Math Model: how do you compare teams who played different difficulty opponents??

tool21 · 04-17-08, 07:07 PM

I always like to include examples of what i'm talking about and this one is best described as one. I just want the BEST math behind this and the right way to do it. I have tried about 4 different methods and i have my favorites but i want to know how you guys would do it. This is just something small that i need to zero in on for accuracy.

Answer this question and how do you explain how to do it.

If Team A's offense averages 78.2 points against teams that give up on average 71.96667 and Team B's defense gives up 66.1 points against teams that put up on average 75.0875 then what is a good estimation about how many points Team A will put up based on the given information?

Team A
78.2
71.9667

Team B
66.1
75.0875

The problem i'm coming up with is with the denominator. If team A difference is +6.233 and Team B difference is -8.98 then i come up with -2.75 against both teams opponents but which denominator (opp) do i add this to? Do i take the average of the denominators and then add it to get 70.78 points for Team A?

Another method is this: Team A averages 78.2 against teams that give up on average 71.9667 but Team B doesn't give up 71.9667 they give up 66.1 so Team A off will have 71.825. On the other side, Team B gives up on average 66.1 against teams that average 75.0875 but Team A doesn't put up 75.0875 they put up 78.2 so Team B def. will give up 68.84. Then take the average of 68.84 and 71.825.

Obviously there IS a distinct difference in the difficulty of the opponents and there is a difference between 68.84 points and 71.825 points in the 2nd example. Should i just stick to games in which the opponents difficulty is fairly the same? Divisonal games in NCAAB, and NBA mid-season? NBA isn't that much of a problem because the teams play all of each other. The denominators are fairly close which leads to less error but IMO NBA lines are a lot sharper than college lines. College is where the problem lies. Take a team like Davidson playing Kansas. Slight variations in opponents difficulty.

Another problem between these 2 ways that i found is when two teams both like to run. Say you have Phoenix playing Denver. If you use the 2nd example method you are taking the averages which leads to a false total. Say team A is +6 and team B defense is +4 then you would have +10 to the teams they've played against. In example 2 the average would be taken between the two numbers and a false total would come out. So instead of having a 117-112 score (method 2) you should really have a 124-119 (method 1) type score. But like i asked before, which denominator do I add the numbers to? So I more or less just need help with the first example because the 2nd gives false totals....sorry for all the info just thinking on paper.

Just throwing out some ideas, i have various others that tweak this....what would you guys do?

BuddyBear · 04-17-08, 08:39 PM

I didn't read the entire message b/c it's too long, but based on the first paragraph and the title of this thread you are interested in controlling for the quality of competition.

I do not think the way you are considering this issue by looking at point differential is very valid but some might argue that point.

Well, basically, you would need to develop a variable that measures for opponent's strength of schedule and treat that as a stand alone primary variable in your model or you could simply treat it as a control variable...it would be up to you how you wanted to use it but I certainly agree that it is important.

IMO, this is an extremely difficult variable to construct because there are several validity issues that need to be considered before doing this. The biggest obstacle with this variable is that it is ALWAYS varying and the earlier in a season the more volatile it is going to be. Really, it won't become stagnant until you get deep into the season and everyone has played everyone.

There are a number of reputable websites that offer strength of schedule measures that you can consider. It will save you a ton of time.

Ganchrow · 04-18-08, 08:22 AM

BuddyBear brings up some very good points.

He's particularly correct that without prior knowledge of the sport in question (in other words, additional data and/or theory), point estimates of scoring expectations are by themselves generally insufficient to provide the answer for which you're looking.

Off-the-cuff I'd say that to really go about this theoretically correctly ("BEST math", as you say) you'd first want to model the pdf of each team's scoring (for or against as applicable) conditioned on some measure of its opponent's performance. You'd further want to model a marginal (i.e., a "prior") probability distribution for each teams' (offensive or defensive) performance (assuming a one-to-one relationship existed between a team's marginal performance and it's scoring). Finally you'd utilize Bayesian inference to combine all your estimates in order to determine a posterior pdf for scoring by team A conditioned on the joint realization of your predetermined performance measures. By integrating across all possible performance measure realizations you'd then be able to calculate a final expectation. This would be what I'd consider a serious project, with likely considerably overreaching complexity.

If, however, you'd settle for roughly estimating Team A's win probability vs. Team B, then there is a straightforward (if slightly involved) methodology that may be employed to this end. Still necessary, however, is the assumption that the opponents of Team A and B have themselves played against directly relatable opposition (consider that there'd be no way one would be able to draw meaningful conclusions about the Knicks' likely performance against a Division III team based solely on the two teams' results within their respective leagues).

I'll warn you, however, that the results at best represent a first-order approximation far better suited to Rotisserie League play than to professional advantage betting. Nevertheless, the general techniques involved may be instructive and could potentially prove useful within alternate contexts.

