# Which of these two methods provides the correct answer for this probability riddle?

First, I know this riddle has been asked (many times) before. The question I want answering is why is a tree diagram not a correct method for determining the probability in this case.

There are two children, equally likely to be Boy or Girl. If we know one (or more) is a Boy, what is the probability that there is a Girl in the pair?

The sample space looks like this:

BB
BG
GB
GG - not possible

Therefore the probability is 2/3, but if I draw a tree diagram: The probability of a girl in the pair seems to be $$\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$$

Why does a tree diagram fail?

Your probiabilities are not correct. Note that if we call the two children $X_1$ and $X_2$, then \begin{align*} \def\P{\mathbf P}\P[X_1 = B \mid X_1 = B \vee X_2 = B] &= \frac{\P[X_1 = B]}{\P[X_1 = B \vee X_2 = B]}\\ &= \frac{\frac 12}{\frac 34}\\ &= \frac 23 \end{align*} So, the probiabilities at the first node should be $\frac 23$ and $\frac 13$, respectively, giving $$\frac 23 \cdot \frac 12 + \frac 13 = \frac 13 + \frac 13 = \frac 23$$ in total.
In your tree diagram, the $G-B$ link should also have a probability of $\dfrac12$,
and the $G-G$ link should be crossed out.
So $Pr = \dfrac{1/4 + 1/4}{3/4} = \dfrac23$