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You call to someone's house and asked if they have two children. The answer happens to be yes. Then you ask if one of their children's name a William. The answer happens to be yes again.(We assume William is a boy's name, and that it's possible that both children are Williams) What's the probability that the second child is a boy?

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closed as off-topic by Jam, KReiser, NCh, Saad, choco_addicted Dec 11 '18 at 7:07

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    $\begingroup$ Potentially a couple of interesting caveats to take care of in this formulation of the problem beyond the run-of-the-mill arithmetic. What are your thoughts? Where are you getting stuck? $\endgroup$ – gt6989b Dec 10 '18 at 22:44
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    $\begingroup$ If a couple has two children, what's the probability that both are boys? That exactly one is? $\endgroup$ – Ben W Dec 10 '18 at 22:49
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    $\begingroup$ Well, kudos to you for asking this in an unambiguous manner. This is a famous example but most of the time it is asked in an ambiguous manner. (Namely if one (not specified which) what is the prob of "the other" (not specified which). $\endgroup$ – fleablood Dec 10 '18 at 23:07
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    $\begingroup$ @ebramos I think your formulation looks good. Except, the probability that someone has two children named William is very small. $\endgroup$ – Doug M Dec 10 '18 at 23:53
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    $\begingroup$ Related: math.stackexchange.com/questions/1893041/… $\endgroup$ – Henry Dec 11 '18 at 1:32
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The probability is $\dfrac{1-p}{2-p}$, where $p<0.5$ is the probability that a child’s name is William. Represent the sample space of one child’s gender and name by $I$, a unit interval where numbers from $0$ to $0.5$ represent girls and numbers from $0.5$ to $1$ represent boys. Assume that like names are contiguous within the gender range, so there is a subinterval of width $p$ within $[0.5,1]$ that represents Williams.

enter image description here

The two children correspond to some point $(x,y)$ in $I\times I$. Knowing nothing other than that the family has (exactly) two children puts no restriction on $(x,y)$. However, if you know one child is a William, you must be in the colored cross-shaped region shown in the picture. Within that region, the red area represents one-boy-and-one-girl families, and the blue area represents two-boy families.

The probability that you are in the blue region, given that you are in the cross-shaped region, is the quotient of areas blue/cross, or $\dfrac{p-p^2}{2p-p^2}$.

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  • $\begingroup$ Thanks. To be fair, I based it from an answer I gave a couple of years ago to a variation on the same question: math.stackexchange.com/questions/1582465/… $\endgroup$ – Steve Kass Dec 11 '18 at 0:59
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    $\begingroup$ Well, estimating $p$ would be a great open-ended data science question! Good guesses for $p$ probably depend on when and where the encounter occurred, among other things. But in the real world, the probability that one family has two boys both named William is almost certainly not $p^2$, so this question is not very realistic. $\endgroup$ – Steve Kass Dec 11 '18 at 1:40
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    $\begingroup$ @Steve Kass: I suggested renaming duplicate Williams, not removing their whole families! In the case of maximum p that will mean every family with a boy has a William and 1/3 of them have another boy. You can't just remove the little square from your diagram as the total area will no longer be 1. Assuming two-William families only rename one of them, the measure of the little square will be redistributed over the rest of the blue cross but the total measure of the blue region will not change. Nor does the red region, so the final ratio is also unchanged. However ... $\endgroup$ – David Hartley Dec 12 '18 at 17:33
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    $\begingroup$ there are two problems with that argument. First, the answer has supposedly not changed, but it's a function of p, and p has changed since the overall number of Williams has been altered. Second, if duplicate Williams have their names changed, then so should duplicate other names. Some of those could get renamed William and that alters everything. I should have stuck with my original approach to the case where two Williams in the same family is not allowed, which won't fit in this comment ... $\endgroup$ – David Hartley Dec 12 '18 at 18:14
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    $\begingroup$ Suppose two William's in the same family is not allowed. Let p be the probability that a boy who doesn't already have an elder brother called William, is named William himself. Simple use of Bayes' theorem gives the answer $(2-p)/(4-p)$ to the original question. This matches the answers which allow duplicates, in that p corresponds to 1/m in fleablood's answer and to 2p in Steve's and Jam's. That's what prompted my original comment that barring duplicates made no difference. You get the same expression in p, but of course p is defined in a different way. $\endgroup$ – David Hartley Dec 12 '18 at 18:28
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For sake of argument suppose $1$ in $m$ boys (but no girls) are named william. And for sake of argument in every family has two children and one of the children is soaked in skunk urine to tell it apart for the other.

