Question

Eight players $P_1,P_2,....P_8$ play a knock out 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 player $P_4$ reaches the final ?

• A. $\frac{4}{35}$
• B. $\frac{5}{35}$
• C. $\frac{6}{35}$
• D. $\frac{1}{35}$

Answer 1

The correct option is A $\frac{4}{35}$

The number of ways in which $P_1,P_2,...P_8$ can be paired in four pairs
$\frac{1}{4!}\left({{}^{8}C}_2\right)\left({{}^{6}C}_2\right)\left({{}^{4}C}_2\right)\left({{}^{3}C}_2\right)=\frac{1}{4!}\times \frac{8!}{2!6!}\times \frac{6!}{2!4!}\times \frac{4!}{2!2!}\times 1$
$=\frac{1}{4!}\times \frac{8\times 7}{2!\times 1}\times \frac{6\times 5}{2!\times 1}\times \frac{4\times 3}{2!\times 1}=\frac{8\times 7\times 6\times 5}{\mathrm{2.2.2.2}}=105$
Now, at least two players 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 against each other in between $P_1,P_2,P_3$ and remaining players will play against one of the players from $P_5,P_6,P_7,P_8$ and $P_4$ play against one 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 $i=2,3$ 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 final round from the second $=\frac{1}{3}$
Therefore probability the $P_4$ reach the final is $=\frac{12}{35}\times \frac{1}{3}=\frac{4}{35}$