Let's assume that four of the players are in the left bracket and the other four players are in the
right bracket.
The number one ranked player always makes it to the finals. The other players, depending upon
the draw, that could make it to the finals are players number two, three, four, and five.
Let's assume that player number one is in the left bracket (the analysis is the same if this
player is in the right bracket) ...
Where is player number two?
If this player is in the right bracket, this person will make it into the finals; if not, this player will
be defeated by player number one.
After player number one is put into the left bracket, there are seven slots open for player number
two; four of them in the right bracket (and makes it into the finals and three in the left bracket and
loses to player number one).
The probability that this player is in the right bracket is 4/7.
We need to multiply this probability by the rank: (4/7) x (2) = 8/7 = 1.143.
If both player number one and player number two are in the left bracket, where is player number three?
There are six slots left and four are in the right bracket, so the probability that this player is in the right
bracket is 4/6.
The probability is (3/7) x (4/6) = 12/42 = 2/7
[We need to include 3/7 because we have to have player number two in the left bracket.]
Multiplying this probability by the rank: (2/7) x (3) = 6/7 = 0.857
If player numbers one, two, and three are all in the left bracket, where is player number four?
The probability that players one, two, and three are all in the left bracket and player number four
is in the right bracket is (3/7) x (2/6) x (4/5) = 4/35.
Multiplying this probability by the rank: (4/35) x 4 = 0.457
If players number one, two, three and four are all in the left bracket, player number five will be
in the right bracket: (3/7) x (2/6) x (1/5) x (4/4) = 1/35
Multiplying this probabiliy by the rank: (1/35) x 5 = 0.143
Adding these answers together: 1.143 + 0.857 + 0.457 + 0.143 = 2.6