*Originally posted by EinsteinMind*

**sorry for the double post guys. need to give a more detailed explanation of proving his theory.
**

you see, in any tournament, the points won has to equal the points lost.

From this we can derive a simple equation: w=l (win points equals loss points.)

Now he wants me to prove that It is impossible that losses are greater than wins, or this equatio ...[text shortened]... asically a law of chess tournaments, every player cannot lose more than they can win, only some.

I don't think you are being asked to prove that losses are greater than wins, well at least I can't understand your logic which doesn't surprise me if you check out the hat thread. You are being asked to prove that in a round robin tournament (X) with a certain number (

*n*) of players there exists a clear winner (A) and that the maximum cumulative points total of those players who defeated (A) is less than the minimum number of the cumulative points total of those players who lost to (A).

If there is 1 player in the tourney then it's absurd. If 2 players then the winner can't lose a game and still win, if 3 players each player plays two games so yet again you can't lose a game and be a clear winner so we are only being asked to prove this when

*n* > 3.

Example: 8 player round robin.

It's an all-play-all so the number of games shall be 7+6+5+4 etc which can be written as

*n*(

*n*-1)/2 (Like the triangular number series only the n value isn't counted: we start from

*n*-1, and hence

*n*(

*n*-1)/2 and not

*n*(

*n*+1)/2.) So in an eight player tourney there will be 28 games and a total of 28 points available. If we take the minimum number of points needed by A (in order to maximise the number of players that beat A, and their cumulative total) we find that A needs 4 points to be outright winner (we don't count draws because they will merely take a player out of the loop.)

If A wins 4 games (

*n*/2) then A had 3 losses therefore there are three players in the group of players who beat A (

*n*/2-1). The maximum that each member of this group can score and still be a runner-up is a half point less than the score by the winner, therefore 3.5 (

*n*/2-0.5).

The total points that are available is 28 so if we subtract the winner's points we have 24. The maximum that the group of players who beat A can muster is 3.5*3 ([

*n*/2-1][

*n*/2-0.5]) =10.5. If we subtract 10.5 from the 24 we are left with 13.5 therefore the group of players that lost to A have a higher cumulative total than those who won against A.

A tourney with

*n* players:

For a tourney with n players there are

*n*(

*n*-1)/2 games available and n(n-1)/2 total points. If we subtract the wins of player A we get:

*n*[2]/2-

*n*/2-

*n*/2 [

*n*[2] means n squared]

=

*n*[2]/2 -

*n*
The maximum amount of points won by the group of players who beat A is:

(

*n*/2-1)(

*n*/2-1/2)

=

*n*[2]/4 - 3/4

*n* + 1/2

We must prove that

*n*[2]/2-

*n* is more than double

*n*[2]/4-3/4

*n*+1/2

We know that

*n*[2]/2 is double

*n*[2]/4 so we only need to show that

*n* is greater than 3/4

*n* +1/2.

But we said at the start that we are proving this for values of

*n*>3 and it's clear that even substituting for the lowest value of

*n* (that is 4) we get 4 > 3+ 1/2.

I am crap at maths so excuse the laboured explanation I am probably missing something mundane in the earlier explanations. I am not sure if this explanation is exactly the same for odd numbers of participants.