Lets say the Warriors will be at home vs Atlanta.
The Warriors have a home win % of 75 for the season and Atlanta has an away win % of 20 also for the season.

Is there a way to estimate statiscally the likelyhood of each team winning when they play each other based on this info ?

The log5 method was developed/derived to answer questions like this.

Answer = [.75 - (.75*.20)] / [.75 + .20 - (2*.75*.20)]

= .60 / .65

= .923

Don't we have to assume that each team's winning % is against a random team for that to work?

To put it another way, do we really the Warriors have the same chance of winning against the Hawks as they do against the Celtics?

I googled Log5 and found this :
Is this the same formula just written in a different way ?
Seems different to me. Which one is correct ?

The “5″ in “log5″ and a More General Formulation

The “5″ part of “log5″ was in reference to the fact that teams were being compared to .500, the average winning percentage. But when we’re dealing with individual matchups, the league average isn’t always .500 (for a batter/pitcher matchup to estimate the probability of a hit, we would need to use the league-wide batting average). To deal with this we need to add another term to the formula representing the league average probability (or odds).

In the odds ratio formulation, this is easy. We just divide by the league average odds (Odds_{LG}). When the league average probability is .500, the odds are \{.500}{(1 - .500)} = 1, so the term can be omitted without consequence.

OddsRatio_{AvB} = {Odds_A}/{Odds_B}}/{Odds_{LG}}
Converting this to a probability, we have what I find to be the clearest formulation of the generalized log5 formula:

p_{AvB} = {Odds_A}/{Odds_A + (Odds_B * Odds_{LG})}

My above formula uses league odds, which could be the overall 50% or maybe the away teams league average of 42% ??

=0.75/(0.75+(0.2*0.5)) or
=0.75/(0.75+(0.2*0.42))

No, but its a generic starting point thats pretty important imo

Edit: Actually, Boston's win% will be much higher than 20%. So this will account for the difference in talent, no?

Is this the formula for a 75% against a 20% shot? If so it is wrong in this case because the home field advantage is not evaluated properly and is essentially "counted twice". (To show this assume that all teams were 75% at home. That would mean that a 75% shot vs a 25% shot was 75% which I assume is a lot smaller than the log formula gives. Or for a more real life example, most of you realize, I am sure, that if a college basketball team that wins 49% of its away games plays a team that wins 51% of its home games that first team will actually usually be the favorite when it plays on the road against the second team.)

Yes, the plain vanilla Log5 formula posted in post #2 above assumes all you know is the two team's respective win percentages (i.e., it ignores home/road factors, etc.).

I believe OP posted a more general Log5 formula in post #4 that incorporates an additional factor such as home/road splits.

So does his formula give 75% in my first example and, in my second example, make the 49% road team the favorite against the 51% home team when playing at the latter's home court, given the average team wins 55% at home?

I was surprised at how poor the info is online after spending more time researching Log 5. I'm thinking the following equation may be most accurate?? However, I'm confused as to which league average (LA%) to use...

{(Awin% * Bwin%) / LA%}

/

{(Awin% * Bwin%) / LA%} + [{(1-Awin%) * (1-Bwin%)} / (1-LAwin%)]


Using my original example....

ATL is away with win % of 20.
League away avg % is 42% so far this year.
GSW home win % is 75%
League home avg % is 59% so far this year.

The "59" and "42" need to add to 100.

Interestingly, even though there is a logic flaw in using the original formula, the answer appears to be almost perfect. Because if you win 75% at home you are almost four points better than average and if you win 20% on the road you are about four and a half points worse. This would make the better team at home almost 12 points better which is statistically between 93 and 95 %.

Suppose Team A wins 75% of its home games.

Suppose Team B wins 20% of its road games.

Suppose overall league-wide the home team wins 60% of all games.

Then I thought that the more "general" Log5 formula in this case predicts that Team A at Home vs Team B on the Road will win with probability:

= (.75*.80/.60)/[(.75*.80/.60) + (.25*.20/.40)]

= .888

Admittedly, I have not thought about this stuff in years, so maybe I don't know what I am talking about.

It's not everyday that David Sklansky replies to a post of mine, so first let me say ty and how cool it is that you take the time to be an active member of the forums. 2+2 has been, and will be, a huge wealth of knowledge because of the contributors.

Back to the issue at hand...
There are 3 "LA"'s % in the formula. Idk where to use the 52 and where to use the 49.

Point taken about the numbers adding up to 100 btw

I gave the correct formula (in my opinion) in the post right above yours.