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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?

  1. That a girl has blue eyes if her mother has blue eyes
  2. That the mother has blue eyes, if her daughter has blue eyes
  3. 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?

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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%

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  • $\begingroup$ Thank you for this answer. I like your point about an n-child policy (+1). When the number of children is random (but independent across mothers and independent of eye color), can you prove that Pr[A|B] and Pr[B|A] will converge as the population size increases? That seems to be what's happening in your simulation, right? $\endgroup$
    – Adrian
    Commented Oct 26, 2016 at 7:19
  • $\begingroup$ Thank you! Yes, I believe that's what's happening, but I can't think of how to prove it. $\endgroup$
    – MikeP
    Commented Oct 26, 2016 at 13:27

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