Knockout tournament. P1, P2, P3, P4, P5, P6, P7, P8, P9, P10, P11, P12, P13, P14, P15, P16 are 16 players who play a knockout tournament. In any match between P(i) and P(j), P(i) wins if i is less than j.
Find the probability that P6 reaches the final.
I tried making cases, but they seem endless.
We know for sure that P1 will win the tournament and P16 will be eliminated in the first round.
But there are many other cases for the first round only. Please help.
 A: WLOG we can first assign P6 a spot on the tournament tree, then choose 5 spots for P1,...,P5, and the rest of the choices are irrelevant for this question.
P6 makes the final if and only if P1, P2, ... P5 are all assigned positions on the side of the tree of the tree opposite to the one where P6 is assigned.  Once P6 is assigned her spot, there are 15 remaining, of which 8 constitute the "other side".  There are
${8 \choose 5}$ ways to choose 5 of the "other side" spots for P1,...,P5, 
out of a total of ${15 \choose 5}$.  Thus the probability is
$$ \dfrac{8 \choose 5}{15 \choose 5} = \frac{8}{429}$$
A: Consider the intial draw sheet, expressed as a tree.

The half of the draw containing P6 has $7$ other players, so there are $\text{C}(15,7)$ choices for those $7$ players, all equally likely.

P6 reaches the final if and only if those $7$ players are all weaker than P6. There are $10$ such players.

It follows that P6 reaches the final with probability $\dfrac{\text{C}(10,7)}{\text{C}(15,7)} = \dfrac{8}{429}$.
