Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to understand the boy and girl paradox. The paradox states that if a family has two children and one of them is a boy, then the probability of the other being a girl is 2/3. When you write out the set of possible outcomes { bb, bg, gb, gg } it makes a little more sense. My question is why does age/order matter? The two possible outcomes boy/girl and girl/ boy are the same right?

share|improve this question

marked as duplicate by M Turgeon, user127.0.0.1, TMM, T. Bongers, egreg Feb 2 at 0:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

@ChristianBlatter It's even more complicated than that, because it depends on how and why you were told this information. (To be more precise, it depends on what the probability that you would be told this information in various situations, even if you assume that you would not be told it if it were false.) –  Trevor Wilson Sep 22 '12 at 19:25

7 Answers 7

It's not really the case that age or order matters. Suppose that the children were twins, and were somehow born in parallel rather than in series. Forgetting about the information that there is at least one boy, there are exactly three distinct possibilities, "two boys", "two girls", and "one of each". However, the mere fact that there are three possibilities does not imply that all three possibilities have the same probability of $1/3$.

To give another example of this, if I flip a coin it will either come up heads, come up tails, or land balanced on its edge. However, it would be a mistake to conclude that these events all have probability $1/3$.

share|improve this answer

The sample space is $\Omega := \{(B,B), (B,G), (G,B), (G,G)\}$, where the 1st element of each pair denotes the gender of the first child, and the 2nd element denotes the gender of the second child. Let us assume that each pair is equally likely, which is a reasonable assumption.

If you tell me that one of the children is a boy, then the sample space is reduced to

$$\Omega' := \{(B,B), (B,G), (G,B)\}$$

after I incorporate the information you gave me ("One of the children is a boy"). Since it is reasonable to assume that all three pairs are equally likely, the probability that the other child is a girl is given by

$$\mathbb{P} \left(\{(B,G), (G,B)\}\right) = \mathbb{P} \left(\{(B,G)\}\right) + \mathbb{P} \left(\{(G,B)\}\right) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$$

share|improve this answer
What if I've been shown the boy? This implicitely makes him child #1 and the other child is child #2. This makes the sample space ${(1:B, 2:B), (1:B, 2:G)}$. It seems to be a matter of definition and the real-world sitation allows for both interpretations? –  Gerenuk Sep 22 '12 at 20:22
@Gerenuk: By "first child" and "second child" I meant "1st child that was born" and "2nd child that was born", i.e., I was referring to chronological order. The order in which you observe them does not change their dates of birth ;-) –  Rod Carvalho Sep 22 '12 at 20:27
I wasn't actually refering to someone's notion of first and second :) I was merely suggesting that instead of tell "one is a boy" the situation could be that I'm shown a boy and told there is another child behind the door. In that case I'd conclude there is a 0.5 change. Why is that discrepancy? –  Gerenuk Sep 22 '12 at 22:20
@Gerenuk: One picks the uniform probability mass function (pmf) because it maximizes entropy. Thus, if we have no information at all, we assume a uniform pmf over $\Omega$. If you tell me that one of the children is a boy, then one of the points in the sample space is eliminated and the pmf is updated as we assume a uniform pmf over $\Omega'$. This is a Bayesian interpretation of the problem. Would be interesting to think of a Frequentist interpretation. –  Rod Carvalho Sep 22 '12 at 22:28
I think it's more important to prove alternatives wrong. You only repeated your argument. Can you explain what is wrong with my notation where it directly and naturally follows that the sample space is different to start with? I propose that the very start of proposing a BB,BG,GB,GG sample space might be incorrect (probably depending on the method the samples where determined!). –  Gerenuk Sep 23 '12 at 8:47

My question is why does age/order matter?

The order bg or gb matters, mathematically because when defining all events (outcomes) for a random variable, you must assure that the sum of probabilities for all outcomes must equal 1. And indeed, as you know the probability associated for every outcome is $\frac{1}2*\frac{1}2=\frac{1}4$ and multiplied by $4$ outcomes you get 1.

Intuitively this is the same as a coin toss (2 throws, where e.g. B means heads and G tails).

The two possible outcomes boy/girl and girl/ boy are the same right?

