I am no mathematician, just a programmer and gamer who thought about this problem. I reckoned it's more relevant here than one of the gaming-SEs, as it's not really about any particular game, but "game" as an idea.
Consider a game (probably considered "fair") where you have one player as the winner and one player as the loser. Every player has a history of games he played which is used to calculate his "win percentage", that is e.g. played 523 games, won 312 --> 59.65%. There can be quite a large amount of players and games.
The thing that I wonder is: In a system like this, is it possible for ALL players of the game to have a win % over 50?
Rephrased: For every player above 50%, does there need to be a player that has the same amount of percentage below 50% to "balance" it?
Maybe this doesn't make a lot of sense or I am missing a simple solution, but I just can't wrap my head around the math of how to think about this...
Let $N \geq 1$ be the number of played games, $P_1$ the number of games the first player won and $P_2$ the number of games the second player won.
As every game has a unique winner we have $P_1 + P_2 = N$. The percentage of games the first player won is given by $P_1/N$. Similarly the percentage of games the second player won is $P_2/N$. If we had $P_1/N > 0.5$ and $P_2/N > 0.5$ then $$ \frac{P_1}{N} + \frac{P_2}{N} > 0.5 + 0.5 = 1. $$ But we must have $$ \frac{P_1}{N} + \frac{P_2}{N} = \frac{P_1 + P_2}{N} = \frac{N}{N} = 1. $$ So this is not possible.
Notice that instead of just two players we could also look at $n$ players, $n \geq 2$, and still get the same result: Instead of just $P_1$ and $P_2$ we have $P_1, \dotsc, P_n$ with $P_1 + \dotsb + P_n = N$. If we then had $P_1/N > 1/n, \dotsc, P_n/N > 1/n$ then $$ \frac{P_1}{N} + \dotsb + \frac{P_n}{N} > \underbrace{\frac{1}{n} + \dotsb + \frac{1}{n}}_{n} = 1, $$ but we must have $$ \frac{P_1}{N} + \dotsb + \frac{P_n}{N} = \frac{P_1 + \dotsb + P_n}{N} = \frac{N}{N} = 1. $$
It is maybe interesting to notice that this only works if every game has at most one winner. If a game can have multiple winners then it could happen that every player has won more games than lost: Suppose for example that we have three players $A$, $B$ and $C$, and that $A$ and $B$ have won the first game, $B$ and $C$ have won the second game, and $A$ and $C$ have won the third, and final, game. Then every player has won $2$ out of $3$ games!
But of course in this case “more than average” does not longer mean having won more than one third of the games, but having won more than $2$ games, as everyone has won $2$ games on average.