[element1286]

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?

[Ganchrow]

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.

[element1286]

Thanks Ganch!

Just wondering what formula that uses.

[Ganchrow]

Originally Posted by element1286

Thanks Ganch!

Just wondering what formula that uses.

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%
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 Home

Define the logit function of a probability p as the log of the "fair" fractional payout odds p/(1-p) so that:

lg(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%) ≈ -0.6190392084
lg(B) = ln(49%) - ln(1-49%) ≈ -0.0400053346
lg(H*) = ln(50%) - ln(50%) ≈ 0

So from Bayes' Theorem:

lg(P) = lg(A) - lg(B) - lg(H*)
lg(P) ≈ -0.6190392084 - -0.0400053346 - 0 ≈ -0.5790338738

Solving for P we have:

exp(lg(P)) ≈ exp(-0.5790338738) ≈ 0.5604395604
P = ^0.5604395604/_{(1+0.5604395604)} ≈ 35.915%

So 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%.

[element1286]

Thanks again!

[element1286]

Originally Posted by Ganchrow

lg(P) = lg(A) - lg(B) - lg(H*)

Should it be lg(P) = lg(A) - lg(B) + lg(H*)?

[Ganchrow]

Originally Posted by element1286

Should it be lg(P) = lg(A) - lg(B) + lg(H*)?

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%
B = 50%
H* = 50%

This would yield P=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%.

[element1286]

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?

[Ganchrow]

Originally Posted by element1286

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?

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.

[element1286]

Originally Posted by Ganchrow

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.

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.

[Data]

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.

[Ganchrow]

Originally Posted by Data

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?

If I'm remembering correctly then log5 is either an approximation of this method or simply a restating of identical results.

I'll refresh my memory tomorrow.

[Ganchrow]

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.

QED

[Wheell]

data, you are in fine ballbusting form

[dcbt]

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.

[Data]

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.

[Ganchrow]

Originally Posted by dcbt

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.

The equation you're using above is just log5, which as I demonstrated is identical to the logit() formulation assuming H* = 50%. As I said, I just happen to find it easier to remember and more intuitive that way.

[Ganchrow]

Originally Posted by Data

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.

I think Wheell was just busting your balls.

[dcbt]

Originally Posted by Ganchrow

The equation you're using above is just log5, which as I demonstrated is identical to the logit() formulation assuming H* = 50%. As I said, I just happen to find it easier to remember and more intuitive that way.

Ah, gotcha, thanks!

[Justin7]

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.

[Justin7]

Hmm. I assumed your win percentages are for the same game.

[dwaechte]

Originally Posted by Justin7

Hmm. I assumed your win percentages are for the same game.

I did the same before seeing the responses and re-reading the post.

[dcbt]

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'...

[Ganchrow]

Originally Posted by dcbt

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!

Glad we could help.

Originally Posted by dcbt

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.

Well all of those building blocks remain very important. It's just an issue of honing your statistical/probabilistic intuition to the point where you can readily apply similar logic to advantage betting.

[dcbt]

Can the aforementioned formulas be applied to 3-way, soccer-style betting with the draw option? Or is that not feasible given the parameters?

[Ganchrow]

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.

[dcbt]

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.

[Ganchrow]

Let A_n = Team A neutral field win probability = 60%
Let B_n = Team B neutral field win probability = 45%
Let H* = expected home win probability between two equally matched teams = 58%

then:

lg(p) = lg(A_n) - lg(B_n) + lg(H*)

or equivalently:

lg(p) = lg(A_n) - lg(B_n) - lg(1-H*)

Solving for p:

p ≈ 71.685%

Multipurposing the alpha version home/away calculator:

Home Team Win Prob: 60%
Away Team Win Prob: 45%
League Avg Home Win Prob: 1-58% = 42%

Clicking calculate yields:

Home Win Over Away Prob: 71.685%

[dcbt]

Ahhh, yes, thanks!