Power Rankings - obvious error

[SumZero]

King Yao may have been glossing over this with the "This is a simple addition and subtraction method. You may feel a different formula more accurately describes the probability one team has of beating the other." but the effect of using:

prob(a wins) = Arank% - Brank% + 50%

leads to wrong answers that are more wrong the further from even strength you get. The correct formula to use is:

prob(a wins) = (Arank% * (100%-Brank%)) / ((Arank% * (100%-Brank%)) + ((100%-Arank%) * Brank%)))

this should be familiar to baseball saber folks as bill james's log5 formula, or as one of the forms of the Pythagorean formulas and isn't too complicated (it is more obvious with numbers than with "Arank%" and the like).

Intuitively you can think that it models A playing B by actually having A play the fictitious "average team" and B play the fictitious "average team". If both A and B lose or both A and B win this is a non-event and you repeat their games against the "average team". If only one of them wins then that is the team that wins the match between A and B. Thus the numerator is the probability A wins against the "average team" and B loses against the "average team" and the denominator is the probability that EITHER A wins against the "average team" and B loses against the "average team" OR A loses against the "average team" and B wins against the "average team".

For the values of A=55% and B=45% the fake method in the article yields A wins 60% where as the real calculation gives ~59.9% which isn't too much of a big deal (the formula is (.55 * (1-.45))/((.55 * (1-.45))+(.45 * (1-.55)) which simplifies to .55^2/(.55^2+.45^2) hence the Pythagorean name). But consider A=75% and B=25%. The fake method says A wins 100% where the real method gives 90%. And consider A=80% and B=20%. The fake method says A will win 110% where the real method gives ~94.1%. Finally consider A=100% and B=0%. The fake method says A will win 150% where the real method gives 100%.

The real method also has the reasonable property that if A is 100% against the "average team" - i.e., it can never lose, it is the perfect team - then it will have a 100% win probability against any B that isn't also a perfect 100% (if both are 100 then it is and in determinant 0/0) and, likewise, if A is 0% against the "average team" - i.e., it can never win, it is the worst team possible - then it will have a 0% win probability against any B that isn't also a complete 0% (which again will be 0/0). A final sanity check is any teams of equal strength (other than 100 and 0) will result in 50% as the answer and all answers for any combination of teams (other than 100 v 100 and 0 v 0) will result in an answer between 0 and 100%, and A's win probability plus B's win probability will always equal 100%.

The effect of "adding" in the home field advantage is similarly flawed as HFA will matter differently depending on the neutral field result. Trivially, if HFA is +8 you can show it is wrong if there could ever be two teams who's neutral score was less than 8% or greater than 92% since HFA will take you below 0% or greater than 100%.

[King Yao]

I just saw this...nice post. I agree with you.

The simple addition/subtraction method is nice and easy if you have to do it on the fly (i.e., in a Las Vegas sportsbook), but clearly your method is superior with a spreadsheet.