Setting Fair Line
Say I have determined that team A has a 35% chance of winning a game and team B has a 49% chance of winning a game. Team A is playing Team B.
How would I set a fair line for this game?
How would I set a fair line for this game?
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If Team A has a 35% chance of winning a game vs. an average (50/50) opponent and Team B has a 49% chance vs. an average (50/50) opponent, then you can use my (still experimental) Home/Away Probability calculator.
Enter 35% in box 1, 49% in box, and 50% in box 3. Click calculate and you see that Team A has a ~ 35.915% probability of beating Team B.Comment -
Thanks Ganch!
Just wondering what formula that uses.Comment -
Here's how the calculator is programmed to solve this:Let A = Team A home win % (taken as an unbiased estimator of A's "true" home win probability versus an "average" opponent) = 35%Define the logit function of a probability p as the log of the "fair" fractional payout odds p/(1-p) so that:
Let B = Team B road win % (taken as an unbiased estimator of B's "true" road win probability versus an "average" opponent) = 49%
Let H* = expected home win probability between two equally matched teams = 50%
We solve for P = expected probability of A beating B playing at A's Homelg(p) = ln(p) - ln(1-p)inverted:p = exp(lg(p)) (1+exp(lg(p)))This gives us:lg(A) = ln(35%) - ln(1-35%) ≈ -0.6190392084So from Bayes' Theorem:
lg(B) = ln(49%) - ln(1-49%) ≈ -0.0400053346
lg(H*) = ln(50%) - ln(50%) ≈ 0lg(P) = lg(A) - lg(B) - lg(H*)Solving for P we have:
lg(P) ≈ -0.6190392084 - -0.0400053346 - 0 ≈ -0.5790338738
exp(lg(P)) ≈ exp(-0.5790338738) ≈ 0.5604395604So based on the above data (which assumes no home field advantage for either team), A's win probability over B would be about 35.915%.
P = 0.5604395604 (1+0.5604395604) ≈ 35.915%Comment -
Thanks again!Comment -
Nope, I believe it's correct as written:lg(P) = lg(A) - lg(B) - lg(H*)
All else being equal a higher expected home win probability between any two evenly matched teams will result in a lower expected win probability between two specified teams.
For example:A = 50%This would yield P=50%.
B = 50%
H* = 50%
But if we raise H* to, say, 51%, then P drops to 49%.
This drop is expected because due to the fact that A's 50% home win rate is worse than average (51%) A is a below average home team and conversely due to the fact that B's 50% away win rate is better than average (49%) B is an above average road team.
Hence, this implies that P needs to go down when H* moves from 50% to 51%.Comment -
So let's say that
A=.3323
B=.6317
h*=.55
lg(A) = -.693298
lg(B) = .559517
lg(H*) = .200671
lg(P) = -.693298 - .559517 - .200671 = -1.45349
exp(lg(P)) = . 233754
P = .233754/(1+.233754) = .189466
Is this correct?Comment -
I'm getting slightly different numbers that look beyond the discrepancy expected due to rounding error.
lg(A) = ln(0.3323) - ln(1-0.3323) ≈ -0.697800796
lg(B) = ln(0.6317) - ln(1-0.6317) ≈ 0.539516774
lg(H*) = ln(0.55) - ln(1-0.55) ≈ 0.200670695
lg(P) = -0.697800796 - 0.539516774 - 0.200670695 ≈ -1.437988265
exp(lg(P)) ≈ 0.237404874
P = exp(lg(P)) / (1+exp(lg(P)) ≈ 0.237404874 / (1+0.237404874) ≈ 19.186%
See alpha version home/away calculator.Comment -
Not sure about the rounding thing, I did do it in a calculator real quick.
I just found it odd that the home team % would go down so dramatically. But now that I thought about it more, it makes sense. As the far superior home teams would win more than 55% of the time, while the far superior road teams would win more than 45% of the time. It threw me for a second.
Thanks.Comment -
The log5 method that Bill James invented back in 1981 seems to be a standard method used by sports statisticians across different sports.
element1286, may I suggest googling log5?
Ganchrow, care to provide your critics of this method?