Given by OP:

Team A's points for = 78.2
Team A's opps' points against = 71.96667
Team B's points against = 66.1
Team B's opps' points for = 75.0875

Let's add a few more variables to the problem specification, namely:

Team A's points against = w
Team A's opps' points for = x
Team B's points for = y
Team B's opps' points against = z

Baseball's Bill James posited that the number of wins expected by a baseball team over some stretch of games could be reasonably estimated as a function solely of total runs scored and total runs allowed. This is known as the "Pythagorean expectation" and is given by the following formula:

Expected MLB Wins = Games Played * RunsFor^n/(RunsFor^n + RunsAgainst^n)

where n is a data-determined exponent equal to roughly 2 for Major League Baseball.

It's certainly possible (although not necessarily advisable) to apply this same methodology to other sports. I'm going to demonstrate an extremely simple exponent estimation technique for the NBA. Disinterested readers won't lose much by skipping ahead to the conclusion.

Using covers.com data one can derive the following data set through the 2006-07 season (also see Sheet1 of the attached spreadsheet):

The goal is to select an exponent, n, for the modified Pythagorean expectation equation

Expected NBA Wins ≈ Games Played * PointsFor^n/(PointsFor^n + PointsAgainst^n)

that best "fits" the actual data set. You should note that by dividing through by "Games Played" we then have an estimator of the probability of a team winning a particular game given the expected number of points scored for and against.

When attempting to match actual frequencies with predicted expectations, one frequently uses what's known as "logistic modeling". Realize that probabilities are necessarily bounded by 0 and 1 and are inherently nonlinear in nature. This can be intuitively understood by considering the difference between a 100% probability event overvalued at 99.01% (US: -10,000) versus a 50% probability event overvalued at 49.50% (US: +102). Both events are overvalued by 1%, but in the former case a growth-maximizing investor should be willing to bet and borrow every penny possibly available to him, while in the latter case such a bettor would invest less than 1% of his bankroll.

Logistic modeling defines the following function of a probability, p:

logit(p) = ln(p) - ln(1-p)

Another way of saying this is that the logit() of a probability is the logarithm of the "fair" fractional odds (i.e., decimal odds-1) associated with that probability. So in other words, the logit() of an event that occurs exactly half of the time (p=50%) would be 0, the logit() of an event that always occurs (p=100%) would be +∞, and the logit() of an event that never occurs (p=0%) would be -∞.

We then proceed by determining the exponent that minimizes the sum of the squared deviations between the logit of the actual win percentage and logit of the Pythagorean expectation. This is a straightforward problem of convex optimization well within the capabilities of Microsoft Excel Solver. If you look at Sheet1 of the attached spreadsheet you'll see that the exponent value that provides the best fit for the data set is roughly 14.02. Ths is frighteningly close to the value of 14 attributed to Dean Oliver by Wikipedia.

Determination of relative goodness-of-fit for this and other sports are left as an exercise for the interested reader (translation: don't expect me to chime in, but please see this paper if particularly motivated).

So we have:

NBA Win Probability ≈ PointsFor^14.02/(PointsFor^14.02 + PointsAgainst^14.02)

Because the OP wasn't so kind as to provide values for the variables w, x, y, and z I'll arbitrarily assign them as follows:

w = 74
x = 73
y = 68
z = 72

Using the Pythagorean expectation formula above we come up with predicted win probabilities as follows (see Sheet2 of the attached spreadsheet):

Team A: 68.44%
Team A opp: 54.98%
Team B: 59.81%
Team B opp: 64.31%

If we assume that Team A & B's opponents have themselves established their win records against "average" opposition we can then impute Team A and B's respective performances against average opposition. See the following post I had made at another site for a brief discussion of the general methodology:

Quote:

Originally Posted by Ganchrow

Quote:

Originally Posted by trytowin

I see on nba.com in the standings section that Boston has a home win percentage of 0.87879 (29-4), and that Utah has a road win percentage of 0.42857 (15-20). The question is, what is the expected probability based on these two numbers only that Boston will win tonight's game?

a) 0.5 - 0.42857 = 0.07143 + 0.87879 ==> 0.95022
b) 0.42857 + 0.87879 / 2 ==> 0.65368
c) (0.87879 - 0.42857) * 2 ==> 0.90044
d) none of the above

I not sure of the correct answer, so please explain. Thx, TTW

The Sabermetrics guys have a methodology for handling this called "log5", but it assumes a league average home team win probability of 50%.

Just thinking about it logically, here's how I'd do it:

Let A = Team A home win % (taken as an unbiased estimator of Boston's "true" home win probability versus an "average" opponent) = 29/33
Let B = Team B road win% (taken as an unbiased estimator of Utah's "true" road win probability versus an "average" opponent) = 15/35
Let H* = expected home win probability between two equally matched teams = 61% (which is roughly the total NBA home win win frequency over the past 25 years)
Let 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:

logit(p) = ln(p) - ln(1-p)

inverted:

p = exp(logit(p))/(1+exp(logit(p)))

This gives us:

logit(A) = ln(29) - ln(4) ≈ 1.981001469
logit(B) = ln(15) - ln(20) ≈ -0.287682072
logit(H*) = ln(61%) - ln(39%) ≈ 0.447312218

So from Bayes' Theorem:

logit(P) = logit(A) - logit(B) - logit(H*)
logit(P) ≈ 1.981001469 - -0.287682072 - 0.447312218 ≈ 1.821371323

Solving for P we have:

exp(logit(P)) ≈ exp(1.246007179) ≈ 6.180327869
P = 6.180327869 / (1+6.180327869) ≈ 86.073%

So based on the above data, Boston's win probability over Utah was about 86.073%, implying fair US payout odds of about -618.

