Probability Paradox If a family has two children, and you know that one of them is a boy, what is the probability that the other is a boy? I get two different answers depending on what sample space I use. If I say the sample space is all combinations of the genders, i.e. $\{bg, bb, gg\}$, then the probability is $\frac{1}{2}$; since we know one is a boy, the possible outcomes are $bg$ and $bb$ (oh, also assuming that all combinations are equally likely). However, my textbook describes the sample space as permutations of the genders, listing the eldest one first. In this case, the sample space becomes $\{bg, gb, gg, bb\}$, and the probability of the family having two boys now becomes $\frac{1}{3}$. This is really, really confusing me; could anyone shed any light on which one is correct and why?
 A: No. The combination of $BG$ is twice as likely as $BB.$ So the correct answer should be $\frac{1}{3}.$ If we know one is a boy, that could mean either $BG$ or $GB.$ Order matters here, because a boy is not the same as a girl!!
A: K. Jiang's answer is correct, but I wanted to add a few more details.
$\{bg,bb,gg\}$ is a fine sample space of unordered pairs, but the problem is the probability isn't uniformly distributed, or in other words, some of the events have higher probabilities than others(specifically bg is more likely than bb or gg). When we use the formula $Pr(E)=\frac{\text{# of outcomes in E}}{\text{total # of outcomes}}$ there is a caveat, namely that the outcomes are all equally likely. Since that isn't the case in our sample space, we would have to use a different method.
An easy way to get around this is to use a uniformly distributed probability space, such as $\{bg,gb,bb,gg\}$ where the elements are ordered pairs, since each event is equally likely. Then we can use our formula to find the probability.
