I have observed that parents having brown eyes had three female offspring with blue eyes, one son with green eyes, while only the youngest daughter had brown eyes.

How can they produce so many blue-eyed offspring?

  • 2
    $\begingroup$ You directly observed it, why are you asking if it's possible? Also, you should rephrase your question to remove all personal aspects. $\endgroup$
    – canadianer
    Commented Jun 26, 2015 at 4:27
  • 4
    $\begingroup$ Eye colour is more complex than one colour being dominant over the other. At least 10 genes and the relationship between those genes account for many factors including hue and saturation. $\endgroup$
    – James
    Commented Jun 26, 2015 at 7:50
  • $\begingroup$ I would recommend to "father" to perform genetic testings in question does he related to these children at all. $\endgroup$
    – Ilan
    Commented Jun 26, 2015 at 12:21
  • $\begingroup$ ""My mother," answered Telemakhos, "tells me I am son to Odysseus, but it is a wise child that knows his own father." - Homer, ca 1000 BCE. $\endgroup$
    – jamesqf
    Commented Jun 26, 2015 at 17:42
  • $\begingroup$ My father's father had blue eyes. My mother has blue-green eyes. Between the two of them they produced four brown-eyed daughters, all with strong family resemblances. Probability is only meaningful with a much larger sample set. $\endgroup$
    – arp
    Commented Apr 18, 2018 at 19:57

2 Answers 2


Possible...yes of course! The answer could stop there but I guess you want to ask how likely it is.

Here is a chart of probability of a given kid to have eyes of a certain color given the parents eyes color. (Note that the second column contain probabilites and not odds contrary to what the figure is stating).

enter image description here

The probability that $k$ kids out of $n$ have eyes of a color which appears with probability $p$ is given by the probability mass function of the binomial distribution $\binom{n}{k} p^k (1-p)^{n-k}$

Knowing that both parents have brown eyes (even for your oldest sister):

  • The probability that exactly one out of 5 offsprings have brown eyes (and the others have eyes of another color) is about 0.015 (3 out of 200).

  • The probability that exactly 3 offsprings have blue eyes (and the others have eyes of another color) is about 0.0021 (2 out of 1000)

You might want to calculate the probability of having $k$ or more kids that have eyes of a given color (a p.value) is obtained by summing probabilities. For example, the probability that 3 or more kids (out of 5) have blue eyes is $\binom{5}{3} 0.0625^3 (1-0.0625)^{5-3} + \binom{5}{4} 0.0625^4 (1-0.0625)^{5-4} + \binom{5}{5} 0.0625^5 (1-0.0625)^{5-5} ≈ 0.0022$

Learn more about Stats

You'll probably want to learn a bit about statistics if some things are unclear. You might want to make sure you understand the concept of p.value and its interest and make sure you understand what a binomial distribution is. You may also want to understand the concept of multiple testing to avoid falling in simple pitfalls.

  • $\begingroup$ Where is that image from? $\endgroup$
    – canadianer
    Commented Jun 25, 2015 at 19:38
  • 1
    $\begingroup$ Hum... I quickly took something from google image. I already saw this kind of image several time and assumed it was trustful. Does it seems wrong to you? There are several images of the kind all showing more or less the same probabilities $\endgroup$
    – Remi.b
    Commented Jun 25, 2015 at 19:40
  • 1
    $\begingroup$ @Remi.b Although I agree that they are important concepts I fail to see what does p.value and multiple testing have to do with your answer? As far as I know you are not performing null distribution hypothesis testing, just computing probabilities. (I see the relation for the binomial distribution though). $\endgroup$
    – ddiez
    Commented Jun 26, 2015 at 4:11
  • 2
    $\begingroup$ @Remi.b I'm not saying it's wrong, l just think you should cite your source both for fairness and so people (such as myself) can read more about how the probabilities were derived. $\endgroup$
    – canadianer
    Commented Jun 26, 2015 at 4:19
  • $\begingroup$ @canadianer ou are right. I'll add a reference if I can find one. $\endgroup$
    – Remi.b
    Commented Jun 26, 2015 at 12:55

After thinking about it for a while I have arrived to a slightly but important different view on this problem. I will try to explain the statistics behind it to the best of my knowledge. I will use the eyes’ color probabilities in @Remi.b’s response, but of course those may not be accurate and the results can change accordingly.

The problem is to determine how likely is to have 3 children with blue eyes, 1 with green eyes and 1 with brown eyes when the parents both have brown eyes. I will assume each child is independently born (e.g. there are not identical twins). We can use the analogy of a bag with balls from which we draw 5 balls independently.

Imagine you have several bags with $n=100$ balls each. The balls can be brown (M), blue (B) and green (G). Depending on the bag, you will have different proportion of balls. For example, in the bag labelled MM (for brown:brown), there are 75 M balls, 19 G balls and 6 B balls (rounding the probabilities in @Remi.b's answer). The bag labelled GB does not have any M balls (0), and has 50 G and 50 B balls. And so on. This gives us for each bag the probabilities defined in @Remi.b’s response.

Although there are many bags in our particular case we only care about the MM bag, so we should only consider those probabilities. Imagine now you pick five balls at random from this (the MM) bag. What is the probability that you will get $m=1$, $g=3$ and $b=1$ (with m, g and b representing the number of M, G and B balls respectively)? Because there are three types of balls (or categories) this is described by the multinomial distribution (and not the binomial, which describes events involving two categories, like the flip of a coin). The multinomial probability mass function is defined in this problem as:

$$ Pr(M=m,G=g,B=b) = \frac{n!}{m!g!b!} p_M^{m} p_G^{g} p_B^b $$

Where $P_i$ are the probabilities described above, n is the total number of childs, and m, g, and b are the number of childs with M, G and B eye color respectively. Therefore, for our family we have the following computation:

$$ Pr(M=1,G=1,B=3) = \frac{5!}{1!1!3!}(0.75)^1(0.1875)^1(0.0625)^3 = 0.000686643 $$

The probability (~0.07%) looks very low, but it is basically about 7 out of 10,000 families with parents having both brown eyes and 5 children. Given the large human population it is difficult but not impossible.

Personally, I knew a family where the parents had both brown eyes and the two children had blue eyes. Always wondered about it but this is the first time I actually tried to think about the probabilities! In this case the probability would be 0.00390625 (~0.4%).

BTW, you can easily compute the same in R with the following code:

> dmultinom(c(1,1,3), 5, c(.75, .1875, .0625))
[1] 0.0006866455
> dmultinom(c(0,0,2), 2, c(.75, .1875, .0625))
[1] 0.00390625

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