$\frac 12*\frac 12$ of all families have two girls. None of them named william.

$\frac 12*\frac 12=\frac 14$ of all families has one girl soaked in skunk urine and a clean boy.

So these families $\frac 1m$ of them have the boy named william.

So $\frac 1{4m}$ have a skunk urine girl and a boy named william and the other a girl.

$\frac 12*\frac 12=\frac 14$ of all families has one girl clean and one boy soaked in skunk urine. Of these $\frac 1{m}$ have a boy named william.

So $\frac 1{4m}$ has skunk urine boy named william and the other a girl.

$\frac 12*\frac 12 = \frac 14$ of all families have two boys.

$\frac 1{m^2}$ or $\frac 1{4m^2}$ of all families, of these both are named william.

$\frac 1{m}\frac {m-1}m$ or $\frac {m-1}{4m^2}$ of all, of these the skunk urine one is called william and the other isn't.

$\frac {m-1}m\frac 1{m}$ or $\frac {m-1}{4m^2}$ of all, of the the clean boy is called william that the other isn't.

So $\frac 1{4m^2} + \frac {m-1}{4m^2} + \frac {m-1}{4m^2}= \frac {2m-1}{4m^2}$ of all families has a child name william and the other a boy.

And $\frac 1{4m}+\frac 1{4m}=\frac 1{2m} = \frac {2m}{4m^2}$ of all families have a child named william and the other a girl.

And there are $\frac 1{4m^2} + \frac {m-1}{4m^2} + \frac {m-1}{4m^2}+\frac 1{4m}+\frac 1{4m} = \frac {4m -1}{4m^2}$ of all families has a child named william.

So the probability of a family with two children, one william, having two boys is

$\frac {\frac {2m-1}{4m^2}}{\frac {4m-1}{4m^2}} = \frac {2m-1}{4m-1}\approx \frac 12$ (depending on how rare william is as a name)

and the probability of having a boy and a girl is $\frac {2m}{4m-1}\approx \frac 12$.

....

There is a well known paradox that if a family with two children have at least one boy what is the probability that "the other" is a boy, or in other words what is the probability that they have two boys.

The answer is $\frac 13$. This is because of the four possible outcomes, BB, BG, GB, GG, the GG is thrown away. SO of the three equally likely outcomes only one is $BB$.

But this question is worded differently. we are told specifically that a certain one of them is a boy, not just any of them.

(Let's assume they aren't both named william).

So of the four possibilities: William is a Boy: Boy, William is Boy: Girl, William is a girl:Boy, William is girl; Girl. Now 2 of them are thrown away and only William is a Boy: Boy, William is Boy: Girl are left. So the probability is now $\frac 12$.

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These questions are actually a lot of fun. Thank you for posting this!

A related question to this question is this:

You call households that have exactly two children. You ask whether one child is a boy. What is the probability that the second child is a boy.

Let A be the event that both children are girls, B the event that the first-born is a boy, the second-born a girl, C the event that the first-born is a girl the second-born a boy. and D the event that both children are boys. Then Events A,B,C,D are mutually disjoint and each occur with probability $\frac{1}{4}$.

Let us assume that, instead of asking if one child is a boy, you ask if the first-born is a boy. They will say yes if either Events B or D occured. Then the probability that the second-born is a boy, given that the first-born is a boy (i.e., it was Event D that occurred given that one of D or B occurred), is 1/2.

