Understanding a solution based on Enumerative Combinatorics 
PROBLEM  STATEMENT :


$(a)$ At  a round robin chess tournament, at least $\frac {3}{4}th $ of the games ended by a draw. Prove that there were two players who had the same final score.

NOTE: A player gets $1$, $\frac {1}{2}$, $0$ points respectively for a win, draw, and a loss.

$(b)$ Now assume the tournament has been interrupted after t rounds,
that is, after each player has finished t games. (We assume, for
simplicity, that the number of players is even.) Is it still true that if
at least 3/4 of the games played ended by a draw, then there were
two players with the same total score?


$(c)$  If the games of the tournament are played in a random
order (there are no rounds; one player can finish many games before
another player starts), and the tournament is interrupted at some
point, could it happen that three 3/4 of the finished games ended
by a draw, but there were no two players with the same total score?


$(d)$. Is there a constant K < 1 such that if we organize the tournament
as in the preceding case, and we interrupt the tournament at a point
when at least K of the finished games ended by a draw, then there
will always be two players with the same total score?


SOLUTIONS

Here are the solutions to part $(a)$ and $(d)$:



After going through the solutions thoroughly, I couldn't understand some part of solution (a), and entire of (d). Parts (b) and (c) I could understand.
Can someone help me in understanding the following parts:
(1) The first rectangular box : I think it's a typo, it should have been $\frac {b_y}{2}$, am I right.
(2) Second figure in red: Why is it valid to assume that k players have a positive score?
(3) I literally couldn't understand solution of part (d) at all. So I would be happy is someone explains it in a simple language: I tired taking examples, and assuming values, but it didn't work much.
Thanks.
 A: *

*You are correct, it should be $\frac{b_y}2$.


*There are $n$ different scores. At least $n-1$ of these are nonzero. There are two cases; either there are at least as many positive scores as negative scores, or there are more negative scores than positive scores.

*

*If there are at least as many positive scores as negative ones, then number of positive scores is at least half, so at least $\lceil (n-1)/2\rceil$, which you can show is equal to $n/2$ when $n$ is even and $(n-1)/2$ when $n$ is odd, which is exactly the definition of $k$.


*If there are more negative than positive scores, then you imagine reversing the results of all the games. This will negate everyone's scores, so now there will be more positive scores than negative ones, so we are in the previous bullet. Since reversing all outcomes does not affect the number of draws, we can complete the proof for the reversed situation, and the result still holds.




*Here is an illustration of their scheme when $n=8$. The symbol $\times$ means that no game was played. Clearly, of the games that were played, the overwhelming majority of the results were draws. Furthermore, you can check all players have different scores. As you $n$ gets larger, the proportion of the games that were played which were draws gets arbitrary close to $1$. Therefore, there does not exists a $K$ such that when you stop the games as soon as the proportion of draws exceeds $K$, there will exist $2$ people with the same score.
Edit: There was a mistake in my table, corrected entry is red.
The table is now exactly as the book described. As you note, it is not true that that each $A_i$ has $i/2$ wins. However, it is true that $A_i$ has $(i-1)/2$ wins, so this is likely just an off-by-one error on the part of the author. The important thing is that all of the scores are different.
$\newcommand{\d}{\text{draw}}$
$$
\begin{array}{c|cccccccc}
\text{vs.} & A_1 & A_2 & A_3 & A_4 & A_5 & A_6 & A_7 & B
\\\hline
A_1 & \times & \times & \times & \times& \times & \times & \times & \text{B wins}
\\
A_2 & \times & \times & \times & \times& \times & \times & \text{draw} & \text{B wins}
\\
A_3 & \times & \times & \times & \times& \times&  \text{draw}  & \text{draw} & \text{B wins}
\\
A_4 & \times & \times & \times & \times& \d&  \text{draw}  & \text{draw} & \color{red}{\text{B wins}}
\\
A_5 & \times & \times & \times & \d& \times&  \text{draw}  & \text{draw} & \d
\\
A_6 & \times & \times & \d& \d& \d&  \times  & \text{draw} & \d
\\
A_7 & \times & \d & \d& \d& \d&  \text{draw}  & \times & \d
\\
B & \text{B wins}& \text{B wins}& \text{B wins} & \color{red}{\text{B wins}}& \d& \d& \d
\end{array}
$$
