Dear Uncle Colin,

I have a question from my childhood. I know the answer, but not how to get it. Can you explain? Here it is:

Oxford and Cambridge Universities decided to join forces and send a combined rugby union squad on a tour. Enough players were chosen from both universities to ensure the squad had adequate reserves. On the eve of the first match the captain put the names of all the players in to a hat and drew out 2 names at random. He automatically gave those 2 players a place in the team for the first match. Someone noticed that the probability of both players being Oxonians was a half. So how many players were in the squad?

- Solution Could Require University Maths

Hello, SCRUM, and thanks for your message!

This question is, as I think you realise, easy to answer but difficult to solve. By this I mean, it’s not hard to come up with numbers that work ((I’ll spoiler the answer in the footnote: of the 21 players, 15 were from Oxford.)), nor is it difficult to see that this is the only solution with reasonable numbers. However, it’s a bit unsatisfactory: there’s one equation in two unknowns; there are (it turns out) infinitely many solutions, and generating them is much trickier.

So let’s tackle it from the ground up.

Let’s say there are $x$ Oxford students out of a total of $t$. The probability of the first player being from Oxford is $\frac{x}{t}$; the probability of the second also being from Oxford is $\frac{x-1}{t-1}$ – there’s one fewer Oxonian, and one fewer player to choose from.

We need to solve $\frac{x}{T}\cdot \frac{x-1}{t-1} = \frac{1}{2}$, or $2x(x-1) = t(t-1)$.

To solve this in general, we will need some heavy machinery.

First up, I have a quandary: I want to complete the square on both sides, but I don’t like fractions. The easy solution? Multiply everything by 4.

• $2 \left(4x^2 - 4x\right) = 4t^2 - 4t$
• $2 (2x -1)^2 - 2 = (2t-1)^2 - 1$
• If I let $X = 2x-1$ and $T = 2t-1$, this simplifies to $2X^2 - T^2 = 1$

This is a Pell’s equation. We know how to solve Pell’s equations, don’t we? Well, I do. It’s all to do with continued fractions. The solutions to equations of the form $a m^2 + n^2 = \pm 1$ are $m$ and $n$ such that $\frac{n}{m}$ is a convergent of $\sqrt{a}$, with the sign alternating.

Here, we’re looking for convergents of $\sqrt{2}$ – in particular, the slight underestimates.

The square root of 2 is one of the easier ones to generate convergents for((The continued fraction is $[1; \bar{2}]$)): the numerators and denominators both follow the recurrence relation $a_{n+2}=2a_{n+1} + a_n$. The difference is the starting conditions; for the numerator, the first two terms are 1 and 1; for the denominator, it’s 0 and 1.

And we can reel these off:

• $(1,1)$ is an underestimate, giving a nonsense solution.
• $(3,2)$ is an overestimate, which we ignore.
• $(7,5)$ is an underestimate, giving $t = 4$ and $x=3$ – a valid solution, if a bit short-handed for a rugby tour.
• $(17,12)$ is an overestimate.
• $(41,29)$ is an underestimate, giving $t=21$ and $x=15$, which is the solution I imagine we want.
• $(99,70)$ is an overestimate.
• $(239, 169)$ is an underestimate, giving $t=120$ and $x=85$ – even with the modern game’s penchant for replacements and its reasonable concern about concussion, 120 players is a bit of a large squad for a tour.

We could carry on in the same vein. If it wasn’t rapidly approaching my bedtime, I might examine how to solve that recurrence relation and get explicit solutions; instead, I’ll just hint that it’s possible and leave it as an exercise.

Hope that helps!

- Uncle Colin

* Thanks to Phil for the question.