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we are organizing a regatta, there are 12 competing crews and we have six boats. Therefore the regatta is organized in so-called fleets, each consisting of two races at which six crews (using the six available boats) are participating. We assume to be able to race 10 to 15 fleets (i.e. 20 to 30 races), but of course that depends on the weather, especially the wind. However, of course we intend to have balanced fleets during the entire regatta with respect to the crews racing directly against each other, that means we want to mix the fleets/races to achieve as little difference between the number of direct matches between each of the competitors.

Overall there are ##\frac{1}{2} {12\choose 6}=462## possibilities to assign the twelve crews to two races at with six boats. After 462 fleets every crew had the same number races with each other competitor, if there is only one fleet organized that's of course not possible. Now there is my question: How is it possible to determine how balanced the number of direct encounters between every combination of two crews is after a certain number of fleets raced (10 to 15 in my case).

I tried to write a small code, and for 15 fleets I obtain combinations of (some) two crews which meet 5 times in the same race after 15 fleets, where the maximal number of encounters between (some) two crews is 9. I'm not able achieve smaller differences than 4. The average would be 15 races each crew participates at multiplied with 5 other boats per race divided by the eleven other crews:

$$ \frac{15\;races\cdot 5\;boats}{11\;opponents}\approx 6.8\;encounters$$

Therefore after 15 fleets every crew should meet every other crew directly in a race 6.8 times, which means 6, or more probable, 7 times. With my code I obtain between 5 and 9 times. Are my numbers really the best mixing achievable or is it possible to decrease the difference between the lowest and highest number of encounters? Could you help me to find the this difference mathematically?

Thanks and have a nice morning/afternoon/evening/night - wherever you are,

stockzahn