Math CentralQuandaries & Queries


Question from Elizabeth, a parent:

I am trying to sort out golfing for 16 guys for 7 rounds. However the twist is that the teams are broken down into 8 old boys vs. 8 young boys. Ideally, each of the players would only play each other once within their own team and then twice within all of the matches. Is that possible? Thanks for your help, my brain is hurting!


The short answer is that you can't do it exactly. The number of pairs of players you need to cover is 28 + 28 + 2x64 = 184 -- there are 28 pairs of old boys, 28 pairs of young boys, and 64 pairs consisting of one old boy and one young boy. Each foursome covers 6 pairs of players. Unfortunately 184 is not a multiple of 6, so an exact solution is not possible.

There is a well-known schedule where every pair of 16 golfers play together exactly once over 5 rounds. You can find that by searching the archive for golf 16. A good way to go would be to use that schedule for the first 5 rounds, and then in the last 2 rounds try to put together players who'd most like to be together twice.

Good luck!

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