Question

In a knock-out chess tournament, eight players $P_1,P_2,.....P_8$ participated. It is known that whenever the players $P_i$ and $P_j$play, the players $P_i$will win j 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. $\frac{31}{35}$
• B. $\frac{4}{35}$
• C. $\frac{8}{35}$
• D. none of the above.

Answer 1

The correct option is B

$\frac{4}{35}$

Explanation for the correct answer:

Find the probability that the player $P_4$ reaches the final.

Given,

eight players $P_1,P_2,.....P_8$ participated, whenever the players $P_i$ and $P_j$play, the players $P_i$will win j if $i<j.$

Now,

$P_1,P_2,.....P_8$ can be paired into four pairs.

So,

The number of different combinations is

$$\begin{gathered} =\frac{1}{4!}\left[\left({}^8{C}_2\right)\left({}^6{C}_2\right)\left({}^4{C}_2\right)\left({}^2{C}_2\right)\right] \\ =\frac{1}{4!}\left[\left(\frac{8!}{6!2!}\right)\left(\frac{6!}{4!2!}\right)\left(\frac{4!}{2!2!}\right)\left(\frac{2!}{0!2!}\right)\right] \\ =\frac{1}{4!}\left[\left(\frac{8\times 7}{2}\right)\left(\frac{6\times 5}{2}\right)\left(\frac{4\times 3}{2}\right)\left(\frac{2\times 1}{2}\right)\right] \\ =\frac{1}{4!}\left[28\times 15\times 6\times 1\right] \\ =105 \end{gathered}$$

Now, at least two players reach the second round out of $P_1,P_2\mathrm{and}{P}_3$. And $P_4$ can reach in final if definitely two players against each other in between $P_1,P_2,P_3$ and the other players will play with one of the players from $P_5,P_6,P_7,P_8\mathrm{and}{P}_4$ play against one of the other three from $P_5,P_6,.....P_8$.

This can be possible in

$\begin{array}{rcl}{}^3{C}_2{\times }^4{C}_1{\times }^3{C}_1& =& 3\times 4\times 3\\ & =& 36\mathrm{ways}\end{array}$

Therefore the probability that $P_4$and exactly one of $P_5,P_6,.....P_8$ reach the second round $=\frac{36}{105}=\frac{12}{35}$.

If $P_1,P_i,P_4,and{P}_j$ where $i=2,3\mathrm{and}j=5\mathrm{or}6\mathrm{or}7$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 $=\frac{12}{35}\times \frac{1}{3}=\frac{4}{35}$

Hence, the correct option is B.