8 top seeds, numbered 1 to 8 enter the quarter-final of a tourney. Seed number 1 is the top ranked seed, followed by seed number 2, then seed number 3, and so on...
Pairing in both the quarter-final and the semi-final are made randomly among the players entering that round.
A higher seeded player defeats the immediate next lower seeded player with a probability of 75%, and always defeats all other lower seeded players.
The probability that seed number 1 reaches the final of the tourney can be expressed as , where and are coprime positive integers. Find .
As an explicit example, this is how the seeds would perform during a match:
Seed number 1 defeats seed number 2 with 75% chance and always defeats seed numbers 3-8.
Seed number 2 defeats seed number 1 with 25% chance, defeats seed number 3 with 75% chance and always defeats seed numbers 4-8.
Seed number 3 always loses to seed number 1, defeats seed number 2 with 25% chance, defeats seed number 4 with 75% chance, and always defeats seed numbers 5-8.
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We need to consider 3 scenarios
1/7 chance of meeting 2 in QF and 3/4 chance of defeating = 3/28. Once he defeats 2, seed 1 is 100% entering final
Faces 3 with 1/7 chance in QF and always beats him. Now 2 will 100% enter SF.
Seed 1 meets seed 2 in SF with 1/3 chance and defeats with 3/4 chance= 1/4
Seed 1 doesn't play seed 2 with 2/3 chance and always wins
so total of 1/4 = 2/3 = 11/12
Total chance is 11/84 of reaching Final
seed 1 meets 4-8 with 5/7 chance in QF and defeats always
Now the question is does seed 2 go through to SF
Chance of Seed 2 not reaching semis... 1/5 chance seed 2 meets seed 3 and loses with 1/4 chance = 1/20
So chance of seed 2 going to SF is 19/20
Probability of seed 1 going to the final then is 5/7 {(1/20) + (19/20)(11/12)}= 221/336