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 ${1 \over 2}$.
I can't read Czech, but the answer is indeed 1/2.
Here's the explanation. One child, Jezebel, is a girl. Let the other child be B. Since the probability of a child being boy or girl is 50/50, and since the sex of one child is independent of the sex of the other child, the probability that B is a girl is 50%.
(Both these statements are idealizations; in the real world, they are not perfectly true.)
The point of this question is to contrast with the other example: the probability that both children are girls (given that one is a girl). The best way to see the difference between the two examples is to make a table (see below)
Let's name the two children A and B, since Jezebel is not a neutral name. I believe the point of mentioning Jezebel is not to consider the probability of parents choosing this name. Rather, the idea is to contrast knowing that a specific child is a girl with knowing only that one of the children is a girl. So I think it makes the issues clearer if we label the children A and B rather than Jezebel and 'unnamed other child'.
Here's the table: $$\begin{array}{|c|c|c|c|}\hline\\ A & B & \text{one child is a girl} & \text{A is a girl}\\ \hline G & G & \text{true} & \text{true} \\ G & B & \text{true} & \text{true} \\ B & G & \text{true} & \text{false} \\ B & B & \text{false} & \text{false} \\\hline\end{array}$$
If we are told that one child is a girl, then we're restricted to the first three rows. Since the other child is a girl in one out of the three rows, it follows that the probability that the other child is a girl is 1/3. On the other hand, if we're told that A is a girl, that restricts us to the top two rows. In one of the two, the other child is a girl, so the probability is 1/2.
I can't resist adding the relation with Bose-Einstein statistics. Instead of two children, say we have two particles. Instead of boy and girl, say we have two quantum states, say spin-up and spin-down. If the particles are distinguishable and non-interacting (say an electron and a proton, far apart), then the probability that they are both spin-up is 1/4; if we are told that the proton is spin-up, the probability that the electron is spin-up is 1/2. If we're merely told that one particle is spin-up, but not which one, then it's like the boy-girl problem: the probability is 1/3 that the other is also spin-up.
(I had to say "non-interacting" because magnetic effects, for example, can cause the spin of one to affect the spin of the other, if they're close enough.)
But if we have two photons, these are in principle indistinguishable. So you can't name them. So the table now looks more like this: $$\begin{array}{|c|c|}\hline\\ \text{photon-photon} & \text{one photon is spin-up}\\ \hline UU & \text{true} \\ UD=DU& \text{true} \\ DD & \text{false} \\\hline\end{array}$$
If one is spin-up, then we're in the top two rows, and the probability that the other is also spin-up is 1/2. This is even without any direct interaction, like through magnetic forces.
So with classical statistics, if we're given that one particle is spin-up, the probability that the other is also spin-up is just 1/3. But with Bose-Einstein statistics, the probability rises to 1/2.
In general, if you have a bunch of indistinguishable particles of the type known as bosons, they "like" to get into the same quantum state. Photons are bosons.
For another type of indistinguishable particle, so-called fermions, two particles can't get into the same quantum state (the famous Pauli exclusion principle).
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 $1/2$.
