as an oldest child a boy we are talking about two different scenarios. The events are suddenly related. If you want it spelled out, the matrices are, The childbirth options are this, BB BG GB GG However, for the man it's been stated that his oldest child is a boy, so it becomes, BB BG XX XX For the woman (who has as a condition 'at least one boy, but not necessarily the eldest') it is, BB BG GB XX (Where 'B' is Boy, 'G' is Girl and 'X' is a non-option).
jfarmer 18y ago
If the FIRST round is black, only two of the four possibilities are left, one of which leads to red. Similarly, if Atwood's problem stated that the FIRST child is a girl then the probability that the second is a boy is 1/2. The problem, however, states that one of the children is a girl -- it could be the first or the second. In this case the probability that the other child is a boy is 2/3.

randallsquared 18y ago

It's true, as a commenter on this article mentioned, that successive children are (more-or-less) independent events, but that doesn't mean what that commenter (or our author) apparently thinks. If you specify that the older child is a girl, then the chances of the younger child being a girl are 1/2. If you don't specify which child is a girl, only that one is, then it's 1/3. This and the Monty Hall problem are just examples of the same limited-information situation.

paul 18y ago
Nope. Announcing the gender of one child does not magically alter the gender of the other child. It's like a print statement in code. If you don't believe me, try writing some code to simulate this. I will bet you an arbitrary amount of money that I'm right :) Here's another way of looking at it: By your logic, if I announce that one of the children is a girl, then the other child only has a 1/3 chance of also being a girl. Likewise, if I announce that one of the children is a boy, then the other child only has a 1/3 chance of being a boy. Therefore, by your logic, the act of my arbitrarily announcing the gender of one of the children increases the probability that the other child is of the opposite gender from 1/2 (what it was before I spoke) to 2/3, regardless of whether I said it was a girl or boy. Hopefully you can see why this is not correct.
- randallsquared 18y ago
  "Announcing the gender of one child does not magically alter the gender of the other child." Nor is that what I said. :) Assuming you meant "chance of being a boy" where you said "chance of being a girl", I agree that announcing the gender of one of the two children increases the probability of the other child being the opposite gender from 1/2 to 2/3. The reason this is so is that some of the original probability was sunk in a case you've now eliminated: the case where there were two of the gender you didn't announce. [Edit: Jeff already wrote some code, and while I haven't bothered to review it, I assume the lack of outcry about it indicates its correctness. Care to share where he got that wrong?]
  - 18y ago

cduan 18y ago

Isn't the probability still 2/3 even in the proposed alternative? The fact that we "arbitrarily announce the gender of one of the children" provides new information for calculating the probability, which is ignored if you assume that the original 50/50 distribution is unchanged by the announcement. In an extreme example, consider this algorithm, analogous to the one in the article: 1. Choose a random parent that has exactly two children 2. Announce the gender of _both_ children 3. Ask about the odds that the parent has both a boy and a girl In this situation, it would be ridiculous to say that the odds calculated at step 3 are still 50/50. Because of step 2, it is either certainly true (100%) or not true (0%) that the parent had both a boy and a girl. Granted, on average with repeated trials, half the time the step-3 value will be 100% and the other half 0%. But still, the step 2 information changes the value at step 3.

trominos 18y ago
No. You've gained no useful information. The person who announces the gender of one child can say one of two things: "this child is a girl" or "this child is a boy." However, these two statements are symmetric in their effect on P(one child is a girl and one is a boy). There's a huge difference between announcing the gender of one child and announcing the gender of both children. If the announcer announces both children's genders, he can say that "both are boys," that "both are girls," or that "one is a boy and one is a girl." But the last statement is not symmetric to the first two in how it affects P(one child is a boy and one is a girl).

