What is wrong with the logic/math here?

Question

Logically, in a win/loss scenario with x number of teams, the average win/loss ratio for x# of teams shoul equal 1 (1 win for 1 loss).

But:
Team 1 = 10-5
Team 2 = 40-10
Team 3 = 2 -15
Team 4 = 3-25
(the fact that the total number of w/l for each team is different doesn't matter...look at it as more of a points for and points against scenario)

The total number of wins is 55 and the total number of losses is 55. The overall ratio of wins to losses is therefore 55:55 or 1:1 or 1.0 wins for every loss.

However, if you take the perspective that Team 1 has a 4.0 (4 wins for every 1 loss) Team 2 has a 2.0, Team 3 has a .13 and Team 4 has a .12.

The average of these ratios (4, 2, .12, .13) is not equal to 1 but it is greater.

I would like some help understanding why this is.

Answer 1

There is really no reason why the win/loss ratios should have an arithmetic mean (average) of 1 despite the fact that the total wins must necessarily equal the total losses.

Indeed, it can be expected that the mean will usually be greater than 1 because an excess of wins will necessarily increase the ratio more than a similar excess of losses will decrease it.

To see how this works, take the simple case of only two teams, with team 1 being victorious v times and defeated d times. In that case, team 2 will be victorious d times and defeated v times. If either v or d is zero, one of these ratios will be zero and the other will be undefined, so we insist that both v and d be greater than zero.

If we let r = v/d, then the average of the two win/loss ratios becomes (r + 1/r) / 2. A little experimenting will show that for any value of r greater than zero except 1, or we can do it mathematically like so:

If r > 0 and r ≠ 1, then
(r - 1)^2 > 0
Expanding the polynomial on the left, we get
r^2 - 2r + 1 > 0.
Adding 2r to both sides gives
r^2 + 1 > 2r
Dividing through by 2r will not change the sense of the inequality because r is necessarily positive. Doing the division then gives
(r + 1/r) / 2 > 1
The expression on the left is our average for the win/loss ratio, which always be 1 for any legal value of r other than 1. By a similar process, we can show that the average will be 1 only in the special case that r = 1.

Answer 2

Each W/L ratio is a fraction. When you add the ratios and then average, what you are really doing is adding the numerators (total number of wins) and dividing by the least common denominator of losses. So if you put each W/L fraction into a common denominator (I used 75, but it doesn't really matter), you get the following:

300/75 150/75 10/75 9/75

You will find those W/L fractions are the same as your scenario. So by adding those four fractions you get

469/75

which is the same thing you get when you add the W/L ratios as decimals, 6.25 or so (and when you divide by four (taking the average) you get something a little larger than 1.5 (which you observed).

But the point is that you added up all the wins (numerators) but only used the common denominator for the losses, since the denominators don't add when you sum fractions. So in effect you are not really summing the losses when you average the W/L ratios, you are only summing the wins and dividing by something related to the losses.

Maybe this would be clearer if I explained it using an interpretive dance.

Answer 3

There is no reason the average of the ratios should equal one. This is similar to asking why the average of 2 and 1/2 isn't one. There is no reason for it to be.

Answer 4

Thou shall not average averages.