It's common for some games (Hearthstone, Clash Royale...) a tournament in which you play until you had won $12$ matches or lost $3$.

In this kind of tournaments, you are matched with the player the most similar to you in terms of wins and loses.

Taking that into account, what's the minimum amount of players to ensure that one will get $12$ wins?

And more general, to ensure that $n$ players will get $12$ wins?

  • $\begingroup$ I don't think there is any amount to ensure that one will get 12 wins. If you have any number of players it could happen that all of them lose 3 matches before winning 12. Anyway, a precise answer can be given only knowing how the tournament is played. $\endgroup$
    – Crostul
    May 12 '17 at 11:17
  • $\begingroup$ @Crostul if you are paired against someone with the same number of wins as you, with enough players, at some point one would get 12 wins. $\endgroup$
    – Masclins
    May 12 '17 at 11:20
  • $\begingroup$ How can you prove it? I am not so convinced about that. $\endgroup$
    – Crostul
    May 12 '17 at 11:32
  • $\begingroup$ Supose $4096$ players. The $2048$ winners play against each other. From those, $1024$ will win and play against each other... Then $512$, $256$... At the end there will be a single player at $12-0$. $\endgroup$
    – Masclins
    May 12 '17 at 11:33
  • 1
    $\begingroup$ A fun problem that got lost. $\endgroup$ Aug 19 '18 at 0:38

It depends on what happens when you have an odd number of people in a round. I will make the rule that one of the players with the worst record is given a bye to the next round and gets to count it as a win so that all the players have the same number of games. Other rules are quite possible. I believe we want to have the losers take as many wins with them as possible, so if we have an odd number of teams with no losses we will have one of them play a one loss team and assume the one loss team wins the match. Similarly if there is a one loss team left over we pair it with a two loss team and assume the two loss team wins.

Under these assumptions we can show that a player must win if we start with $256$ players. For the first eight rounds each player can play someone with a matching record and we are left with $1$ player at $8-0$, $8$ at $7-1$ and $28$ at $5-2$. In the In the ninth round the $8-0$ loses to a $7-1$, one of the $7-1$s loses to a $6-2$ and $13\ 6-2$s are eliminated, leaving $5$ at $8-1$ and $19$ at $7-2$. The tenth round has $2$ at $9-1$ and $13$ at $8-2$. The eleventh round has $1$ at $10-1$ and $7$ at $9-2$. These eight players play an elimination tournament and somebody has to win $3$ games, getting to $12$ wins. We can probably get by with a few less.


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