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I know this has been answered a dozen times, and I know entirely how to get the 2 possible answers. My question revolves around the phrasing of this question, aren't (1) and (2) the exact same question?

A neighbor of yours has two children. Assuming that the gender of a child is like a coin flip, it is most likely that the neighbor has one boy and one girl, with probability of a half. The other possibilities are a quarter each (either two boys or two girls).

(1) Suppose that you ask the neighbor whether she has any boys, and she said yes. What is the probability that one child is a girl?

(2) Instead, suppose that you happened to see one of her children passing by, and it was a boy. What is the probability that the other child is a girl?

In both questions, it states that we have knowledge of the neighbor having a boy. The first question, she says yes (she has at least one boy). The second, we see at least one boy. Both asks the probability of the other child being a girl. Clearly, as I've read the other questions on this topic, as well as the wiki this is a play on words. So I guess I'm having trouble depicting which one is 2/3 and which is independent of each other and thus 1/2.

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    $\begingroup$ Knowing that the first child you happen to see is a boy is different that knowing that one of the two children is a boy. Similarly, if someone flips a coin twice and you look at one specific result and see H, that is different information than if you learn that one of the results is H. $\endgroup$ – Steve Kass Oct 26 '18 at 22:22
  • $\begingroup$ The first case amounts knowing that at least one child is a boy, resulting in 2/3 chance that the other is a girl. In the second case, you are given that a randomly chosen child turned out to be a boy (NOT "at least one boy" out of two). So the second problem gives probability 1/2 of the other child being a girl (unless there is some unstated reason to think that the seen child was not randomly chosen among the two children). $\endgroup$ – Ned Oct 28 '18 at 19:37
  • $\begingroup$ @Ned The seen child is not random. If gives us the same information as asking. It tells us there is at least one boy on the family. Also, the probability that the woman has one boy and one girl is $\frac{1}{2}$. To say that the probability is still $\frac{1}{2}$ says that seeing the boy gave us no additional information. $\endgroup$ – John Douma Oct 29 '18 at 20:02
  • $\begingroup$ No John, not unless there is reason to believe you are certain to see the boy from a boy-girl pair. You are seeing a random child unless there is more to the story than given. $\endgroup$ – Ned Oct 29 '18 at 22:02
  • $\begingroup$ @Ned I believe you are adding more than given. How could you possibly calculate the probability of seeing one of her sons? Also, how did the observer know it was her son? This observation simply removes the outcome $GG$. Please see my answer for an explanation. By the way, what is the probability that the woman tells the truth in case 1? The information in both cases serves to show that she has at least one boy. $\endgroup$ – John Douma Oct 30 '18 at 4:45
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This problem demonstrates the difference between facts and events. And to understand it, you need to be painfully exact with procedure.

  1. Identify every independent outcome in your sample space, even those that may contradict what you will learn.
  2. Assign a probability to each outcome.
  3. An "event" is a set of outcomes defined by some common property. In these questions, the first event you need is the set of outcomes where you obtain the knowledge that there is at least one boy. It isn't necessarily the set where the fact is true, it is the set where you obtain the knowledge by the procedures described.
  4. The second event is the subset of the first where there is also a girl.
  5. The answer is found by adding up the probabilities for the outcomes in each set, and dividing the sum for the subset in #4 by the sum for the set in #3.

In Question #1, there are four outcomes: BB, BG, GB, and GG. Each has a probability of 1/4. The mother will answer "yes" for the set {BB, BG, GB}, and "no" otherwise. The subset {BG, GB} also has a girl. So the answer is (1/4+1/4)/(1/4+1/4+1/4)=2/3.

In Question #2, there is another factor needed to define the outcomes. Which child did you see? So there are eight outcomes: BB1, BB2, BG1, BG2, GB1, GB2, GG1, and GG2; where the number indicates which child you see. The probability of each is 1/8.

You saw a boy, meaning the set for #3 is {BB1, BB2, BG1, GB2}. Notice how some cases, where the fact "there is a boy" is true, that are not included. The subset where there is also a girl is {BG1, GB2}. The answer is {1/8+1/8)/(1/8+1/8+1/8+1/8)=1/2.

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The second scenario is like you ask "Is your older child a boy?", since you are considering a fixed child. This is why the answer is $1/2$, instead of the $2/3$ you get if you ask "Do you have any boys?"

