Future Win Totals Thinking

Earlier today, Jonny approached me with a question regarding win totals for the upcoming college football season. BetUS has regular season win totals up for most of the major conference teams for 2009. I was asked to compute win expectancies based off the juice. It seems straightforward at first, but I made some assumptions along the way and I want to make sure they are right.

Here is the methodology: first, find the actual juice adjusted probabilties from each line by adjusting each outcome's odds to probabilties then dividing through each outcome's probability by the sum of the probabilities of the two outcomes.

Then, if the total is a whole number, multiply the over juice adjusted probability by the total plus one and the under juice adjusted probability by the total minus one. If the total is a half number, the procedure is the same, except instead of adding/subtracting by one, I added or subtracted by one-half. Two examples follow.

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Example 1.
Team A o10.5 -130/u10.5 +110

p(-130) = 1.3/2.3 = 56.52%
p(+110) = 1 - (1.1/2.1) = 45.73%

.5652 + .4573 = 1.0414

56.52%/1.0414 = 54.27%
45.73%/1.0414 = 45.73%

E(A) = 11*.5427 + 10*.4573 = 10.54 wins



Example 2.
Team B o10 -130/ u10 +110

E(B) 11*.5427 + 9*.4573 = 10.08 wins

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So, the assumptions I want to check are: the juice adjustment to the probabilities and the calculation of the win expectancies by adding one or one-half.

Obviously, one assumption is that the probability distribution is symmetric, which may not be valid for Florida (o/u 11), for example.

I decided on adding one or one-half because of my SftC experience with soccer odds. When calculating the probabilities for three outcomes (Team A, Team B, or Draw), they are proportional to the Team A pk/Team B pk odds. In other words, If Team A is 35% to win, Team B is 35% to win, and there is a 30% chance of a draw, the odds end up at Team A pk 50% and Team B pk 50%. So, by adding one or one-half, I am calculating the odds of the outcomes that can actually happen, since there is no push option.

The reason why I am questioning myself, beyond never really thinking about it before today, is because the cutoff point for moving a line is different if the total is an integer or a half-number. For a whole number, over -200/under +150 equals a quarter of an expected win. Any higher, and you would expect the book to move it.

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Example 3.
Team C o9 -200/u9 +150

p(-200) = 2/3 = 66.67%
p(+150) = 1- 1.5/2.5 = 40%

66.67%/1.0667 = 62.5%
40%/1.0667 = 37.5%

E(C) = 10*.625 + 8*.375 = 9.25

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If you do the same calculation for a half number line, you don't get to the value to move the line.

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Example 4
Team D o9.5 -200/u9.5 +150

E(D) = 10*.625 + 9*.375 = 9.625

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Obviously, the mathematical reason for this is the larger spread on the endpoints (8 and 10 vs. 9 and 10). But is this right? Should the books be more willing to move the total off an integer than a half-number?

I'm finding myself questioning this outcome, because I think win expectancy should be normally distributed, and therefore the same amount odds should persuade the book to move a number, whether the total is a whole number or a half number.

On the other hand, there is also the push factor that has to be thought about. By moving the line off a half-number, the book creates an opportunity for a push, which I assume would be an undesireable outcome. Does the math prove that or did I make a mistake somewhere? I assume the latter.

In any case, assuming the math is right, Jonny should have win expectancies posted on his blog sometime soon.