Proving that mean KDR in a videogame is one

Question

This is not related to schoolwork. A friend of mine challenged me to prove that the mean KDR (assuming players can only die at the hands of other players) must always be equal to one. I have gotten through the logic part, and am now faced with the math part (which I am less capable of doing), and would like some help.

My logic is as follows: Every kill event is also a death event, and every death event is also a kill event. Therefore, the total number of kills must always equal the number of deaths. We know then that TOTAL kdr = 1. For a more explicit expression of this, we can represent the game as a directed graph whose nodes are players and whose edges represent kills. In such a graph, the in-degree of a node is its deaths and its out-degree is its kills.

I don't think that will be necessary. Instead I'm just sticking with that kills and deaths are distributed a certain way among players, and that each player's kdr = kills/deaths for that player. Proving what I have written in the following picture should finalize the whole proof. If it isn't possible to prove this, maybe the graph abstraction will help us.

EDIT: Forgot to mention earlier that if a player has 0 deaths, the denominator of their KDR is 1, not 0 (to prevent infinite KDR, widely used in games).

Answer 1

This is not true: consider for example $2$ players $A$ and $B$, $A$ kills $B$ once, $B$ kills $A$ $m>1$ times, then we have:

$$\frac{m+\frac{1}{m}}{2}\not=1$$

Answer 2

This is clearly false,if there are two player: Alice and Bob and Bob kills alice 20 times and Bob kills alice once then Alice's ratio will be $\frac{1}{20}$ and Bob's ratio will be $\frac{20}{1}$, the mean of these two ratio's is $\frac{401}{40}>20$

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Construction to increase mean kdr arbitrarily for any number of players:

Pick 1 player $v$ and make him kill all other players once. Then make all other players kill $v$ $k$ times. What is the sum of all the kdr now?

$\underbrace{\frac{n-1}{(n-1)k}}_{\text{player v}}+\underbrace{(n-1)\frac{k}{1}}_\text{the other n-1 players}\geq(n-1)k$.

So the arithmetic mean of the kdr's is at least $\frac{(n-1)k}{n}$. Increasing $k$ arbitrarily increases the kdr mean arbitrarily.