peakok 18y ago

You are right about the question ("Announcing the gender of one child does not magically alter the gender of the other child"), but wrong about the statistics, even in your second formulation the probability is 2/3, not 1/2. This problem, the way Jeff has formulated it, has nothing to do with the Monty Hall problem. But still, the end probability is the same : 2/3 (that's the only common point). Tell me if there is a mistake : The question is : What are the odds that the person has a boy and a girl, if we already know that one of the child is a girl ? (If we agree on the question, then we must agree on the following probabilities). Possibilities are : 1/ boy / boy 2/ boy / girl 3/ girl / boy 4/ girl / girl Since one of them is a girl, we must remove possibility number 1. That leaves us with 3 possiblities, and 2 of them have a boy. Probability : 2/3 (edited for correction, we search the probability of having a boy, not a girl :p)

timr 18y ago
Ordering is irrelevant here. Options 2 and 3 are identical.
- randallsquared 18y ago
  Options 2 and 3 each still have the same probability weight as each of options 1 and 4.
  - deleted 18y ago
    [deleted]
  - timr 18y ago
    No, they don't. Again, ordering is irrelevant in this problem. We want to know only the probability that there will be a boy/girl pair, not the probability that the boy/girl pair was born in a particular order. But the same result can be obtained when taking ordering into account -- the key observation is that if you subdivide options 2 and 3 to account for sibling ordering, then you also must subdivide options 1 and 4 to account for sibling order, resulting in 6 total possibilities, of which 2 are M/F sibling pairs, 2 are M/M pairs, and 2 are F/F pairs: 1) M/m 2) m/M 3) M/F 4) F/M 5) F/f 6) f/F Given the knowledge that one sibling is female, you then exclude 2 of the 6 possibilities (the m/M and M/m pairs), to obtain 2/4 = 50% probability that the pair of siblings is of mixed sex. The mistake you're making is that you're including ordering on the mixed-sex pairs, but not including ordering on the same-sex pairs.

mixmax 18y ago

In Jeff's post the question was posed to show how bad humans are at probabilities, and it got the point across very well. I know that I fell right into the trap... This post takes what was meant to be an enlightening and refreshing take on how bad we as humans are at probabilities and turns it into a numbers-duel. No the original questions doesn't clearly state the definition of who is asked, it doesn't take into account that slightly more girls than boys are born, and it doesn't answer the question about whether this all happened in a cult where all boys are killed at birth. It was a simple metaphor. And it worked... Here is a TED talk about the subject of how we are fooled by probabilities: http://www.ted.com/index.php/talks/peter_donnelly_shows_how_stats_fool_juries.html http://www.ted.com/index.php/talks/peter_donnelly_shows_how_...

jerf 18y ago
Really, a (not the, but a) problem is that almost everybody is right. We are certainly fooled by probability. But also, if someone volunteers that one out of their two children is a girl, then it's effectively-100% that the other is a boy, in real life. Slightly amending a quote in a similar context from "Principles of Economics, Explained" (http://www.youtube.com/watch?v=VVp8UGjECt4 http://www.youtube.com/watch?v=VVp8UGjECt4), nobody says "I have a girl. I have another girl. I have another girl." One way of interpreting this perennial debate is as proof of how just how fuzzy English can be. People aren't just arguing about the answer, they argue about the question too.
- randallsquared 18y ago
  "But also, if someone volunteers that one out of their two children is a girl, then it's effectively-100% that the other is a boy, in real life." Well, yes, because saying one is implies that the other isn't, in speech. But we could talk about sets of two of anything that has a 50% probability.
- mixmax 18y ago
  Really, a problem is that almost everybody is right. Yes - this is why I pointed out that in Jeff's post this riddle was used as a metaphor. Otherwise we could spend all our time bickering about the correct spelling of "colour" or other trivialities instead of addressing the interesting questions. The interesting question in Jeff's post was the poor human understanding of probabilities.

cedsav 18y ago

Manolis comments on Paul's post summed it up pretty well for me. What feels counter intuitive is that announcing the gender of one child seems to increase the chances of the other child being of a different gender, but it's actually the opposite. "Family with 2 children" -> chance of having at least one boy: 75% "but I have at least one girl" -> chance of having at least one boy: 66% "but the first one is a girl" -> chance of having at least one boy: 50% "but they're both girls" -> chance of having at least one boy: 0%

Andi 18y ago

It's all a question about the correct precodition definition. I think people who heard about conditional probability know that the correct answer is "2/3". Most people will just imagine the first precondition "they have two children" and forget the second one. What about going in the opposite direction and simply ask "What is the probability that a mother has a boy and a girl?" I think many people would answer "1/2", too, simply adding the precodition "she has 2 children" ;) The origin for this is - I think - an implicit context people are thinking of without real awareness of it. You think quicker with implicit context.