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  • $\begingroup$ I do not consider the second scenario to be equivalent to asking the gender of the older child, despite the probabilities working out to be the same. The second scenario differs from the first because the likelihood of seeing a boy pass by is different for the 1B1G and 2B cases, whereas the likelihood is the same for the answer to be yes. $\endgroup$ – Dean Oct 26 '18 at 22:30
  • $\begingroup$ These two scenarios are exactly the same. In both cases, we know that the woman has one boy. We don't know if the boy is the oldest or the youngest so there is no difference between the two cases. $\endgroup$ – John Douma Oct 28 '18 at 16:36
  • $\begingroup$ i disagree John. If you flip two coins and one happens to slide off the table and you observe it to be heads, the chances of two heads is 1/2, not 1/3. Not the same as knowing "at least one head", unless you know the head would always the one to slide off from a HT pair. $\endgroup$ – Ned Oct 29 '18 at 21:56
  • $\begingroup$ @Ned $BB$, $BG$, $GB$ and $GG$ are the equally likely outcomes. When we see the boy, which of these are eliminated? $\endgroup$ – John Douma Oct 30 '18 at 4:52
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    $\begingroup$ @JohnDouma That's just not true John. If you flip two coins and randomly pick one to peek at, and it turns out to be H, then the probability of HH is 1/2, not 1/3. The point is, you HAVE eliminated one of HT or TH, but you don't know necessarily which one. Not the same as finding out (precisely) that there is at least one head. My last attempt to explain this: Take an urn with four coins: 1 two-headed, 1 two-tailed, and 2 ordinary. Pick one coin at random, flip it and suppose you see H on top. The probability that the other side of that coin is H is .... 1/2, not 1/3. Draw the tree. $\endgroup$ – Ned Oct 30 '18 at 20:20
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It is not a trick. There are four ways to have two children. They are, as you point out, $BG$, $BB$, $GB$ and $GG$. Each of these is equally likely. We are told there is at least one boy so our sample space is reduced to $BB$, $BG$ and $GB$. In two of these, there is a girl so there is a $\frac{2}{3}$ chance of having a girl.

This part is added to answer the edited question.

They are the same. Think in terms of a coin toss. If I see you toss a head, but I don't know whether it was the first toss or the second, the probability of having tossed a tail is $\frac{2}{3}$. It is only $\frac{1}{2}$ if I know which toss I am seeing. The same is true for the boys. If you knew her oldest was a boy then the probability of a girl would be $\frac{1}{2}$.

There is no difference in this problem.

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  • $\begingroup$ So knowing that at least one of the kids is a boy, gives the possibility of the girl being 66% chance (2/3). That's the answer to the first question, correct? The second question, we see the boy, but that's independent of the other kid, so that's still a probability of 50% for number 2. Am I understand that correctly? $\endgroup$ – Speakmore Oct 26 '18 at 22:26
  • $\begingroup$ @Speakmore No. They are both the same. The possible outcomes are $BB$, $BG$ and $GB$ in both cases because we don't know if the boy we saw was the youngest or the oldest. If we knew he was the youngest, for example, then the outcomes would be $BB$ and $BG$. Then we would get a probability of $\frac{1}{2}$. $\endgroup$ – John Douma Oct 26 '18 at 22:28
  • $\begingroup$ I see, so my initial suspicion was actually correct and they're both 2/3 considering we don't have an order. I guess it was indeed a silly question to ask. Not much to do with probability, more to do with English it seems... $\endgroup$ – Speakmore Oct 26 '18 at 22:33
  • $\begingroup$ @Speakmore, I think you are right to ask the question. Clearly it makes no difference to be told that the child you saw was the oldest! (Or youngest, or tallest, or heaviest, or any other factor). In fact they could be twins! John Douma seems to be confused here. The nature of the data for the two cases are different, in that for case (2), the likelihood of seeing a boy child under the 2B hypothesis is twice that under the 1B1G hypothesis and that explains why the posterior probabilities are different for cases (1) and (2). $\endgroup$ – Dean Oct 29 '18 at 17:18
  • $\begingroup$ @Dean Suppose I visit the woman. I ask if she has any boys? If she answers yes you say that is one probability. If she says nothing but simply points to the boy watching TV, you think that is another probability. Is that correct? Also, the fact that you believe knowing whether or not the boy is the oldest is the same as knowing his weight is disturbing. Saying that it clearly makes no difference is ridiculous. $\endgroup$ – John Douma Oct 29 '18 at 19:07
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Even simple problems like this one are best worked out by using Bayes theorem. In part (2), it is important to realize that the data favors the hypothesis that there are 2 boys compared to the hypothesis that there is just one boy. In part (1) the data does not favor one hypothesis over the other.

For (1): $$ P(1B1G|data) = \frac{P(data|1B1G)P(1B1G)}{P(data|1B1G)P(1B1G)+P(data|2B)P(2B)+P(data|2G)P(2G)}=\frac{1\cdot\frac{1}{2}}{1\cdot\frac{1}{2}+1\cdot\frac{1}{4}+0\cdot\frac{1}{4}}=\frac{2}{3} $$

For (2): $$ P(1B1G|data) = \frac{P(data|1B1G)P(1B1G)}{P(data|1B1G)P(1B1G)+P(data|2B)P(2B)+P(data|2G)P(2G)}=\frac{\frac{1}{2}\cdot\frac{1}{2}}{\frac{1}{2}\cdot\frac{1}{2}+1\cdot\frac{1}{4}+0\cdot\frac{1}{4}}=\frac{1}{2} $$

