I've rewritten the question to better illustrate why knowing the name and gender of one of the children does not affect the odds:
Two copper Lincoln pennies are flipped sequentially (and neither lands on edge).
1. What is the probability that both pennies landed heads up?
2. Given that one of the pennies landed heads up, what is the probability that both pennies landed heads up?
3. Given that one of the pennies landed heads up
and was minted in 1943, what is the probability that both pennies landed heads up?
Even with the knowledge that
1943 copper pennies are extremely rare, it is easy to recognize that knowing the date of the penny that we know landed heads up does not affect the probability that that penny was flipped first or second or the probability of the other penny landing heads up.
Now, back to children:
Ignoring the possibility of twins and assuming the probability of any particular child being female to be 1/2, given a family with two children, the possibilities of gender that make up our sample space are:
{BB, BG, GB, GG}.
Each possibility is equally likely, i.e. P(BB) = P(BG) = P(GB) = P(GG) = 1/4.
1. What's the probability of both children being girls?
Only GG satisfies that condition, so the probability is 1/4.
2. Given that one of the children is a girl, what is the probability that both children are girls?
Since we are given the information that one of the children is female, eliminating the possibility of BB, our sample space becomes:
{BG, GB, GG}.
Each possibility is equally likely. Since only GG satisfies our condition of both children being girls, the probability is 1/3.
3. Given that one of the children is a girl named Florida, what is the probability that both children are females. Here, our sample space doesn't change from the previous question. (Note, this is where the author of the problem,
Leonard Mlodinow, erred in his logic.) To understand why, it might be helpful to consider each possibility in the sample space to be a subset of the sample space:
BG = {BfGf BfGnf BnfGf BnfGnf};
GB = {GfBf GfBnf GnfBf GnfBnf};
GG = {GfGf GfGnf GnfGf GnfGnf},
where f implies the child has the name Florida, and nf implies not named Florida. So,
{BG, GB, GG} = {{BfGf BfGnf BnfGf BnfGnf} {GfBf GfBnf GnfBf GnfBnf} {GfGf GfGnf GnfGf GnfGnf}}.
Each subset of the sample space has a total probability of 1/3. Eliminating the possibilities that do not contain a girl named Florida, we are left with:
{{BfGf BnfGf} {GfBf GfBnf} {GfGf GfGnf GnfGf}}
We can further simplify the sample space by disregarding the name of the possible male child, leaving:
{{BGf} {GfB} {GfGf GfGnf GnfGf}}
Eliminating possibilities within the subsets of the sample space does not affect the total probability of any subset, only the possibilities within that subset. (In fact, we use the same logic to correctly solve the
Monty Hall problem.) So, we have:
P(BGf) = 1/3
P(GfB) = 1/3
P(GfGf)+P(GfGnf)+P(GnfGf) = 1/3
Since only the possibilities in the last subset of our sample space contains two girls, the probability of both children being girls, given one child is a girl named Florida, is 1/3.