# Probability question re: boy/girl riddle **URL:** \ **Category:** Factual Questions\ **Created:** [July 24, 2008, 11:25pm UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676 "2008-07-24T23:25:02Z")\ **Posts on this page:** 18\ **Page:** 2
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 25, 2008, 7:35am UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/21 "2008-07-25T07:35:32Z")
Sure, in some sense, although that’s not really it. In another sense, there’s very specifically only one child whose sex we don’t know: the child who didn’t answer the door. Though, we don’t know whether that child is the older or the younger one… Symmetrically, though, we might consider the case where we know of a family that the older child is a son, but we don’t know what the gender of the younger child is, nor whether the older child is the one who just left to answer the door or if that was the younger child. There’s nothing distinguished about categorizing children according to “older”/“younger” vs. “door-answerer”/“non-door-answerer”. The key difference between the OP’s case 2 and case 1 is that there is _more_ information available in case 2: the information about the gender of who answered the door, on top of the information that the family has at least one son. Concerns about rigid designators and so on are an orthogonal issue. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 25, 2008, 7:53am UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/22 "2008-07-25T07:53:36Z")
To perhaps illustrate it: suppose you already know of Family X that they have two children, and at least one son (but know nothing more). Before ringing their doorbell, then, conditionalizing on your available information (against the conventional distribution of families), the probability that they have two sons is 1/3. After ringing their doorbell, suppose a son shows up to answer the door. Have you learnt nothing new, since you already knew they had at least one son? No, you’ve learnt plenty new; now, the probability that they have two sons is 1/2[sup]\*[/sup]. The fact that it was a son who answered the door rather than a daughter is significant new evidence to conditionalize upon, even though you already knew there was a son in the house. And nothing in this reasoning involves any invocation of relative ages. \*: Specifically, we can write it out using Bayes’ Theorem as P(two sons | son answers door) = P(son answers door | two sons) \* P(two sons)/P(son answers door) = 1 \* (1/3)/P(son answers door) = (1/3)/[P(son answers door | two sons) \* P(two sons) + P(son answers door | only one son) \* P(only one son)] = (1/3)/[1 \* 1/3 + 1/2 \* 2/3] = 1/2. [This deduction is based, of course, on the model where one of the two children is selected at random to answer the door] ---
**Author:** ![Princhester](https://avatars.discourse-cdn.com/v4/letter/p/3e96dc/32.png) [@Princhester](https://boards.straightdope.com/u/Princhester)\ **Post date:** [July 25, 2008, 8:37am UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/23 "2008-07-25T08:37:36Z")
I think you miss my point. In the post to which I was responding, **Tazman** was talking about the first situation and suggesting there were two options. I was pointing out that one of those options was really two ie three total. Sorry if I didn’t explain myself in detail. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 25, 2008, 8:54am UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/24 "2008-07-25T08:54:52Z")
My apologies; I think I did misinterpret your intent. I’d say, yes, there’s three options, but the OP’s not wrong that there are two options too. And his confusion is the question as to _why_ he should split one of his options up into two, when, in other cases, he’d be perfectly fine to leave it as it is. I think the issue of “How many possibilities?” has been suitably addressed already, but I’d point out, also, that in the OP’s first case (“the riddle”), we are not primarily told of any designated child that they are male. That is, we are told “Family X has at least one son”, rather than “The child of Family X who [is oldest/answered the door/whatever…] is male”. I guess this may have been your point; that there is just information about the family, but not about any child in particular. (This sort of thing is why I take issue with a traditional, but cheatingly broken, phrasing of the riddle: “Family X has at least one son. What’s the probability that their other child is a boy?”. The incoherence of this type of phrasing is illustrated by thinking about the analogous “Adam and Eve have at least one son. Is their other child a murderer?” or “The famed Nintendo plumber siblings include at least one male. Does the other one wear red?” or so on.) ---
