Six players decide to play a simple tournament, everyone facing everybody else once, sides chosen randomly. This makes a total of 15 games.
Show that no matter how the games end, there is always at least one triplet of players among the six whose mutual games either all ended as a draw, or all ended in a victory for one player or the other.
Solution.
Start with one player, say, player A. She faced five opponents, so at least three of those ended the same way - all draws, or all non-draws, say, vs. players B, C and D. Consider three players whose games with player one ended similarly. If even one of their mutual games has the same result.. e.g. B vs D.. the threesome with similar results is found, A-B-D. If none did, then B-C-D is the threesome sought. Either way, the threesome exists.
A follow-up question might be.. how many players are needed for there to necessarily be six players between whom all 15 games ended up the same way? A British fellow called Frank Ramsey looked into those, and far as I know, no one has yet found out the minimum number of elements for a six-element clique.