# Players on a 3-losses-out tournament

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?

• 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. May 12 '17 at 11:17
• @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. May 12 '17 at 11:20
• How can you prove it? I am not so convinced about that. May 12 '17 at 11:32
• 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 $12-0$. May 12 '17 at 11:33
• A fun problem that got lost. Aug 19 '18 at 0:38

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.