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  • $\begingroup$ What is the $data$? You are trying to solve the the probability that the woman has a girl given that she has one boy. There is no difference between the cases because seeing the boy and being told about the boy is the same. $\endgroup$ – John Douma Oct 28 '18 at 16:29
  • $\begingroup$ The data for (1) is that the mother says that she has at least one boy. The likelihood of that data is the same for the 1B1G hypothesis and the 2B hypothesis. The data for (2) is that you observe one of the children, and it happens to be a boy. The likelihood of that data is twice as large for the 2B hypothesis than the 1B1G hypothesis. Because the likelihood ratio is different for case 2, the posterior probabilities work out to be different as shown above. $\endgroup$ – Dean Oct 29 '18 at 17:07
  • $\begingroup$ There is no probability associated with the data. The outcomes are $BB$, $BG$, $GB$ and $GG$. In either case we can only eliminate $GG$ because we know there is at least one boy. That leaves three outcomes that are equally likely and so the probability that the woman has a girl is $\frac{2}{3}$. If we knew that the boy we saw was the youngest or the oldest then we could eliminate either $BG$ or $GB$ which would make the probability $\frac{1}{2}$. I hope that clarifies it. $\endgroup$ – John Douma Oct 29 '18 at 18:52
  • $\begingroup$ The concept of likelihood (the probability of seeing data given a hypothesis) needs to be understood. In this problem it is the likelihood ratio that is needed. To make it obvious, suppose the two hypotheses left under consideration are: (Ha) 1000B and 0G; (Hb) 1B and 999G. Getting the answer to the question in case (1) does not help distinguish these possibilities. However, if you happen to see a randomly selected child of the 1000, and you identify the child to be a Boy, that data strongly supports (Ha). This is a basic application of Bayes theorem. The same applies here. $\endgroup$ – Dean Oct 30 '18 at 19:49
  • $\begingroup$ How do you know the probability of seeing the boy given that the woman has one boy and one girl is $\frac{1}{2}$? $\endgroup$ – John Douma Oct 30 '18 at 19:51
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Assuming that the gender of a child is like a coin flip

...

(2) Instead, suppose that you happened to see one of her children passing by, and it was a boy. What is the probability that the other child is a girl?

This is the easier one to answer: according to the assumption, the sex of the one child simply has no bearing on that of the other. If I flip heads, what's the probability that my next flip is tails? Or that my previous flip was tails?

For variety, here's another boy/girl puzzle that can be solved with the same observation. Consider a country where families keep having children until they have a son (so every family has exactly one son). What is the ratio of boys to girls?

This can be worked out through an infinite sum, but the easier approach is to see that the choice of whether or not to continue having children has no impact on the probability of any other child being a boy or girl; the ratio is therefore $1:1$.

Incidentally, the above country is a case where the answer to your question would in fact be different: since every family has exactly one son, then if we see a girl in a family with two children, the chance of the other child being a boy is of course $1$.

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Think about the questions before clicking on the spoiler for the answers. We can model two child families by coins in different ways.


There are four coins in a bag: two are fair, one is double-header, and one is a double-tailer. I select one at random, noting which it is, and secretly toss it, covering it with a cup before you see the result.

I assert that each side has a equal and independent probabilty for being a head; do you agree?

In senario A: I tell you that at least one side is a head. What is the probability that they both are?

There were three coins in the bag with at least one head, one of them has two heads, so $1/3$

In senario B: I lift the cup to show that the up-side is a head. What is the probability that they both are?

The upside has equal chance of being any of the four heads that were in the bag; two of these had a head on the other side of their coin. $1/2$.


This time I have two fair coins and two cups (opaque). I toss each and quickly cover each with a cup.

I assert that each coin has a equal and independent probabilty for showing a head; do you agree?

Senario C: I peek under the cups and tell you at least one coin shows a head. What is the probability that they both do?

$1/3$ as above $\tfrac 14/(\tfrac 14+\tfrac 24)=\tfrac 13$

Senario D: I shuffle the cups, and lift one to fairly reveal that it show a head. What is the probability that they both do?

You know now that this coin shows a head, so the probability that the other does is $1/2$. But it is just happenstance that this coin was the one revealed, so let's check to be sure: They could have both been head with probability $1/4$ and when given that then you would have certainly veiwed a head, or the could have been a head and tail with probability $1/2$ and when given that you could have viewed the head with probability $1/2$. So the conditional probability of them both being heads given that you viewed a head is $\tfrac 14/(\tfrac 14+\tfrac 12\tfrac 12)$ or again $\tfrac 12$.

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Problem 1:

(A) Family has at least one boy. (B) Family has at least one girl.

$ Pr(B|A)=Pr(A\cap B)/Pr(A)=(1/2)/(1-1/4)=2/3 $

Problem 2:

(C) The child you saw is a boy. (D) the other child is a girl.

$ Pr(D|C)=Pr(C\cap D)/Pr(C)=Pr(C)*Pr(D)/Pr(C)=1/2$

, because C and D are independent.

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