**Author:** ![don\_t\_ask](https://avatars.discourse-cdn.com/v4/letter/d/e68b1a/32.png) [@don\_t\_ask](https://boards.straightdope.com/u/don_t_ask)\ **Post date:** [July 25, 2008, 9:31am UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/25 "2008-07-25T09:31:25Z")
It’s the equivalent of something like this. Your best friend invites you to a Christmas party at a workmates place. You figure if you go you will take a token gift for any kids so you ask if the workmate has any kids. “Yeah two.” “Boys, girls or one of each?” “I don’t know but one is named Dave so he’s a boy. The other is Tony or Toni, don’t know which though. But at least one boy.” When you arrive there a child answers the door and is obviously a boy. So it’s not Toni. If he is Dave then there is still only a 50:50 chance that the kid you haven’t met is Toni. It’s just as likely that the unmet kid is Tony. But if it is Tony standing in front of you then the other kid is Dave. And there is no Toni. Before you got there the door kid/unmet kid could have been: Dave/Tony…Dave/Toni…Tony/Dave…Toni/Dave seeing the kid at the door eliminated the last combination. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [July 25, 2008, 11:57am UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/26 "2008-07-25T11:57:05Z")
References: [ul] [li][Monty Hall Problem](http://www.straightdope.com/classics/a3_189.html) from Cecil Adams also includes a discussion of the boy/girl problem.[/li][li][comment on On “Let’s Make a Deal,” answer](http://boards.straightdope.com/sdmb/showthread.php?t=459767) (really a discussion of the boy-girl problem)[/li][li][Must Comment on Monty Hall sub riddle/logic trick](http://boards.straightdope.com/sdmb/showthread.php?t=403560) (boy-girl again)[/li][li][Wrong reasoning and/or conclusion in a column](http://boards.straightdope.com/sdmb/showthread.php?t=397008) (boy-girl again)[/li][li][Boy/Girl probability in Monty Hall](http://boards.straightdope.com/sdmb/showthread.php?t=383492) (self-explanatory)[/li][li][Are there any statisticians in the house?](http://boards.straightdope.com/sdmb/showthread.php?t=384757) (ambiguity and phrasing in the boy/girl problem)[/li][li][Monty Hall [edited title]](http://boards.straightdope.com/sdmb/showthread.php?t=366732) (title should have been edited to mantion the boy-girl problem)[/li][li][Monty Hall Question](http://boards.straightdope.com/sdmb/showthread.php?t=357158) (boy/girl question, really)[/li][li][Monty Hall Problem](http://boards.straightdope.com/sdmb/showthread.php?t=2095) (segues into boy/girl problem; scroll down)[/li][/ul] ---
**Author:** ![NineToTheSky](https://avatars.discourse-cdn.com/v4/letter/n/b5a626/32.png) [@NineToTheSky](https://boards.straightdope.com/u/NineToTheSky)\ **Post date:** [July 25, 2008, 12:56pm UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/27 "2008-07-25T12:56:04Z")
[QUOTE=don’t ask] Before you got there the door kid/unmet kid could have been: Dave/Tony…Dave/Toni…Tony/Dave…Toni/Dave seeing the kid at the door eliminated the last combination. [/QUOTE] My problem with this (and the other similar explanations) is that I really don’t get the difference between Dave/Tony and Tony/Dave. As they’re both boys, how does the order difference make a difference. As we already know there is one boy (Dave), that leaves only one other choice: boy (two boys) or girl (one of each.) In total: two choices. ---
**Author:** ![vetbridge](https://avatars.discourse-cdn.com/v4/letter/v/b5e925/32.png) [@vetbridge](https://boards.straightdope.com/u/vetbridge)\ **Post date:** [July 25, 2008, 1:06pm UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/28 "2008-07-25T13:06:16Z")
Of course, if the possibility of ambiguous genitalia is included, then the riddle becomes tricky. ---
**Author:** ![NineToTheSky](https://avatars.discourse-cdn.com/v4/letter/n/b5a626/32.png) [@NineToTheSky](https://boards.straightdope.com/u/NineToTheSky)\ **Post date:** [July 25, 2008, 1:29pm UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/29 "2008-07-25T13:29:26Z")
Ah! I’ve got it. In [this](http://boards.straightdope.com/sdmb/showthread.php?t=459767) thread, \*\*Zut \*\* says > [@](#): > > 1. In a small city, there happens to be 1000 families with two children. > 2. In half these families, the first child is a male. In the other half, the first child is a female. Likewise, the second child’s sex is split evenly. > 3. That leads to three groups of families: > Group A (250 families) has a female first child and a male second child. > Group B (250 families) has a female first child and a female second child. > Group C (250 families) has a male first child and a female second child. > Group D (250 families) has a male first child and a male second child. > > OK, with me so far? > > 1. Now, some families have a daughter. In fact, 750 have