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I was doing a probability homework and the question says:

A conventional knock-out tournament (such as that at Wimbledon) begins with 2n competitors and has n rounds. There are no play-offs for the positions 2,3,...,2n-1, and the initial table of draws is specified. Give a concise description of the sample space of all possible outcomes.

So I don't understand no play-offs for the positions 2,3,...,2n-1. I don't understand the game. What is the initial table of draws? How does this tournament work?

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When they say "no play offs", they mean, you don't get a second chance if you lose.

(The terminology here is not conventional American English, but from the reference to Wimbledon the writer is probably British, so maybe this is the terminology in British English.)

Anyway, the way such a contest works is this. Say we have 8 contestants. We pair them off into 4 contests. In each contest one person wins and one person loses. The losers are then eliminated from the tournament. That leaves 4 winners. Now we break the 4 winners into 2 contests. That gives 2 losers and 2 winners. The 2 winners then compete for the championship. One wins and one loses. The winner is declared the winner of the tournament.

So "no play offs" means that once you lose a game, you're out. You don't get to play another game. If the second best player has the bad luck to be paired against the best player in the first round, he'll be eliminated in the first round. The final round is not necessarily between the 2 best players. Who ends up in second place depends on who they got paired against.

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    Conventional means the number of contestants is halved every round and only consists of the winners from previous rounds. This can't be done if there's not a 2^n number of teams/players. Consider the difference between this and world cup football where there are group stages where the top two or three progress to the knockout stage (this also increases the number of games played (which increase the number of tickets sold / sponsorship etc etc) – Smock Jul 25 at 16:01
  • RE: The final round is not necessarily between the 2 best players. Who ends up in second place depends on who they got paired against. True – sort of. Most of these tournaments are not seeded randomly, but are seeded such that, if the favorited player won every round, the last round would be a 1 vs. 2 pairing (i.e., they are seeded like this). – J.R. Jul 25 at 16:08
  • @J.R. But if you know in advance exactly how all the players (or teams) rank, and rely on that knowledge, than the tournament itself is superfluous. :-) Anyway, I don't doubt that in professional tournaments they carefully arrange the matches like this. I've never been in a professional sports tournament (because I'm an overweight, balding old man in poor health), but when I was a kid I was in tournaments of various kinds in school, and in those, as I recall, the pairings were always random. Either drawn out of a hat, alphabetical order, or some such. – Jay Jul 26 at 14:23
  • @Jay - No doubt, baseball tournaments for preteens are frequently seeded at random. However, the OP's example mentions Wimbledon, where I'm pretty sure they don't draw names out of a hat. However, I'll concede that when I said, "Most of these tournaments are not seeded at random," I was assuming professional-level athletics. As you point out, many tournaments at the amateur level use luck of the draw. – J.R. Jul 26 at 14:34

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