The Log5 method seems to be what you're after:
http://home.ptd.net/~artnpeg/Log5.pdf
Only winning percentages are considered.
p(x,y) = (x - x*y)/(x + y -2*x*y)
Art
I would like to find a formula to calculate the probability that Team A will beat Team B taking into account just their records to date.
For example, suppose these 6 teams have these records against the rest of the league, which includes 30-40 other teams. Assume that each of these teams have played each of the other teams more or less the same number of times.
Is there a good formula for calculating the probability that each of these teams will beat each of the others?Code:Team Wins Games % A 40 50 80.00% B 30 50 60.00% C 25 50 50.00% D 20 50 40.00% E 10 50 20.00% F 0 50 0.00%
If P(x,y) is the probability that Team X will beat Team Y, where x & y are the winning percentages from the table above, then P should have the following properties:
I have tried a couple of formulas, but they all have flaws. I'd appreciate suggestions.
- 0 <= P <= 1
- P(x,x) = 0.50 (Two teams with the same record should have 50% odds against each other.
- P(x,y) + P(y,x) = 1 (The complementary probabilities should sum to 1.)
- P(x,0.5) = x (Since Team X's record is against the entire league and the average of all the wins and losses for all of the teams in 50%, if Team X plays a team with a record of 50%, it should be like playing the entire league, which is what its record is. I am not sure about this one,m but it seems right to me.)
Two that I tried are:
- P(x,y) = (x-y+1)/2
- P(x,y) = x/(x+y)
The Log5 method seems to be what you're after:
http://home.ptd.net/~artnpeg/Log5.pdf
Only winning percentages are considered.
p(x,y) = (x - x*y)/(x + y -2*x*y)
Art
Jennifer Murphy (02-23-2016), Outlier (02-22-2016)
Thanks, Art. This does appear to be exactly what I was looking for. I even attempted this very derivation, but fell short. I see now where I went wrong.
Thanks so much.
How do you know about this?
Do you know why they call it the "Log5" method?
Are you sure? I don't see where it says that?Only winning percentages are considered.
Jennifer, you're welcome. I learned of the log5 while doing internet searching for
a solution. I don't know why the name log5 since it has nothing to do with
logarithms. When I say "only winning percentages are considered" I mean that
the analysis is simplistic since it doesn't allow for known significant factors
such as home field advantage (several percent), etc. If you internet search
along these lines you'll find formulas that do take such factors into
consideration.
Art
This one
". . .provide a terriﬁc estimate. . ."
kinda' jumped out at me.
How does this author define "terriﬁc"? (I thought stats was quantitative, not qualitative - objective, not persuasive).
I have not read the posts above, but I also recall seeing a program that does the same thing, but incorporate if you had teams head-to-head records as well.
Stop cowardice, ban guns!
Tweet |