There are 56 teams in a knockout tournament, then how many matches has to be played to select the champion? ''There are 56 teams in a knockout tournament, then how many matches has to be played to select the champion?''
I found this in a question paper, and I am stuck to solve this problem.
I have proceeded to solve it like this:
Round 1: $\frac{56}{2}=28$ matches
Round 2:  $\frac{28}{2}=14$ matches
Round 3: $\frac{14}{2}=7$ matches
Now I can't find how many matches will be played in Round 4.
PLEASE HELP!
 A: The interesting thing about this question is that it doesn't matter how the tournament is structured. For example:


*

*We could randomly assign a bye in the first round to 8 of the teams. At the end of the second round, there would be 24 teams left who had played, and the 8 who'd had a bye, leaving 32 teams. In each subsequent round, all remaining teams will participate, until one remains.

*We could have every team play each round until the fourth round, at which point only 7 teams remain. One of these teams is randomly assigned a bye in the fourth round, and the rest play, so that in the fifth round, 4 teams remain, and every team plays each subsequent round until one remains.

*We could number the teams from 1 to 56. The first round is a match between teams 1 and 2. The second round pits team 3 against the winner of the first round, the third round pits team 4 against the winner of the second round, and so on (more generally, for any integer $1\le n\le54,$ in round $n+1,$ team $n+2$ plays against the winner of round $n$).


Of course, these arrangements aren't equally "fair," whatever that means, but regardless, it turns out that the same number of games will be played! This is because each game will result in one loser, who is then removed from the set of possible winners, until only one team remains. Hence, 55 games must be played (which you can verify to be the case in the three disparate examples I gave above).
