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Say I have a game in which $i$ players at a time can play, and I want to organize a tournament, inviting $n$ people. We play in rounds, so of course $i$ has to divide $n$ because winning by default is boring. Each round then consists of $n/i$ games played simultaneously, and of course no player plays more than one game at any one time.

Now, to make iot interesting, I choose $n$ so that $(i-1)$ also divides $(n-1)$, and let the tournament consist of $\frac{n-1}{i-1}$ rounds in total. Is it possible in general to seat players so that everyone plays against everyone else exactly once during the tournament?

The background is that yesterday I took part in a tournament in a three-player game, with 15 participants. After five rounds I'm pretty certain no one had met anyone else twice, and I was curious whether it would possible to keep that record up for two more rounds. However, the tournament ended there, and I didn't pester the host with my petty theoretical mathematical questions.

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    $\begingroup$ It seems that a very similar problem was already posed here $\endgroup$ – Alex Ravsky Apr 14 '13 at 18:37
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In answer to your specific case of the $3$-player $7$-round tournament. The answer is yes. In fact, this is the famous Kirkman's Schoolgirl Problem. In general, you're seeking a resolvable BIBD (balanced incomplete block design).

Fifteen young ladies in a school walk out three abreast for seven days in succession: it is required to arrange them daily so that no two shall walk twice abreast.

Here's the solution on Wikipedia:

A solution to the Kirkman's Schoolgirl Problem

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