Crew selection: ranking rowers by letting them race against each other

Question

Seat selection is a common practice in competitive rowing and I would be curious about more solid statistical underpinnings: there are more rowers in a team than the 8 seats in the crew boat. So the coach splits the team repeatedly into two smaller crews of 4 and lets them race against each other - noting both who wins and the margin of winning (say: seconds). There is prior knowledge from land training about relative strength. The goal is to find the strongest 8 rowers from a team of 12 or so and typically only 2 or 3 seats are really contentious.

• I am curious about how to model this but don't have a strong background in statistics and hence look for a starting point. I was attracted by the simplicity of an answer for Measuring individual player effectiveness in 2-player per team sports

• The number of races is limited because they are tiring. This suggests the partioning of the team for races should be designed carefully to learn the most.

• I would like an approach where each race adds knowledge about the ranking (or relative contribution). A naive approach would simply use the fraction of races won but I would expect that more can be learned by taking prior knowledge into account: was the win to be expected or surprising?

Answer 1

You should look into the design of experiments. Say there are 12 rowers, and 8 seats in the boat. We start with the model $$ Y= \beta_0 + \sum_{i=1}^{12} \beta_i I_i + \epsilon_i $$ where $I_i$ are inclusion indicators for the rowers, and $\sum_{i=1}^{12} I_i=8$. I would start with optimal experimental design for this model. The number of possible teams are $\binom{12}{8}=495$, so trying all teams would be prohibitive.

You could just generate all 495 teams, and use as input for some optimal design algorithms, like in R package AlgDesign which implements the Fedorov exchange algorithm using the criterion of D-optimality. I would guess there are theoretical results for this specific kind of model, but cannot find references.

... and typically only 2 or 3 seats are really contentious

With some prior information, you could look into Bayesian design of experiments. Not much on this site, see https://en.wikipedia.org/wiki/Bayesian_experimental_design.

Answer 2

I don't have a fully general answer but it is a start. This assumes:

• 8 rowers (identified as 1 to 4 and A to D); 1 to 4 always row on one side, and A to D on the other.
• 2 boats with 4 rowers each: two from 1 to 4 and two from A to D.
• 6 races of 2 boats; distance 1000m, for each boat the time is taken.
• rowers are placed into boats according to a swap matrix
• from the time of a race the average power of the crew is computed. The general connection is $P=k*v^3$ with $k=2.8*4$ a typical drag coefficient for boats with 4 rowers and $v$ the speed of the boat calculated from distance and time.

The problem now becomes finding a solution for the following equation system (with some crew power values as an example):

$$ \begin{bmatrix} 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 & 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 & 0 & 1 & 1 & 0 \\ \end{bmatrix} \times \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\\ x_A\\ x_B\\ x_C\\ x_D\\ \end{bmatrix} = \begin{bmatrix} 1303.02 \\ 1350.62 \\ 1354.82 \\ 1274.37 \\ 1305.56 \\ 1304.04 \\ 1304.49 \\ 1256.89 \\ 1252.69 \\ 1333.14 \\ 1301.95 \\ 1303.47 \\ \end{bmatrix}\\ $$ or more succinctly as \begin{equation} \label{eq:sx=p} S x = P \end{equation} where $S$ is the swap matrix, $x$ the power of each rower, and $P$ the observed crew power. The swap matrix $S$ identifies who is racing together in a crew.

Solving $Sx=P$ for $x$ is not straight forward: $S$ is not square and therefore has no inverse. The left inverse $S'$ with $S'S=I$ does not exist because $\textrm{rank}(S)= 7 < 8$. But $S$ has a unique generalised inverse $S^+$ which we can use to describe all solutions.

\begin{align} S^+ &= \frac{1}{48} \begin{bmatrix} +7&-5&-5&+7&-5&-5&-5&+7&+7&-5&+7&+7\\ +7&+7&+7&+7&+7&+7&-5&-5&-5&-5&-5&-5\\ -5&-5&+7&-5&-5&+7&+7&+7&-5&+7&+7&-5\\ -5&+7&-5&-5&+7&-5&+7&-5&+7&+7&-5&+7\\ +7&+7&-5&-5&-5&+7&-5&-5&+7&+7&+7&-5\\ +7&-5&+7&-5&+7&-5&-5&+7&-5&+7&-5&+7\\ -5&+7&+7&+7&-5&-5&+7&-5&-5&-5&+7&+7\\ -5&-5&-5&+7&+7&+7&+7&+7&+7&-5&-5&-5 \end{bmatrix} \end{align}

All solutions for rower power $x$ given crew power $P$ are given by

\begin{equation} \label{eq:x=sp} x = S^+ P + c \begin{bmatrix} +1&+1&+1&+1&-1&-1&-1&-1 \end{bmatrix}^T \end{equation}

for an arbitrary constant $c$. Possible solutions are

\begin{equation} \begin{array}{rrrr} rower & c=0 & c=10 & c=30 \\ 1 & 293.40 & 303.40 & 323.40 \\ 2 & 343.42 & 353.42 & 373.42 \\ 3 & 334.14 & 344.14 & 364.14 \\ 4 & 332.80 & 342.80 & 362.80 \\ A & 331.67 & 321.67 & 301.67 \\ B & 334.53 & 324.53 & 304.53 \\ C & 342.74 & 332.74 & 312.74 \\ D & 294.82 & 284.82 & 264.82 \\ \end{array} \end{equation}

Note that the difference in power between rowers within one side is independent of $c$. This is a feature of the swap matrix $S$ and now can be used to rank rowers per side.

• The strongest rowers on starboard are 2, 3, 4, 1 in this order.
• The strongest rowers on port side are C, B, A, D in this order.
• We can't tell who is the strongest rower across sides because power shifts between sides are not detected by the method. Parameter $c$ basically models this shift. Across all solutions, the difference in power between rowers of one side remains constant and leads to an order independent of $c$.
• With $c=0$ the combined power of rowers of 1 to 4 and A to D is equal. This makes it somewhat more likely to be the true scenario than a solution with a large parameter $c$.