Visualizing the Playoffs as a tree,
I devised a necessary condition for 'x' to be a winner.
Let's consider the case of ones, as the case of zeroes is analogous.
Define skill(x) = s(x) .
Define n(l) = No. of players with skill less than x, who lost at lth (one)level from the top.
At the highest one, x defeated a player x1 with s(x1) < s(x). [n(1) = 1 = 2^0]
Now, consider the second highest one, both x and x1 must have defeated players x2 and x3 with skill s(x2)<s(x) and s(x3)<s(x1). This also implies that s(x)>s(x3), s(x2). So at this level, we get 2 more players with smaller skills than x. [n(2)= 2 = 2*n(1)]
Inductively, we can see that [n(h) = 2*n(h-1) = 2^(h-1)], where h is the number of ones in the binary string given.
So, We get that x is greater than a set of numbers with cardinality at least 1 + 2^1 + 2^2 + ... + 2^(h-1).
Similarly in zeroes, We get that x is smaller than a set of numbers with cardinality at least 1 + 2^1 + 2^2 + ... + 2^(z-1). where z is the number of zeroes in the binary string given.
