# Math(s) can be confusing…

I meet a man walking down the street, who I know has two children. He introduces me to his son, what is the probability that his other child is a girl?

On a first read through the question, the answer seems obvious, $$\frac{1}{2}$$. Sadly, that is not right, it is $$\frac{2}{3}$$.

This is the point where you shout at the screen, tell me I’m an idiot, and storm off. Hopefully I’ll be able to convince you pretty quickly that it is actually $$\frac{2}{3}$$, just consider all the possibilities:

BB          GB          BG          GG

(Where B stands for boy, and G for girl). What we have presented here is all the (ordered) possibilities of having two children. So BB means having two boys, whereas GB means have a female child first and then a boy, whereas BG is the other way round (boy first, then girl).

Given that we know that this man has a son, but no other information (i.e. the son’s ‘order’ (eldest/youngest child)), we can only eliminate from the above list the ‘GG‘ option. Leaving us with:

BB          GB          BG

In which case the probability that the other child is a girl (G) is simply how many times a girl appears as one of those three options, which is 2 out of 3. But be careful:

I meet a man walking down the street, who I know has two children. He introduces me to his eldest child, who is a boy, what is the probability that his other child is a girl?

This is a subtly different question, the key bit of information is we now know that his eldest child is a boy. This now leaves us only with two options:

BB          BG

And so the probability that his other child is a girl is $$\frac{1}{2}$$.

Probability can be confusing if you don’t think the problem through entirely.

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### 4 Responses to Math(s) can be confusing…

1. Neil Tarrant says:

I’m not sure your example works –

Can’t I choose to differentiate the two children anyway I choose? So you said (correctly) ‘He introduces me to his eldest child,’ that changes the probability to being 1/2 – there is nothing special about it being the ‘eldest’ child, it could be the ‘tallest’ child, ‘fattest’ child, or indeed ‘smartest’ child.

Now, I am going to choose to base my probability on listing his children by ‘closest’ child – I initially have four options: BB, BG, GB, GG. As his son is now the ‘closest’ child of his to me, I can exclude GB and GG, therefore the probability of his other child being a daughter is 1/2.

As normally stated, the problem is something of the form of ‘I have two children, at least one of whom is a boy’ – that makes the 1/3 probability correct.

• Sean says:

Ah, fair comment. I think that my example does work because of (the hidden assumption) that by BB, GB, BG, GG we mean the first letter represents the first born child. You are right that there is nothing special about it being the ‘eldest’ child, but it is ‘special’ when considering my ordering which is an order of age.

As per your second comment, you are exactly right, it really does start to screw with people. You provide an excellent example.

2. Neil Tarrant says:

Further to the above, if you have people who are familiar with the original problem, then add information – it really starts to screw with people.

Imagine, I have two PhD research students, at least one of who is a male student, born on a Wednesday. What is the probability that the other is a female student?

We now have 16 possibilities. I’ll only list probabilities for those where at least 1 student is male and born on a Wednesday, it’s superficial to work out the others.

M(W), M(W), P = 1/2 * 1/2 * 1/7 * 1/7 = 1/196
M(W), M(NW), P = 1/2 * 1/2 * 1/7 * 6/7 = 6/196
M(NW), M(W), P = 1/2 * 1/2 * 6/7 * 1/7 = 6/196
[M(NW), M(NW)]
M(W), F(W), P = 1/2 * 1/2 * 1/7 * 1/7 = 1/196
M(W), F(NW), P = 1/2 * 1/2 * 1/7 * 6/7 = 6/196
[M(NW), F(W)]
[M(NW), F(NW)]
F(W), M(W), P = 1/2 * 1/2 * 1/7 * 1/7 = 1/196
[F(W), M(NW)]
F(NW), M(W), P = 1/2 * 1/2 * 6/7 * 1/7 = 6/196
[F(NW), M(NW)]
[F(W), F(W)]
[F(W), F(NW)]
[F(NW), F(W)]
[F(NW), F(NW)]

So P(two males) = (1+6+6) / 196 = 13/196
and P(one male, one female) = (1+6+1+6) / 196 = 14/196

So, the probability of the second student being female is 14/(13+14) = 14/27, or just slightly more than 50%.

3. Jack Shaw says:

Well this rings a bell…

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