The Jezebel probability I came across this brain teaser that has two parts.

The first part is already answered here: In a family with two children, what are the chances, if one of the children is a girl, that both children are girls?

The second question is: In a family with two children, what are the chances, if one of the children is a girl called Jezebel, that both children are girls?

SPOILER

The answer is supposedly cca (source [czech])
$(2*2*4*8)/(4*4*4*4)$

There is an explanation, but it sounds like troll logic.
The first part is straightforward, the possibilities are $BG, GB, GG$, hence the probability of both being girls is ${1 \over 3}$.

The second part is also straightforward, assuming that both girls are not named Jezebel. This time let $J$ represent Jezebel and $G$ represent a girl not named Jezebel. Then the possibilities are $BJ, JB, JG, GJ$, hence the probability of both being girls is ${1 \over 2}$.
Well, there are 9 possibilities, where G denotes a girl not called Jezebel:

BB 
BG 
GB 
GG 
BJ 
JB 
GJ
JG
JJ

What are the probabilities of each event? Let's suppose the probability of the parents naming a girl child Jezebel is $p$ if they haven't already got a child called Jezebel, and $q$ if they have. Then the five relevant probabilities are:

BJ : $p/4$

JB : $p/4$

GJ : $(1-p)p/4$

JG : $p(1-q)/4$

JJ : $pq/4$

The sum of these is $p - p^2/4 = p(1-(p/4))$, so the conditional probabilities are:

BJ given $\exists$J : $1/(4-p)$

JB given $\exists$J : $1/(4-p)$

So the probability of them both being girls is $1-(2/(4-p))$ = $(2-p)/(4-p)$. So if $p$ is very small, then the probability is almost $1/2$. If $p$ is very large (i.e. nearly $1$),  then the probability is just over $1/3$. (This is the "one of them is a girl" case).

Since Jezebel is not in the top 1000 girls' names in the US (source: http://www.thinkbabynames.com/meaning/0/Jezebel), $p$ is very small and the probability of them both being girls is almost $1/2$.