Tversky and Kahneman eye color problem There is a probability problem in Causal Schemata in Judgements Under Uncertainty (1977) by Amos Tversky and Daniel Kahneman that goes like this:
Problem 1: Which of the following events is more probable?


*

*That a girl has blue eyes if her mother has blue
eyes 

*That the mother has blue eyes, if her daughter has
blue eyes

*The two events are equally probable 


The authors state that "since the distribution of height or eye color is
essentially the same in successive generations , the correct answer
to both problems is 'Equal'". The logic is simple: we are comparing $\Pr[A | B]$ and $\Pr[B | A]$. Under the assumption that $\Pr[A] = \Pr[B]$ (i.e. the distribution of eye color is the same in successive generations), we must have $\Pr[A | B] = \Pr[B | A]$.
Is it really that simple?  This problem has always bugged me because it does not describe how mothers and daughters are sampled. Consider the following population:


*

*Mother $M_1$ has blue eyes. She has three daughters: $D_{11}$ (blue eyes), $D_{12}$ (blue eyes), and $D_{13}$ (green eyes)

*Mother $M_2$ has green eyes. She has a single daughter, $D_{2}$, also with green eyes.


That's it -- that's the entire population. The fraction of mothers with blue eyes is $\frac{1}{2}$. In the daughter's generation, the fraction with blue eyes is $\frac{2}{4} = \frac{1}{2}$, i.e. the assumption that "the distribution of eye color is the same in successive generations" holds, assuming we weight uniformly within each generation.
What about the conditional probabilities $\Pr[\text{mother has blue eyes} | \text{daughter has blue eyes}]$ and $\Pr[\text{daughter has blue eyes} | \text{mother has blue eyes}]$?
Suppose we select a mother uniformly at random, and then select one of her daughters uniformly at random. In that case


*

*$\Pr[\text{mother has blue eyes} | \text{daughter has blue eyes}] = 1$

*$\Pr[\text{daughter has blue eyes} | \text{mother has blue eyes}] = \frac{2}{3}$

*$\Pr[\text{mother has blue eyes}] = \frac{1}{2}$

*$\Pr[\text{daughter has blue eyes}] = \frac{1}{2} \, \frac{2}{3} = \frac{1}{3}$


If instead we sample a daughter uniformly at random (i.e. each of $D_{11}$, $D_{12}$, $D_{13}$, $D_2$ has a 0.25 probability of being chosen), and pair the sampled daughter with her mother, we have


*

*$\Pr[\text{mother has blue eyes} | \text{daughter has blue eyes}] = 1$

*$\Pr[\text{daughter has blue eyes} | \text{mother has blue eyes}] = \frac{2}{3}$

*$\Pr[\text{mother has blue eyes}] = \frac{3}{4}$

*$\Pr[\text{daughter has blue eyes}] = \frac{1}{2}$


Under both sampling assumptions, the two conditional probabilities of interest are not equal.

Question: what assumptions are needed so that "the distribution of eye color is the same in successive generations" implies $\Pr[\text{mother has blue eyes} | \text{daughter has blue eyes}]$ equals $\Pr[\text{daughter has blue eyes} | \text{mother has blue eyes}]$?  Would we need to assume that each mother has exactly one daughter?
In other words, I feel like the original problem is not correctly stated (since it does not rule out the sort of situation I described above). What does it take to fix it?
 A: Great question!  So, 
P(Blue-Eyed Mom AND Blue-Eyed Daughter) = P(Blue-Eyed Mom | Blue-Eyed Daughter) * P( Blue-Eyed Daughter) = P( Blue-Eyed Daughter | Blue-Eyed Mom ) * P( Blue-Eyed Mom )
If P( Blue-Eyed Mom ) = P( Blue-Eyed Daughter ), then the conditional probabilities should, indeed hold.
However, I think your example excellently illustrates how P( Blue-Eyed Mom ) = P( Blue-Eyed Daughter ) is not a simple matter of counting!
These have to be equal regardless of how we sample them.  So one way to guarantee that is (as you stated!) enforce a one-child policy.  However, I think it would also work with an n-child policy (i.e. as long as every mother had the same number).
Perhaps, we could get away with this being approximately true with a very large number of Moms and some number of Daughter variance.  I ran a quick simulation with 10,000 Moms, each having 1 to 3 daughters (uniform distribution), with a 1/3 chance of blue eyes among Moms and Daughters and a 1/2 chance of Blue implying Blue from a Mom to Daughter perspective.  It looks like the rule held up reasonably well.  On 4 runs the M|D s where 49.08 to 50.66%
