Here is a knockout formula
a = 2 (n – R)
I always like a good real-world application of mathematics, in this case algebra.
If you’re organising a knockout tournament for something and the number of players or teams is not a neat power of two, how many players have to be drawn in round 1, so that round 2 is a neat power of two? I figured this out by solving this pair of equations:
a + b = n
0.5a + b = R
a is the number of players to be drawn in round 1,
b is the remainder of players to be drawn in round 2,
n is the total no. of players and
R is the largest power of two less than or equal to
Solving the equations
0.5a + (n-a) = R
a + 2n - 2a = 2R
2n - a = 2R
a = 2n - 2R
a = 2(n - R)
91 players enter our tournament, how many must be drawn in round 1?
n = 91, R = 64
a = 2 x (91 - 64)
a = 54
So we draw 54 players in round 1, which produces 27 winners to meet the remaining 37 players in round 2.