# A parent has 2 children. Given that one is a boy, what is the probability that the other is a girl? (Look for specifics below)

So, I've been browsing through the net about this (including this site) and I've stumbling upon different answer. In this site, most of the answer were 2/3 but the arguments here are that BG and GB are different since the order of birth are of consequence.

However, what if one would not consider the order of birth but only the existence of the said children? What is the probability that there is one boy and one girl in a family with a children of 2 given that one is a boy and having a boy as a child is as likely as having a girl as a child?

Same case as another post here, I argue that it is 1/2.

It is not as if I'm omitting the GB and/or BG, but rather, I'm combining them since order is not necessary/important in this scenario. BG = GB. Making the original sample space (with their likeliness); { BB (25%), GB/BG (50%), GG(25%) }. And given the condition (one is a boy), the sample space then is reduced to {BG (25%), GB (25%)} as having a boy child is as likely as having a girl.

Is it an invalid assumption?

• Your last paragraph is confusing. Why did you eliminate $BB$ from your sample space. It is possible under the condition that one of the children is a boy. And furthermore, if the reduced sample space is as you show it (containing only $BG$ and $GB$, then the conditional probability of the other child being a girl would be $100%$, not $50%$. Commented Aug 26, 2016 at 14:27

The sample space is not reduced to $\{BG (25\%), GB (25\%)\}$, the sample space is reduced to $\{BB (25\%), GB/BG (50\%)\}$, or equivalently to $\{BB (25\%), BG (25\%), GB (25\%)\}$. Reduced to this case, the probability of a girl as the other child is $2/3$.

Your assumption, or better, your model is valid: the sample space $\Omega=\{BB,BG,GG\}$ with probability distribution $p$ given by $1/4,1/2,1/4$ is adequate. What would be wrong, would be to assign in this sample space the uniform probability $1/3,1/3,1/3$: this would be a model unfit to describe the empirical data. But from this you can understand why people prefer the model $\{BB,BG,GB,GG\}$ even when there is no concept of order of biths involved in the problem: the point is that they can assign a uniform distribution in that space and this is also the most transparent way to explain why the model $1/3,1/3,1/3$ itself would be inadequate from the point of view of the modeling. In fact, the assignement of the probability to $\{BB,BG,GB,GG\}$ is immediate on the basis of the empirical assumptions of

1) $1/2$ probability to be a girl at each birth

2) independency between different births

In the equivalent model $\Omega=\{BB,BG,GG\}$, the right assignement $1/4,1/2,1/4$ needs some thought (which in practice amounts to reason in terms of the model $\{BB,BG,GB,GG\}$ ).

All very interesting. As soon as we are told one is a boy, we are presented with four possibilities:-

The boy has an older or younger sibling, which can be either a boy or a girl. Each of these outcomes is equally probable (1/4).

Hence the answer is and will always be, given we are told one is a boy, his sibling has a 50% chance of being a girl. QED

• Maybe you should articulate the statement you want to show?
– User
Commented Sep 29, 2017 at 1:16

How does the sex of the given child [boy] affect the probability of the 2nd child. It doesn't. It doesn't matter whether the first child is a boy or a girl. There are still only two possible outcomes for the sex of the unknown child: B or G. The probability is simply 1/2. It's similar to this question: A coin is flipped two times. Given that the first flip is heads, what is the probability that the second flip is heads. Thinking that what already happened will affect what happens next in events of chance is called the Gambler's Fallacy. They could have 15 heads in a row, the probability of the 16th coin being tails is still 1/2.

• While the sex of the first born clearly does not affect the sex of the second born, it deeply affects what we are asked to compute, because it makes the difference between an unconditional probability and a conditional probability. Commented Aug 26, 2016 at 15:10
• That would be true if we knew the first born was the boy. Commented Aug 26, 2016 at 15:11