⛵️ Familiar Faces in Far-Off Places
This year I've been participating in training for the Clipper 2025-26 Round the World Yacht Race. Learning to sail and live aboard a seventy-foot racing yacht is a fun challenge and the situation quickly fosters team bonds and friendships. At some point during the two weeks of training I've done so far, the question arose:
What is the probability I'll be in the same race team as someone in the training crew?
In this post, we'll calculate an answer to this and some follow-up questions. But first, it's important to state an assumption: race crews are randomly allocated from the pool of all participants. We know this is untrue; crews are chosen to be balanced in age, experience, and skills. For instance, two medics that train together will probably have a lower chance of racing together. While this assumption might change the exact probabilities, it won't affect our overall conclusions.
There's one more caveat: I'm signed up to all of the eight legs of the race, but many crew swap in and out of the boat between legs. We won't consider this in the calculations, but it can be done with some adjustments. I'll point out when this becomes relevant.
I will do the calculations algebraically, but plug in numbers along the way to get a feel for things. We'll use:
- \(N=750\) as the total number of participants,
- \(B=11\) as the number racing crews,
- \(n=8\) as the number of people in a training crew (I had crews of 7 and 9 for my two weeks so far).
That means there are roughly \(\frac{N}{B} = 68\) people in each race team, although only those on the same legs will race together.
To start, let's consider the probability of myself and another specific person to be in the same race team. This is given by the number of people in the team, excluding me, divided by the total number of people \(N\), excluding me: \[ P_1 = \frac{\frac{N}{B}-1}{N - 1} = \frac{\frac{750}{11} - 1}{750-1} = 9.0\% \] (It's pretty much one in eleven as \(N\) rises above 150, as you'd expect. The value of \(N\) doesn't really impact these calculations much, so long as it's large.)
Nitpicky maths starting from an exact formula
The formula for \(P_1\) above is an approximation that applies when \(N \gg B\). The exact form is given by counting all the pairs of people that could be in the same team, divided by the total number of ordered pairs of people: \[ P_1 = \frac{\sum_{k=1}^B s_k \left( s_k - 1 \right)}{N\left( N - 1 \right)} \] where \(s_k\) is the number of people in the \(k\)-th team. Let's define \(q=\big\lfloor\frac{N}{B}\big\rfloor\) and \(r=N\ \mod B\). Assuming round-robin team allocations, there are \(r\) teams of size \(q+1\) and \(B-r\) teams of size \(q\), so the sum in the numerator is: \[ \begin{aligned} \sum_{k=1}^B s_k \left( s_k - 1 \right) &= r\left( q+1 \right)\left( q \right) + \left( B-r \right)\left( q \right)\left( q-1 \right) \\ &= Bq^2 - Bq + 2rq \end{aligned} \]
Now let's look at what happens when \(N \gg B\). In this case \(q\) is close to \(\frac{N}{B}\) and \(r\) at most \(B-1\), which is small. So we can approximate the sum as \(\frac{N^2}{B} - N\). Finally we divide by the total number of pairings and cancel out the common factor of \(N\): \[ P_1 \approx \frac{\frac{N}{B} - 1}{N - 1} \]
Now the probability of not racing with that person is \(1-P_1 = 91\%\). As is often the case in these sorts of probability puzzles, it's easier to first compute the inverse of the target scenario then subtract it from one. Under our assumption of random independent allocation, the probability that nobody from my training crew is also on my race team is given by \( P_0 = (1-P_1)^{n-1} = 52\% \).
Therefore the final probability of racing alongside someone from my training crew is: \[ P_{\geq 1} = 1 - P_0 = 48\% \]
It's a surprisingly high chance given the large numbers of participants and crews. For leggers, the probability is lower since both \(N\) and \(n\) decrease, but it's still surprisingly good. If I meet thirty other people over all four training weeks, there's a 94% probability I'll already know someone on my boat during one leg (my poor memory for names notwithstanding). In fact, I have to select only eight other people on my race leg(s) to have a 50% chance of being on the same boat as at least one of them.
I find this calculation interesting because it reminded me of the Birthday Problem, although it turns out to be simpler in its construction. Relatedly, we could ask what the probability of any two people in a training group being on the same race team is: it's approximately \( 1 - \left( \frac{11}{11} \times \frac{10}{11} \times \cdots\ \times \frac{4}{11} \right) = 97\% \), for large enough \(N\). We'd need a sample of just five people for there to be a greater than 50% chance of at least two of them racing under the same team name.
The overall conclusion here is that it's often more likely than we might think to end up racing alongside those we've trained with. The Birthday Problem is often called a ‘paradox’ because it feels like the probability should be diluted by the large number of possible pairings. But trust the maths; the bonds we develop during the build-up to the race – down below, in the pub, or on a long night watch – may well be as valuable as the sailing skills we learn.