a daughter: groups A, B, and C. > 2. Of those families that have a daughter, how many also have a son? In this case, only groups A and C have a son: 500 families, in other words. > 3. So, you can say, out of those families who have a daughter, 2/3 also have a son. > > The fact that, when unordered, groups A and C are indistinguishable doesn’t reduce the number of families in those groups. and **C K Dexter Haven** > [@](#): > > How about trying it this way, instead of “A” and “B” use youngest and oldest. This stuff about “order” is being misunderstood. Of course there’s an order, if a family has two children, there’s a younger and older. (I will use “youngest” and “oldest” and avoid grammarians.) > Four possibilities: > (1) Oldest child is female and youngest child is male. > (2) Oldest child is female and youngest child is female. > (3) Oldest child is male and youngest child is female. > (4) Oldest child is male and youngest child is male. > > OK, so if you know that one of the children is female, but you don’t know whether it’s the oldest or the youngest, then the only situation that cannot arise is situation (4), and thus the odds are 2/3 that the other child is male. > > The question of “order” changes the scenario. If you know that the oldest child if female, then you’ve eliminated cases (2) and (4) and you’re left with 1/2 that the other child is male. > > In a nutshell, that’s the essential difference between the two scenarios. It’s not whether there IS an order (of course there’s an order), it’s whether YOU KNOW the order. That does it for me. ---
**Author:** ![Fretful\_Porpentine](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/fretful_porpentine/32/2873_2.png) [@Fretful\_Porpentine](https://boards.straightdope.com/u/Fretful_Porpentine)\ **Post date:** [July 25, 2008, 1:34pm UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/30 "2008-07-25T13:34:32Z")
OK, think about it this way. If you have no information at all about the gender of the kids, there are four equally likely possibilities: 1. BB 2. BG 3. GB 4. GG Since we know one of the kids is a boy, it’s not #4. The other three, however, are still all equally likely. In only one of these cases are both kids boys. ---
**Author:** ![NineToTheSky](https://avatars.discourse-cdn.com/v4/letter/n/b5a626/32.png) [@NineToTheSky](https://boards.straightdope.com/u/NineToTheSky)\ **Post date:** [July 25, 2008, 1:44pm UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/31 "2008-07-25T13:44:34Z")
[QUOTE=Fretful Porpentine] OK, think about it this way. If you have no information at all about the gender of the kids, there are four equally likely possibilities: 1. BB 2. BG 3. GB 4. GG Since we know one of the kids is a boy, it’s not #4. The other three, however, are still all equally likely. In only one of these cases are both kids boys. [/QUOTE] Thanks for your reply (I think our posts crossed), _but_ I’m afraid that your answer doesn’t do it for me, because my stumbling block is the difference (or lack of it) between 2 and 3. What does clear it up for me is this part of **Zut** ’s post > [@](#): > > 1. Now, some families have a daughter. In fact, 750 have a daughter: groups A, B, and C. > 2. Of those families that have a daughter, how many also have a son? In this case, only groups A and C have a son: 500 families, in other words. > 3. So, you can say, out of those families who have a daughter, 2/3 also have a son. > > The fact that, when unordered, groups A and C are indistinguishable doesn’t reduce the number of families in those groups. ---
**Author:** ![The\_Hamster\_King](https://avatars.discourse-cdn.com/v4/letter/t/8edcca/32.png) [@The\_Hamster\_King](https://boards.straightdope.com/u/The_Hamster_King)\ **Post date:** [July 25, 2008, 4:20pm UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/32 "2008-07-25T16:20:02Z")
[QUOTE=NineToTheSky] Thanks for your reply (I think our posts crossed), _but_ I’m afraid that your answer doesn’t do it for me, because my stumbling block is the difference (or lack of it) between 2 and 3. [/QUOTE] Think about it this way. Assume you have no information. Then the probabilities break down like this: BB – 25% BG – 25% GB – 25% GG – 25% Now we can combine BG and GB because for our purposes we don’t care about order: BB – 25% GG – 25% One of each – 50% If you learn that at least one of the kids is a boy that eliminates GG. But note that “one of each” is twice as likely as “boy-boy”. That ratio is unaffected by eliminating “girl-girl”. So if we renormalize: BB – 33% One of each – 66% ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [July 25, 2008, 4:36pm UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/33 "2008-07-25T16:36:10Z")
