Suppose there is a league (your favorite sport here) of N teams that play each other X times throughout the season.
For every team that has W wins, is there always another team that has L losses?
I’ve tried doing this by induction, but I’ve always lost patience once N gets larger than 3. (And I’m convinced that there’s probably an easier way to prove or disprove this conjecture, but I’m lazy. ^^; )
Can you restate your question? I can’t figure out exactly what you’re asking.
No, while professional sports are zero-sum (the wins = the losses for the league) that doesn’t implay that if one team is 10-6 that some other team is 6-10.
Imagine a four team league that plays a 10 game season. Imagine that three teams went 6-4 and one team went 2-8.
A: 6-4
B: 6-4
C: 6-4
Milwaukee Brewers: 2-8
Sorry. Couldn’t resist.
Therefore, since the system balances (the wins for the league equal the losses) and no team is 8-2 (to offset the 2-8) the answer to your question is no.
Note that JC’s answer is also applicable if ties are possible.
I thought we were discussing sports…not hockey.
Anyway, I approached that in an empirical way. I’m sure there are some math geniuses around who could better define it.
Basically, Jonathan Chance already answered the question with a concrete example, but my question, in essense, is this:
In totality, a league’s wins always equals a league’s losses. But individually, is there always at least one (or two) teams were one team’s wins equals another team’s losses? (A team that is, say, 5-5, would count.)
I’m looking for a more abstract mathematic proof, but Jonathan Chance’s answer works, too, I think. (Time to work out a game schedule in excel, whee.)
[math genius]The conjecture is false in general. See Jonathan Chance’s first post in this thread for a counterexample.[/math genius]
The simplest counter-example: a competition of 4 teams, playing each other team once.
A beats B and beats D, getting 2 wins
B beats C and beats D, getting 2 wins
C beats A and beats D, getting 2 wins
D beats no one, getting 0 wins
No team gets 1 win, to match the 2 wins of A, B and C.
No team gets 3 wins, to match the 0 wins of D.
Okay, I’m fairly convinced.
Follow up question:
Are there any situations where you could say with certainty that my conjecture is true (other than the trivial case of 2 teams)? If so, what are they? If not, why not?
My conjecture is that it’s true in the non-trivial case of three teams each playing the other team once (the only possibilities are 1-1-1 and 2-1-0), and it’s not true for more than three teams (just generalise my counter-example above) or for 3 teams playing each other more than once. In the last case, with 3 teams each playing the others 2 times, you can partition the 6 wins as 3-3-0 (A and B consistently beat C, and each of A and B has one win over the other). Given that the conjecture is false for 2 roiunds, I’m sure that it’s false for any larger number of rounds.
The simplest way to disprove this is to imagine a team winning all their games.
Does that mean some team must lose all theirs? Of course not, unless there’s only two teams.
Or if there are only three teams, and each team plays the other only once.
Assume that A beats both B and C. Now, either B beats C (in which case C hgas no wins), or C beats B (in which case B has no wins).
Excluding ties, there will be at least two teams whose records are either the same (8-6 and 8-6) or opposites (8-6 and 6-8). Proof by Pigeon Hole Principle. There are more teams than |wins-losses|. In fact nearly double, but the PHP is not strong enough to say anything that’s both simple and interesting in this case.
No. If you have 4 teams, and 3 rounds, that’s a total of 18 games, with each team playing 9 games.
A wins 7, loses 2
B wins 6, loses 3
C wins 5, loses 4
D wins 0, loses 9