A chicken pecking problem

The following problem was posed in the 2017 Raytheon MathsCounts national competition:

In a barn, 100 chicks sit peacefully in a circle. Suddenly, each chick randomly pecks the chick immediately to its left or right. What is the expected number of unpecked chicks?

The competition is aimed at 13 to 15 year olds, and they have just 45 seconds to answer – the first with the correct solution wins. In this case, a 13 year old Texan boy came up with the answer in just 1 second!


Clearly, this is a probability problem. But you won’t get to the solution by trying to work out what’s happening to all the chicks at the same time: the trick is to consider just one chick, and then use expectation, or expected value, to scale up to all 100. So, what is the probability that our individual chick is unpecked? This will only happen if both the chicks on either side peck the other way, and the probability of this is 0.5 × 0.5 = 0.25 (assuming that by “random” the question means equally likely to peck either way). So the expected number of unpecked chicks is 100 × 0.25 = 25.

It’s interesting to consider whether this answer would change if the chicks were more likely to peck to the left, say, rather than to the right. Presumably it would be different since, in the extreme case, if they all pecked to the left, the number of unpecked chicks would be zero. What if P(pecking left) = 0.8? Then P(pecking right) = 0.2, and the expected number of unpecked chicks would be 100 × 0.8 × 0.2 = 16. We can draw up a table of expected values:

It looks as if the maximum number of unpecked chicks is 25 – can we prove this?

Suppose the probability of pecking to the left is p, then the probability of pecking to the right is 1 – p, and the expected number of unpecked chicks is 100p(1 – p). This is a quadratic function (as you might expect from the symmetrical nature of the values in the table), and we can find the maximum value by differentiation.

            f(p) = 100p(1 – p) = 100p – 100p2

            f ‘(p) = 100 – 200p

For a maximum, f ‘(p) = 0, so 100 – 200p = 0, giving p = 0.5, thus proving that the maximum occurs when the probability is 0.5.


If you feel inadequate because a 13 year old solved this in under a second, don’t worry. The executive director of the competition tells us that competitors can do what they do through endless hours of practice and coaching. And is there a point to all of this? Yes, the competition aims to encourage students into STEM (science, technology, engineering, mathematics) courses where there is at present a worldwide shortage of graduates.

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