Probability Question!! Please Help..!!!

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 ?

Please help me solve my doubt..!!!

#HelpMe! #Advice #MathProblem #Math #Opinions

Easy Math Editor

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Comments

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

Billy Coleman - 7 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..

Rushi Rokad - 7 years, 8 months ago

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).

Lokesh Sharma - 7 years, 8 months ago

No..I Think you are correct.

Rushi Rokad - 7 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$