No, the two outcomes are not the same, but they have the same probability.

share|improve this answer

Well given that the children are different entities, you could say that order matters in this problem, hence the fact that $bg$ is not the same as $gb$. Say you have a big sister, and your aunt has an elder boy and a baby girl. Both families have a boy and a girl but the configuration is different. That is why knowing that one of them is a boy, it leaves $3$ possibilities, $2$ of which have girl, while if you know that the elder is a boy, the you only have $2$ configurations left.

share|improve this answer

Try it with coins: First, flip a nickel and a dime. There quite clearly are four possibilities: (N=H, D=H), (N=H, D=T), (N=T, D=H), and (N=T, D=T). Each, also quite clearly, has a probability of 1/4; so the probability of two heads is 1/4, the probability of two tails is 1/4, and the probability of one of each is 1/2.

Then repeat the same experiment with two absolutely identical quarters. There now seem to be three combinations we can discern: (Both H), (Both T), and (one H and one T). But do you really think the odds change to 1/3 for each combination just because the coins are now identical? I certainly hope not. Even though you can't see the difference between the coins, they still are different coins, and the ways that the different combinations can occur still depend on that difference.

When we count possibilities for two-child families, age-order matters only because we have to acknowledge that the two children are different from each other. Even if we aren't given any information that distinguishes them, that difference still matters.

But as for the probability paradox you mentioned, the answer depends on how you "know" that one is a boy. If it is because you asked if the family had any boys, then every BB, BG, or GB family would answer "yes" and the answer is 1/3. But if you learned that fact incidentally, then you could have learned a BG of GB family has a girl. You can only count half of them, and the answer is 1/2.

share|improve this answer

@Andrew-kudwitt , @rod-carvalho , I have a point here! Evidently the gender of one child shouldnt affect the other..May i state that you are wrong in the very first step of choosing your sample space like @gerenuk said.. You " may or may not consider the order" .If you do it correctly, your answer should be same in both the cases.. For instance , if you donot consider order of birth, your sample space is (xB, xG) because you donot care about the observed child! so probability of the other being a boy is 1/2 ! If you do consider the order, ( take the observed boy to be denoted as B0 ) ,then your sample space is (B0 G, B0 B, G B0, B0 B) , SO still the probability of the other child beig a boy is 2/4=1/2...the order and gender of other child doesnt matter..it shouldnt matter.. because one child being a boy doesnot affect the other's gender (excluding the bio stuffs involved, ofcourse)!

share|improve this answer
gerenuk is right! –  Aishwarya Krishnan Sep 11 '13 at 9:03

python script proof for those of little faith/CS background:

from random import randrange

results = {
"one_boy_born_tuesday":{"B": 0, "G": 0}, "one_boy":{"B": 0, "G": 0}, "first_is_boy":{"B": 0, "G": 0}, "first_is_boy_born_tuesday":{"B": 0, "G": 0}

for i in xrange(1000000):
    house = []
    for j in range(2):
        sex = "B" if randrange(2) == 0 else "G"
        day = randrange(7)
        house.append({"sex":sex, "day":day})

    exp = "one_boy_born_tuesday"
    if (house[0]["sex"] == "B" and house[0]["day"] == 3):
        results[exp][house[1]["sex"]] += 1
    elif (house[1]["sex"] == "B" and house[1]["day"] == 3):
        results[exp][house[0]["sex"]] += 1   

    exp = "one_boy"
    if (house[0]["sex"] == "B"):
        results[exp][house[1]["sex"]] += 1
    elif (house[1]["sex"] == "B"):
        results[exp][house[0]["sex"]] += 1   

    exp = "first_is_boy"
    if (house[0]["sex"] == "B"):
        results[exp][house[1]["sex"]] += 1

    exp = "first_is_boy_born_tuesday"
    if (house[0]["sex"] == "B" and house[0]["day"] == 3):
        results[exp][house[1]["sex"]] += 1

for k, v in results.items():
    print "-"*60
    print k
    print v
    print 1.0*v["B"]/(v["G"]+v["B"])

and the results (tada):

{'B': 250451, 'G': 500196}

{'B': 35337, 'G': 35850}

{'B': 250451, 'G': 249837}

{'B': 66134, 'G': 71911}
share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.