Problem Statement
A chess tournament has 128 players, each with a distinct rating. Assume that the player with the higher rating always wins against a lower rated opponent and that the winner proceeds to the subsequent round. What is the probability that the highest rated and second-highest rated players will meet in the final ?
Original Problem Link: Click here
Solution
Before proceeding to the question, let’s think of the following situation -
Initially, 128 people are divided into 2 groups of 64 people each.
In every round, each group will play within itself (say each guy plays with the guy beside him). How will a group play within itself?
Consider group A with \(A_1, A_2, A_3, A_4, \dots, A_{63}, A_{64}\).
Let’s say that adjacent people play chess with each other. Then the games will be:
From each game, one person (with the higher skill) will win .
Thus, after the first round, we will have 32 players remaining .
This process will go on until only one guy remains in each group , and that person will play with the person from another group to decide the final winner .
Now, try to convince yourself that this model is exactly the same as the situation given in the question.
How? 🤔
- If the initial distribution of people among Group A and B , and the arrangement of people within both groups are all random , it will consider all possible plays between every player.
- Each unique initial arrangement corresponds to a unique gameplay .
- Since by randomizing the initial group allotment , all possible distributions of people in Group A and B are considered, it means all possible gameplays are also considered. This exactly models the given problem.
Key Observation
In our model of the gameplay, the highest-rated and second-highest-rated player will face off against each other only when they are in different groups initially .
This problem is now much simpler, effectively reducing it to:
In how many ways can I arrange 128 people in 2 groups (their position within the group is unique), such that 2 particular people are always in different groups?
Solving the Sub-Problem
First, let’s allot the highest player and second-highest player to each group.
There are 2 ways of doing this:
- Highest -> Group A, Second-Highest -> Group B
- Highest -> Group B, Second-Highest -> Group A
Now, once this is done:
- Out of the remaining 126 people , select 63 people for Group A.
- The remaining 63 go to Group B.
- Now, we must arrange the people within each group itself , thus another \((64!) \times (64!)\) (one for each group).
Our final expression for the favorable cases becomes:
\[n(A) = 2 \times \binom{126}{63} \times (64!) \times (64!)\]where \(A\) is the event described above.
Computing the Probability
To find the probability of this arrangement, we must compute the total number of possible arrangements .
Clearly, this is \(128!\) How 🤔 ?
- Assume a partition after \(P_{64}\) (representing the end of Group A and the beginning of Group B) exists, to seperate groups A and B.
- We can linearly arrange all 128 people in \(128!\) ways. Thus each combination will represent a unique gameplay.
Thus,
\[n(S) = 128!\]The required probability is:
\[P(A) = \frac{n(A)}{n(S)}\]Substituting values:
\[P(A) = \frac{2 \times \binom{126}{63} \times (64!)^2}{128!}\]Expanding the combination formula:
\[\binom{126}{63} = \frac{126!}{63! \times 63!}\]Thus:
\[P(A) = \frac{2 \times 126! \times (64!)^2}{63! \times 63! \times 128!}\]Expanding \(128!\) as \(128 \times 127 \times 126!\):
\[P(A) = \frac{2 \times 126! \times 64! \times 64!}{63! \times 63! \times 128 \times 127 \times 126!}\]Canceling \(126!\):
\[P(A) = \frac{2 \times 64! \times 64!}{63! \times 63! \times 128 \times 127}\]Using:
\[\frac{64!}{63!} = 64\] \[P(A) = \frac{2 \times 64 \times 64}{128 \times 127}\]Simplifying:
\[P(A) = \frac{64}{127}\]Final Answer
\[\boxed{\frac{64}{127}}\]Thus, the probability that the highest-rated and second-highest-rated players meet in the final is \(\frac{64}{127}\).
💡 Did you enjoy this problem? Check out more puzzles in the Problems section!