20 golfers play in groups of four in a round. They rotate partners to try to play with everyone at least once. How many rounds must be played so everyone has played with every other golfer?

Question

20 golfers that play in groups of four each day. How many days will it take so that each golfer plays with all of the other golfers at least once?

Answer 1

If the golfers (A to T) were to play in pairs each golfer the pattern of games would be:

AB ...AT (19), BC...BT (18), CD...CT (17), ...,PQ PR PS PT (4), QR QS QT (3), RS RT (2), ST (1).

The sum of the numbers 1 to 19=190.

There are 190 unique combinations of players.

If two pairs are combined we know that all the possible combinations will be present, and there will be duplications within the groups, but every player will have played at least once with every other player. Those combinations of pairs result in half as many games=190/2=95. However, this appears to be the maximum number of games. 

However, it has been assumed that the exercise of combining two pairs doesn’t include the same player twice. For example, GH can’t be combined with HK. 

Below is a solution containing 58 games (this may not be the minimum).