# Using Bayesian statistics and inaccurate genetics to guess the eye-colour of my next child

My next child is currently in my uterus. I’ve recently had a scan and found out a little bit about its genetic make-up – that bit which has turned it into a girl. That’s the only characteristic I’ll know about for a while, and I can only speculate on the rest. So I’ve decided to speculate on the colour of its eyes.

In GCSE Biology, which I took in 1994, I learned the following model for how genes determine the colour of eyes. As a simplification, it pretends there are only two types of eyes: brown and blue. and that the eye-colour a person has is controlled by a pair of genes – one gene inherited from their mother and the other from their father. The gene determining eye colour can come in one of two variants, a brown variant, which is labelled ‘B’, and a blue variant, labelled ‘b’.

If both genes in the pair are b (written as bb), then you get blue eyes. If they’re both B (written as BB), you get brown eyes. And the B gene always overrides the blue gene, so if you’ve got one B gene and one b gene (Bb) you get brown eyes too. Because the brown-eyed gene always gets its way whenever it’s present, it’s known as a dominant gene, while the blue-eyed gene is called recessive because it meekly submits to its partner’s wishes.

If you know the genes of a mother and father, you can work out what genes their children are likely to have. If both parents have blue eyes then they’re clearly bb, and their children will get a b from both of them, so they’ll all have blue eyes too. If the mother is Bb, and the father is bb, then half the children will get a B from the mother and a b from the father and be brown-eyed Bbs, and half the children will get a b from both parents and be blue-eyed bbs.

As I mentioned before, this model was taught to me in the nineties and is an oversimplification, as many genes have been shown to contribute to eye colour. I’m not sure that it’s still taught today: on BBC GCSE Bitesize, the less happy examples of cystic fibrosis and Huntington’s disease are used to illustrate the theory of dominant and recessive genes. Until today I believed that to claim that the blue/brown eyed model was correct would be an act of disrespect towards the British Royal Family, as I had thought that the Duke and Duchess of Cambridge had blue eyes and their son Prince George had brown eyes, but now having examined many pictures of the Duchess of Cambridge I can’t decide if her eyes are brown or blue, and on revisiting a discussion page on the subject I have realised that the image of the indisputably blue-eyed woman displayed there is not in fact of the Duchess of Cambridge but a professional look-alike named Heidi Agan. Anyway, I’m going to stick the single-gene theory because nothing in my personal experience has happened to contradict it, and it makes the maths simple.

I have brown eyes, but I don’t know whether I’m a BB or a Bb. Both my parents have brown eyes. I’m assuming that my mother is a BB, as she comes from Thailand and blue eyes are rare or absent in that population. I know that my father is a Bb, as my niece (my brother’s child) has blue eyes, so my brown-eyed brother must be a Bb. My husband also has blue eyes, so he’s a bb. In fact I am the only person in my and my husband’s family whose genome with respect to eye-colour isn’t fully known.

If I were to give birth to a blue-eyed child, then it would be clear that I was a Bb, and that the next child would have a probability of 50% of also having brown eyes. If I’d already had 20 brown-eyed children with my blue-eyed partner, then I’d be pretty confident that I was a BB, as the probability of having 20 brown-eyed children if I were Bb would be $1/2^{20}$, or about 0.000001. So I wondered what the chances of the next child also having brown eyes are, given that already have one child, whose eyes are brown.

This is one of the rare opportunities in my life to apply Bayes’ Theorem. (It’s rare not because Bayes’ Theorem isn’t useful, but because I’m not involved in statistics.) Bayes’ theorem can be written as

$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$

Explanation helps, though. In statistics, the raw materials are observations, such as having a child with brown eyes, and what we want to determine is the underlying probabilities. In the formula above B represents an observation, and A is an underlying probability distribution. In the context of this post, A is the probability that I am BB (say) and B is the evidence that I have a child with brown eyes.

• P(A) is the a priori probability – the probability of being BB before the evidence of a brown-eyed child came along
• P(B) is the probability of a brown-eyed child depending on the a priori probability
• P(B|A) is the probability of a brown-eyed child assuming that A has happened – i.e. that I’m BB.
• The result, P(A|B), is the probability that I’m BB based on the evidence.

So let’s evaluate those first three points in turn.

##### P(A)

My parents, being BB mother and a Bb, would have expected to have children who were half BBs and half Bbs. So let’s put the a priori probability, P(A) down as $\frac{1}{2}$.

##### P(B)

Assuming the a priori probability, the chance of my having a brown-eyed child is

$\frac{1}{2} \times \text{probability of brown eyed child with BB} + \frac{1}{2} \times \text{probability of brown eyed child with Bb}$

because I have half a chance of being BB and half of being Bb.

This equals

$\frac{1}{2} \times 1 + \frac{1}{2} \times \frac{1}{2}$

which equals $\frac{3}{4}$

##### P(B|A)

This is, assuming I’m BB, what is the probability of having a brown-eyed child? It’s 1, because all offspring will be Bb.

So, plugging these three values back into the Bayes’ theorem formula:

$P(A|B) = \frac{1 \times \frac{1}{2}}{\frac{3}{4}} = \frac{2}{3}$

So the evidence of having a single child with brown eyes with a blue-eyed partner has made the probability of my having BB rather than Bb jump from $\frac{1}{2}$ to $\frac{2}{3}$.

Now I know the probability of being BB, I can work out the probability of having a brown-eyed child: it’s

$\frac{2}{3} \times 1 + \frac{1}{3} \times \frac {1}{2} = \frac{5}{6}$

I find it interesting to think about how much probability is sometimes just a question of perspective. Both my genome and my child’s are already established – we just haven’t examined our DNA to find out about it – so in a way there’s no probability about it. I said earlier that my parents, as a BB/Bb couple, would have expected to have half BB and half Bb children, but in fact their genomes were not revealed until it became clear that the eyes of their first child would never change from their original baby-blue.