Consider K=1 (2 players), the probability is 1.
Consider K=2 (4 players); first round can be in following ways:
1-2 | 3-4
1-3 | 2-4
1-4 | 2-3
They don't meet in finals in first case (of 3). Thus, for K=2, the probability that they meet is 2/3.
I assure you that this probability gets close to 1/2 for large K. Just notice that if 1 and 2 meet before final, they dont meet at final.

N/2(N-1)

If you worked out conditional probability, that is fine. Here we present the combinatorial solution. Imagine the levels proceed in any random way such that player X and Y appear for final level. This can be imagined as N players getting partitioned into two groups of N/2 players each, with player X topping in first group and player Y in second. The best player of each partition reaches final. We only need to ensure that above partition separates player 1 from 2. The probability that a random partition separates 1 & 2 is (N/2) / (N-1) (how?)

Here is how: for creating a partition to separate them, we pick (N/2) people from N, wishing to take player 1, but not player 2.

Recursion

Let the expected number of HH for n tossed is E(n). So, probability that an (n-1) toss experiments, ends in T is 1/2.
So, E(n) = 1/2*E(n-1) + 1/2*( 1/2*(E(n-1)+1) + 1/2*(E(n-1)))
(The first case when it ends in T. & The second case when it ends in H.
In the second case, if you get an H then, you get 1 more HH. )
So, E(n) = E(n-1) + 1/4, us E(2) = 1/4
So, E(n) = (n-1)/4
For n=10, E(10) = 2.25

The probability that the break point belongs to first half is 1/2. In this case, the expected length of the small part is L/4. Its L/4 in the other case too. Hence E = 1/2*L/4 + 1/2*L/4 = L/4

Linearity of expectation

27

We form the answer for general deck of 2n cards, with n red cards and n black cards. Consider any sequence X1,X2..X2n (of R & B). Now define Y1 = 1;
Yi = 1 if Xi and X(i-1) are of different colors
Yi = 0 otherwise.

We want to find E[Y1 + ..+ Y2n], using linearity of expectation, = E(Y1) +...+E(Yn). Now note that, E[Y1] = 1 and E[Yi] = E[Yj] for the rest (by symmetry)

Also E[Yi] = E(Y2) = P(X2 is diff from X1) = (number of ways first two are RB or BR) / (Total number of orientations) = 2 * [ (2n-2)choose(n-1) ] / [ 2n choose n ] = n/(2n-1)

Ans = 1+(2n-1)*E[Y_i] = n+1

This is 27 for deck of 52 cards.

Try to find the equivalent vertices with respect to distance yet to travel. This should give 4 equivalent merged vertices, with 1st being start & 4th being destination.

Let the expected number of step required to go from (0,0,0) to (1,1,1) be E0.Also let expected number of step required to reach (1,1,1) from (0,0,1)(Or from (0,1,0) or from (1,0,0)) be E1. similarly expected number of step required to reach (1,1,1) from (0,1,1) (Or from (1,0,1) or from (1,1,0)) be E2.Then we can write:
E0=1/3(E1+E1+E1)+1
E1= 1/3*E0 + 2/3*E2 + 1
E2 = 2/3*E1 + 1
solving this we find E0 as 10.