Eight Player $$P_{1},P_{2},P_{3},P_{4}.......P_{8}$$ are going to play a Knock-Out Tournament.It is known that whenever $$P_{i}$$ and $$P_{j}$$ plays , The Player $$P_{i}$$ wins if i<j . Assuming that Players are paired at random in each round, What is the probability that player $$P_{4}$$ reaches the final ??

Also Find the probability of all the other Players in reaching the final ?

What should be sum of all the individual probabilities of each player in reaching the final ?

4 years, 8 months ago

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For first question, I think $$P$$ $$=$$ $$\frac{4}{35}$$.

Reason: Let, integers from $$1$$ to $$8$$ represent respective players. Firstly, in order for $$P_{4}$$ to win, the player $$4$$ has to be paired with a higher number with is one of these: $$5$$, $$6$$, $$7$$ or $$8$$. This can be done with a probability $$\frac{1}{4}$$.

Secondly, in order to win, two large numbers greater than $$4$$ has to be paired to ensure there is a larger number to win from when 4 reaches in the round before semi-final. This can be done with a probability $$\frac{^3C_2}{^6C_2}$$.

Finally, making use of Bayes Theorem, $$P$$ $$=$$ $$\frac{1}{4}\times\frac{^3C_2}{^6C_2}$$ = $$\frac{4}{35}$$.

Though I am a bit hazy about the answer due to the random effect (players are shuffled after every round).

- 4 years, 8 months ago

No..I Think you are correct.

- 4 years, 8 months ago

I was thinking about this problem and i think I've figured it out! P1 has a 7/7 chance, or %100 P2 has a 6/7 chance, or about %86 P3 has a 5/7 chance, or about %71 P4 has a 4/7 chance, or about %51 P5 has a 3/7 chance, or about %43 P6 has a 2/7 chance, or about %29 P7 has a 1/7 chance, or about %14 P8 has a 0/7 chance, or %0 these are all decimals except for P1 and P8

- 4 years, 8 months ago

The chance of $$P_{7}$$ and $$P_{6}$$ in reaching the final is also zero...The probabilities you have given are for Players to reach the semifinal and not the finals..

- 4 years, 8 months ago