Chess Tournament I

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:

$(A_1-A_2),(A_3-A_4),\dots ,(A_{63}-A_{64})$

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.

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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?

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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:

$$n(A)=2\times \left(\genfrac{}{}{0ex}{}{126}{63}\right)\times (64!)\times (64!)$$

where $A$ is the event described above.

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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:

$$P(A)=\frac{2\times \left(\genfrac{}{}{0ex}{}{126}{63}\right)\times (64!{)}^2}{128!}$$

Expanding the combination formula:

$$\left(\genfrac{}{}{0ex}{}{126}{63}\right)=\frac{126!}{63!\times 63!}$$

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:

$$P(A)=\frac{64}{127}$$

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Final Answer

$$\boxed{\frac{64}{127}}$$

Thus, the probability that the highest-rated and second-highest-rated players meet in the final is $\frac{64}{127}$.