I have come up with two methods for answering the question:

If Team 1 scores A ppg and Team 2 allows B ppg in a league where C ppg is average, how many points is Team 1 expected to score against Team 2?

A*B Team 1 Off. vs. Tm 2 Def. = --- (1) C

A*B*(200-C) Team 1 Off. vs. Tm 2 Def. = 200*----------------------------- (2) A*B*(200-C)+(200-A)*(200-B)*C

In both of these equations, the values of points per game can be replaced by points per possession. In the Kalman Filter, Equation 1 is used. Both methods are derived from , equation 1 also using the . Full explanations of the methods are given below.

Let us assume that Team 1 not only scores A ppg, but allows C ppg defensively (in other words, it has an average defense). Let us also assume that Team 2 scores C ppg offensively (it has an average offense to go along with its defense, which allows B ppg). Using the , the projected winning percentages of the two teams are

A^{16.5} Win_{1} = Winning Percentage Team 1 = -------- (3) A^{16.5}+C^{16.5} C^{16.5} Win_{2} = Winning Percentage Team 2 = -------- (4) C^{16.5}+B^{16.5}

Using , we know that team 1 will beat team 2 the following percentage of the time:

Win_{1}(1-Win_{2}) Win_{1v2} = Winning % of Team 1 vs Team 2 = ----------------------- (5) Win_{1}(1-Win_{2})+(1-Win_{1})Win_{2}

We can insert equation 3 and equation 4 into equation 5 to get

A^{16.5} B^{16.5} --------- --------- A^{16.5}+C^{16.5} B^{16.5}+C^{16.5} Win_{1v2} = --------------------------------------------- (6) A^{16.5} B^{16.5} C^{16.5} C^{16.5} --------- --------- + --------- --------- A^{16.5}+C^{16.5} B^{16.5}+C^{16.5} A^{16.5}+C^{16.5} B^{16.5}+C^{16.5}

This can be simplfied (yes, using that 6th grade cancelling procedure for dreaded fractions) to get:

A^{16.5} B^{16.5} Win_{1v2} = --------------------- (7) A^{16.5} B^{16.5} + C^{16.5} C^{16.5}

Now, remember where we are going: we want to find out what Team 1 will score against Team 2. Let's call this number "D"; Team 1 will score D points vs Team 2. Because Team 1 has an average defense and Team 2 has an average offense, we know that Team 2 will score an average number of points against Team 1; in other words, Team 2 will score C points.

This information can then be put into the to get

D^{16.5} Win_{1v2} = --------- (8) D^{16.5}+C^{16.5}

Now, we can set equations 7 and 8 equal to get:

D^{16.5} A^{16.5} B^{16.5} --------- = --------------------- (9) D^{16.5}+C^{16.5} A^{16.5} B^{16.5} + C^{16.5} C^{16.5}

Everything in equation 9 is known except for the value of D. You can apply your powers of algebra to arrive at the result:

A*B Team 1 Off. vs. Tm 2 Def. = --- (10) C

There are two important notes in this derivation. First, the result does not depend upon the exponent of the Pythagorean Method. I use 16.5, others use anything from 12-16, but whatever it is, you end up with the same final result. Second, this derivation relied upon the assumptions that Team 1 had an average defense and Team 2 had an average offense. The result is the same without the assumption, but it is a little more difficult to understand.

Dallas Adams and Bill James came up with a method for answering this type of question, but their solution was built upon the ability to put things in terms of the probability of doing something. That is vague, so let me make it specific. Let's say that Team 1 scores a points per 100 possessions and Team 2 allows b points per 100 possessions in a league where c points per 100 possessions is average (notice that I am switching away from points per game here), but the Adams-James solution does not know how to take this information as it is to determine how Team 1 will do vs. Team 2. In order to do this, we have to convert a, b, and c into probabilities. Fortunately, I already have the concept of , which is the probability that a team will score on a possession. We can convert the values of a, b, and c approximately to floor percentages just by dividing by 200. Why 200? Because I can assume that about 2 points are scored on every scoring possession and because a, b, and c represent the number of points scored per 100 possessions. Then we can just use to get the following equation:

Probability of score for a/200*b/200*(1-c/200) Team 1 Off. vs. Tm 2 Def. = ----------------------------------------------- (11) a/200*b/200*(1-c/200)+(1-a/200)*(1-b/200)*c/200

The expected number of points per 100 possessions is found then by multiplying the above result by 200. Simplifying everything (your teachers told you algebra was useful):

a*b*(200-c) Team 1 Off. vs. Tm 2 Def. = 200*----------------------------- (12) a*b*(200-c)+(200-a)*(200-b)*c

Keep in mind that a, b, and c are , or points per 100 possessions. Replacing these with Points Per Game values A, B, and C is not strictly true, but is is a good approximation. That is what makes equation 2 OK.

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