Based on my own research of NBA, log5 provides a better fit comparing to "fitted" exponential functions which usage, to the best of my knowledge, is traced back to Steven Skiena's Jai-Alai research in the 90s.Comment -
Yeah, standard log5, (which ignores what I've referred to as H*, or rather assumes it to be 50%) is identical to the logit solution above:
log5:
P = A*(1 - B)/(A*(1 - B) + (1 - A)* B)
(1-P) = (1-A)*B/(A*(1 - B) + (1 - A)* B)
P (1-P) = A B * (1-B) (1-A)
taking the log of both sides:
ln(P) - ln(1-P) = ln(A) - ln(1-A) - (- ln(B) + ln(1-B))
hence: lg(P) = lg(A) - lg(B)
So, yes, the two methds are strictly identical in their results.
Personally, I just find the logit() phrasing more intuitive and easier to remember than log5. But then again I may be partial as I'm by no means a Sabermetrics guy and derived the former method with Bayesian inference prior to having read up of the latter.
QEDComment -
data, you are in fine ballbusting formComment -
Interesting...
I use pythagorean as a complement to my main model since my main one only looks at win/loss and quality of opponent, and pythag can help tell me if a team is barely skirting by on wins (see Washingon this year) or killing opponents (a la New England last year.) So I use both to help filter for value.
Should the above formulas be applied to pythagorean?
Currently, I've got it set up as such:
Assume Team A at Team B pythagorean winning percentages are:
Team A = 60%
Team B = 55%
So, from there, I get Team A's chances of beating Team B to be 55.1%, calculated as:
(.6*(1-.55))/(((.6*(1-.55))+(.55*(1-.6))))
Conversely, then, Team B's chances of beating Team A come to 44.9%:
(.55*(1-.6))/(((.55*(1-.6))+(.6*(1-.55))))
But should I be using those logit formulas instead?
Which is more applicable, and just as important, can someone explain why?
Thanks in advance.Comment -
Ganchrow, thank you.
Wheell, I know you mean well but I realy wished you used different wording. I am trying my best to direct my posts to the fellow posters brains and not to go after their balls. If it comes out otherwise then I missed my target. I apologize, that was not my intentions.Comment -
On a 2-way line, it would be
49/35, or 1.4
That's -140 for the favorite, +140 for the dog.
On a 3-way line, divide each percentage by (100 - that percentage)
So favorite: 49 / 51 = 0.96. Invert = 1.04, so +104.
35: 35/65 = 0.54; invert = 1.86 = +186
Tie = 16 / 84 = 0.19; invert = 5.25 = +525.Comment -
Hmm. I assumed your win percentages are for the same game.Comment -
I just reread this thread to apply the home field portion of it to my spreadsheet, and I can honestly say I've learned more about stats just reading this forum than I ever did in my college stats class - and I had a good prof!
It's just too bad the prof's applications were 'odds of 2 people having same bday in this class' or 'odds of rain on your wedding day' or other such stuff I didn't care about. If he had only applied it in terms of 'odds of home team winning outright'...Comment -
Can the aforementioned formulas be applied to 3-way, soccer-style betting with the draw option? Or is that not feasible given the parameters?Comment -
I'd have to think about this some more but my gut reaction is that a solution to this can't be determined from first principles alone.
Of course if anyone has any additional insight on this I'd love to hear it.Comment -
I feel dumb for asking this because I feel like the answer is staring at me, but for some reason I'm having problems getting my head around it. (I'm blaming it on too much beer last night and not enough coffee this morning...)
Let's say I have team A pegged as a 60% chance of winning vs. an avg opp on NEUTRAL field, and Team B at 45% under those same conditions. Further, I have home field pegged as 58% (home team odds with two equally matched teams.)
As I understand the previous discussions, the odds for A and B were home and away, respectively, not neutral field. So how do I go about converting my neutral field odds into home/away based on my HF assessment (let's assume team A is home)?
Thanks in advance.Comment -
Let An = Team A neutral field win probability = 60%
Let Bn = Team B neutral field win probability = 45%
Let H* = expected home win probability between two equally matched teams = 58%
then:lg(p) = lg(An) - lg(Bn) + lg(H*)or equivalently:lg(p) = lg(An) - lg(Bn) - lg(1-H*)
Solving for p:p ≈ 71.685%
Multipurposing the alpha version home/away calculator:Home Team Win Prob: 60%Clicking calculate yields:
Away Team Win Prob: 45%
League Avg Home Win Prob: 1-58% = 42%Home Win Over Away Prob: 71.685%Comment -
Ahhh, yes, thanks!Comment
