Question

Eight players $P_1,P_2,P_3,..........P_8$ play a knockout tournament. It is known that whenever the players $P_i$ and $P_j$ play , the player $P_i$ will win if $i<j$. Assuming that the players are paired at random in each round, what is the probability that the players $P_4$ reaches the final?

• A. $4/35$
• B. $2/35$
• C. $2/21$
• D. $2/7$

Answer 1

Answer: (A)

$4/35$

The number of ways in which $P_1,P_2,...,P_8$ can be paired in four pairs
$=\frac{1}{4!}\left[\left({}^8{C}_2\right)\left({}^6{C}_2\right)\left({}^4{C}_2\right)\left({}^2{C}_2\right)\right]$ ways.
$=\frac{1}{4!}\times \frac{8!}{2!6!}\times \frac{6!}{2!4!}\times \frac{4!}{2!2!}\times 1$
$=\frac{1}{4!}\times \frac{8\times 7}{2!\times 1}\times \frac{6\times 5}{2!\times 1}\times \frac{4\times 3}{2!\times 1}=\frac{8\times 7\times 6\times 5}{\mathrm{2.2.2.2}}=105$
Now, atleast two pairs certainly reach the second round in between $P_1,P_2$ and $P_3$. And $P_4$ can reach in final if exactly two players play against each other in between $P_1,P_2$ and remaining players will play against one of the players from $P_5,P_6,P_7,P_8$ and $P_4$ plays against one of the remaining three from $P_5...P_8$
This can be possible in ${}^3{C}_2{\times }^4{C}_1{\times }^3{C}_1=\mathrm{3.4.3}=36$ ways.
$\therefore$ probability that $P_4$ and exactly one of $P_5...P_8$ reach second round.
$=\frac{36}{105}=\frac{12}{35}$
If $P_1,P_i,P_4$and$P_j$ where $i=2$or$3$ and $j=5$or$6$or$7$ reach the second round, then they can be paired in $2$ pairs in $\frac{1}{2!}\left({}^4{C}_2\right)\left({}^2{C}_2\right)=3$ ways.
But $P_4$ will reach the final, if $P_1$ plays against $P_i$ and $P_4$ plays against $P_j$.
Hence, the probability that $P_4$ reach the final round from the second
$=\frac{1}{3}$
$\therefore$ the probability that $P_4$ reach the final is $\frac{12}{35}\times \frac{1}{3}=\frac{4}{35}$