(You can also see a preview of the following not-yet-ready-for-prime-time version of a calculator that performs these calculations here. Final version to be released soon.)

So we have:

logit(A win prob vs. Avg. Opp) = logit(68.44%) - logit(54.98%) = 0.974158898
logit(B win prob vs. Avg. Opp) = logit(59.81%) - logit(64.31%) = 0.98630472

This then yields:

logit(A win prob vs. B) = logit(A win prob vs. Avg. Opp) - logit(B win prob vs. Avg. Opp) = 0.974158898 - 0.98630472 = -0.012145822

Inverting the logit function yields what's known as the logistic function:

p = 1 / (1+exp(-logit(p))

giving us a final value for the probability of A defeating B of:

p = 1 / (1+exp(0.012145822)) ≈ 49.70%

(If using the alpha-version calculator linked above, you'd enter the team in question's win probability into the "Home Team" box, 1-opponent's win probability into the "Away Team" box, and 50% into the "League Average" box. This will automatically perform the logit calculations. Comments are welcome.)

Data · 04-21-08, 10:24 AM

Quote:

Originally Posted by Ganchrow

If, however, you'd settle for roughly estimating Team A's win probability vs. Team B, then there is a straightforward (if slightly involved) methodology that may be employed to this end. Still necessary, however, is the assumption that the opponents of Team A and B have themselves played against directly relatable opposition (consider that there'd be no way one would be able to draw meaningful conclusions about the Knicks' likely performance against a Division III team based solely on the two teams' results within their respective leagues).

I think that in basketball one can get at least better than meaningless assumptions based on league averages. I would be also interested to hear from european players chiming in on eurocups linemaking.

Ganchrow · 04-21-08, 01:01 PM

Quote:

Originally Posted by Data

I think that in basketball one can get at least better than meaningless assumptions based on league averages.

Please explain how this is possible within the context of the provided example (determining fair market line between the Knicks' and a Division III team based solely on the two teams' results within their respective leagues).

Clearly there would first need to be some predefined means of making inter league comparisons. League averages would by themselves be wholly insufficient.

Data · 04-21-08, 02:10 PM

Quote:

Clearly there would first need to be some predefined means of making inter league comparisons.

My first thought is that in sports like basketball or baseball the game outcome, to the great extent, depends on the sum of some "units" that reflect individual players' skills, while in sports like football or soccer a team-play has much larger effect. Thus, the former sports provide the means for inter league comparisons just by adding those "units".

Quote:

League averages would by themselves be wholly insufficient.

True. Using league averages, which were not mentioned in this thread, while insufficient yet necesary for making this kind of predictions, no matter intra- or inter-league.

Ganchrow · 04-21-08, 02:45 PM

Quote:

Originally Posted by Data

My first thought is that in sports like basketball or baseball the game outcome, to the great extent, depends on the sum of some "units" that reflect individual players' skills, while in sports like football or soccer a team-play has much larger effect. Thus, the former sports provide the means for inter league comparisons just by adding those "units".

True. Using league averages, which were not mentioned in this thread, while insufficient yet necesary for making this kind of predictions, no matter intra- or inter-league.

The whole point of the exercise outlined in this post is to use both Team A and B's averaging scoring (for and against), as well as that of their opponents (established versus league averages), to determine a fair money line.

Data · 04-21-08, 03:21 PM

It seems that you are talking about estimating win% that is essentially can be done using variety of methods, including Pythagorean, that use offensive/defensive efficiencies (ratings, scores etc). The OP and I are talking about estimating those efficiencies/ratings/whatever.

butters · 04-21-08, 05:17 PM

Perhaps this is overly simplistic, but would it be possible to estimate these values using a simple linear regression? Each previous game could be represented as:

Team_A_Off_Rating - Team_B_Def_Rating = Team A's Points For that Game (or points per 100 possessions for that game or something similar)

You could then run a regression to determine the values of each team's offensive and defensive ratings, adjusted for opponent quality. (Note that this would not produce actual points/game or efficiency values, but I think it would produce relative values that could be used to predict offensive efficiency for future matchups.) It seems to me that this would automatically account for points scored/allowed in previous games by both teams, as well as the points scored/allowed by their opponents, and their opponents' opponents, and so on...

Obviously, this wouldn't work for Ganchrow's example of the Knicks playing a DIII team, but I think that it could work for two teams that could be 'indirectly' linked through previous games.

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Math Model: how do you compare teams who played different difficulty opponents??

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