But you are actually calling and asking if one child is a boy, and they will say yes if Events B,C, or D occurred. The only households that will say no are those for which event A occurred. So the probability that the other child is a girl, given that one child is a boy, is the probability of either Event B or Event C happening given that one of Events B,C,D happened. So the probability that the other child is a girl is 2/3.


However, the above question is not equivalent to the original question. That only a small fraction William (as in about every other name). Here households with two boys will say yes at almost twice the rate that households will say yes to the question of "do you have a William", instead of at the same rate before. So as the fraction $p$ of Williams goes to 0, the probability that the 2nd child is a boy goes to 1/2. The calculations were already done in the other answers.

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    $\begingroup$ Unless William is a very common name, or strangely distributed, a family with two boys is roughly twice as likely to have one called William as a family with one boy. So among the families that can answer "yes" to "do you have a William?" there are (roughly) twice as many with two boys as there are among all families with two boys. Of course, overall families with one boy are twice as likely as those with two, so these factors cancel to give a roughly even chance in this case. Hence the result of about 1/2, as the other answers agree. (i.e for small values of p.) $\endgroup$ – David Hartley Dec 11 '18 at 9:45
  • $\begingroup$ Hi @DavidHartley you are correct. I am editting my answer $\endgroup$ – Mike Dec 11 '18 at 16:25
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The crux of this problem is that the probability changes whether we find out that one of the children is William before or after we draw the family. See this paper for details. Assume that both sexes are equiprobable and that the probability of a child being a William is $p$. Denote Williams by $BW$ and male non-Williams by $B'$.

$$\begin{array}{|c|c|c|} \hline \text{Child 1} & \text{Child 2}&\text{P(Both Children)}\\ \hline BW&BW &p^2\\ \hline BW&B'&p\left(\frac12-p\right)\\ \hline BW&G&\frac12p\\ \hline B'&BW&p\left(\frac12-p\right)\\ \hline B'&B'&\left(\frac12-p\right)^2\\ \hline B'&G&\frac12\left(\frac12-p\right)\\ \hline G&BW&\frac12p\\ \hline G&B'&\frac12\left(\frac12-p\right)\\ \hline G&G&\frac14\\ \hline \end{array}\\ $$

In this space, we can see that the probability of child $2$ being a boy after we've found out that child $1$ is a William is $P(B|W)=\frac{P(B\land W)}{P(W)}=\frac1p\frac{p^2+p\left(\frac12-p\right)}{p^2+2p(1/2-p)+p+(1/2-p)^2+(1/2-p)+(1/2)^2}=\frac12$.

However, if we knew that the family had a William but not specifically that child $1$ was a William, we would have the following space.

\begin{array}{|c|c|} \hline \text{Child 1} & \text{Child 2} &\text{P(Both Children)}\\ \hline BW&BW&p^2\\ \hline BW&B'&\left(\frac12-p\right)p\\ \hline BW&G&\frac12p\\ \hline B'&BW&\left(\frac12-p\right)p\\ \hline G&BW&\frac12p\\ \hline \end{array}

So then the probability of the other child being a boy becomes $\frac{p^2+\left(\frac12-p\right)p+\left(\frac12-p\right)p}{p^2+\left(\frac12-p\right)p+\frac12p+\left(\frac12-p\right)p+\frac12p}=\frac{1-p}{2-p}$, which agrees with @Steve_Kass's answer. This is a variant of the well known paradox whose solution depends on the exact phrasing of the problem. In essence, the prior knowledge of changes the system.

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  • $\begingroup$ But you are only taking cases of families with two children with one named william. Families with a boy named william and a sister are twice as common as families with a boy names william and a brother. (think about it). So the probability is 1/3. $\endgroup$ – fleablood Dec 10 '18 at 23:34
  • $\begingroup$ Argh. I'm mistaken. $\endgroup$ – fleablood Dec 10 '18 at 23:50
  • $\begingroup$ @fleablood So was I, to be fair. $\endgroup$ – Jam Dec 10 '18 at 23:53

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