Generally I agree with the consensus here. However, I can see one hypothetical case where “At least one is a boy” does not lead to 2/3 odds. It’s a bit like the Monty Hall case, where Monty Hall knows which door hides the prize. If the person giving the information is known to be equally likely to say, with a given family, that at least one is a boy OR that at least one is a girl, then: (1) With BB, they will say “At least one is a boy.” (2) With GG, they will say “At least one is a girl.” (3) In the other case, they are equally likely to say either. Then, if they say “At least one is a boy”, then there is a 50/50 chance of the other being a girl. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 25, 2008, 6:11pm UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/34 "2008-07-25T18:11:55Z")
Right. Because in that case, you have more information than just that “At least one is a boy”; you also have the additional information that the semirandom selection process transpired so as that the person giving the information came to tell you that at least one is a boy. Knowing this second fact goes above and beyond simply knowing the first fact. But, generally, with these problems, you shouldn’t assume some extra process (and corresponding extra information) like that unless told about one. ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [July 25, 2008, 6:31pm UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/35 "2008-07-25T18:31:00Z")
[QUOTE=NineToTheSky] Thanks for your reply (I think our posts crossed), _but_ I’m afraid that your answer doesn’t do it for me, because my stumbling block is the difference (or lack of it) between 2 and 3. [/QUOTE] I think it confuses people when some minor detail like which child opened the door, or which child was born first, seems to matter when it comes to probabilities. But those things are just a means of distinguishing between the two children. What matters is that the two children are separate entities. That is what makes BG (child X is a boy, child Y is a girl) a different situation from GB, and that’s why the two cases should be considered separately. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 25, 2008, 6:37pm UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/36 "2008-07-25T18:37:35Z")
Eh. You don’t ultimately have to consider the two cases separately, and I am wary of fuzzy concern over invocations of separate entities or such (going with the door-answering version, the two children are door-answerer and non-door-answerer, and the only two respective gender possibilities are BB and BG, since the door-answerer is known to be a boy. No GB here… And yet, if we stipulate the way door-answering works is that a girl cannot answer the door if there is a boy around, and the eldest eligible child always answers the door, then the probability of two boys is still 1/3, even though there are only those two possibilities). I think Pochacco put it well in post [#32](http://boards.straightdope.com/sdmb/showpost.php?p=10038226&postcount=32); you can categorize possibilities however you like, and ignore whatever differences you don’t care about. However, the question remains of how to determine the probability of each possibility (as they may not all be equally likely). One way to do so is to start off with a situation where all possibilities are known to be equiprobable, and then start coalescing them into the coarser categories one cares about. ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [July 25, 2008, 7:13pm UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/37 "2008-07-25T19:13:02Z")
I’m not sure what you’re getting at. I confess that I found your first paragraph rather difficult to understand. But I was just trying to illustrate why BG is different from GB. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 25, 2008, 7:21pm UTC](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676/38 "2008-07-25T19:21:03Z")
What I mean is that BG is different from GB the way BG-with-boy-born-in-June is different from BG-with-boy-born-in-December. It’s not _wrong_ to say that the difference isn’t relevant, and to collapse the two into a single case; after all, if all you knew was “The family has one boy and one girl”, but not which was child X and which was child Y or whatever, you’d still know enough to answer the question “Does the family have two boys?”. One might make the mistake of unwarrantedly assuming that that particular single case (“The family has one boy and one girl”) is as likely as other particular single cases, whatever one takes them to be (e.g., “The family has two boys”, “The family has two girls”). But it’s not, in itself, a mistake to treat it as a single case and not care about the distinction between BG and GB. [Previous page](https://boards.straightdope.com/t/probability-question-re-boy-girl-riddle/457676.md?page=1)