Tournament Riddle
Posted by Rob Herman at July 28th, 2007
I am running a single-elimination tournament with 64 players. The seeding is random. Assume the teams are totally ordered in skill (no two have the same skill, and the “is more skillful than” relationship is transitive) and that the more skillful team wins every game.
The tournament determines the most skillful team, of course. After it’s over, how many games are needed to determine the second best team?
I am feeling a little lazy on the math. Assuming that this has to be run like a standard tournament, there would be 6 rounds of 32, 16, 8, 4, 2, and 1 rounds resppectively for 63 games and that determines the #1 player. According to the rule, only the #1 player can beat the #2 player, leaving 6 possible competitors. That means 3 rounds of 3, 1 and 1 (stupid not power of 2 leaving 3 potentials). Thats 5 more games for a total of 68 games.
I came up with this answer far too quickly, which likely means I misunderstood something in the rules or this answer is wrong (maybe it is a Catan Tournament with 4 players per game, which might be easier becasue we already know who the number one player is and can just make me forfeit in the first round).
No, you nailed it. This one was on the easy side. I included it because I saw it on a list of more difficult riddles somewhere else and thought “maybe it just came really easy to me.”
Nope, it just really is that easy.
Determining more top talents is marginally more work but no more difficult conceptually.
Slightly more work would be to determine the single worst player.
To determine the worst player., just run a tournamanet where the loser advances
To determine the third best player, you have to hold a tournament including everyone the winner has beaten and everyone beaten by any one of those players. Consider the incredible complexity of trying to find the fifth best player. It makes the wisdom of using a regular season, or Swiss tournament, to seed your single-elimination championship very apparent.