edit: never mind. misread something. disregard.
Support MGoBlog: buy stuff at Amazon
Correcting O/U win totals for moneylines
The fact that over/under win totals come with moneylines provide no real complication for the betting public. Convert the moneylines to a return, ask whether the probability is better than the return and bet accordingly. But, what of rest of the public who want to read the over/under’s as Vegas’s expected wins? How can you convert the combination of the moneylines and O/U’s to an expected win value? To be concrete Michigan is at 7.5 (140 over/+100 under) according to the odds just released. Penn State is at 8.5 but (+100/140). How far apart are they? My results suggest Michigan’s expected number of wins is just a little higher at 7.6, while Penn State’s is a ½ win lower than its O/U at 8.0. In 2013, there were a few teams whose moneylines were so extreme that they implied expected wins a full win less than the O/U.
Let p denote the probability of winning more games than the O/U. Then the implied expected number of wins is p E[winswins>O/U] + (1p) E[winswins<O/U] where the expected values can be read as the average number of wins when a team comes above or below the over/under. The probability p is calculated from the moneylines and Table 1 contains estimates of these averages, which depend on the level of the over/under.
The way to interpret the moneylines depends on whether they’re positive or negative. If positive, the line gives you the extra amount you win if you bet $100. If negative, it gives you the amount you have to bet to win an extra $100. Let p(o) and p(u) denote the probabilities read directly from the moneylines, and denote by m as the absolute value of the number in the moneyline. If the moneyline is positive, then the probability is 100/(m+100); if negative it’s m/(100+m). In Michigan’s case, p(o) = 140/(100+140) = 0.583 and p(u) = 100/(100+100) = 0.500. If you think the probability that Michigan wins more than 71/2 games is greater than 58.3%, bet the over; if you think the probability is less than 50%, bet the under.
Note, however that p(o)+p(u) is greater than one. That’s Vegas’s juice to make money. We want a fair probability for our purposes, so we convert them by setting p = p(o)/(p(o)+p(u)). In Michigan’s case, p = 0.583/(0.583+0.500) = 0.538.
Table 1 contains the applicable expected values for various over/under's calculated under two different methodologies. The first method takes actual over/under’s from 20102013. The second method takes Chris Stassen’s data on preseason magazine ratings and converts them as a decent exante estimate of an O/U. Then for each I calculate the average wins when the teams come in above and when they come in below the totals. For my calculations below I then take an average of the two estimates. In Michigan’s case, the applicable expected values are E[winswins>O/U] = 9.20, which is the midpoint of the range and E[winswins<O/U] = 5.75, likewise. Weighting by 0.538 gives 7.6 wins.
Table 1 Conditional Win Averages 

O/U 
E[winswins>O/U] 
E[winswins<O/U] 
3.5 
5.20  4.80 
2.00  2.17 
4.5 
6.69  7.11 
2.86  2.67 
5.5 
7.18  8.00 
3.26  3.52 
6.5 
8.16  8.00 
4.87  4.40 
7.5 
9.33  9.06 
5.86  5.64 
8.5 
9.86 – 10.24 
6.41  6.05 
9.5 
10.73  10.45 
7.53  7.11 
10.5 
11.00  11.33 
9.50  9.50 
11.5 
12.00  12.00 
n.a.  11.00 
To be sure my estimated expected wins in table 1 could be improved upon, and the method as a whole could be augmented by arguing that the expected values should vary with the moneyline.
Table 2 gives expected wins for all Big Ten teams and Notre Dame for those over/under’s provided by Kegs ‘N Eggs. Note that Notre Dame’s expected wins is 0.7 below the O/U.
Table 2 Big Ten Expected Wins 


O/U 
Moneyline 
E[wins] 
Indiana 
5.5 
+145 over / 185 under 
5.0 
Maryland 
7.5 
+100 over / 140 under 
7.3 
Michigan 
7.5 
140 over / +100 under 
7.6 
MichiganState 
9.5 
155 over / +115 under 
9.2 
Nebraska 
7.5 
130 over / 110 under 
7.5 
Notre Dame 
9.5 
+110 over / 150 under 
8.8 
OhioState 
10.5 
140 over / +100 under 
10.4 
PennState 
8.5 
+100 over / 140 under 
8.0 
Rutgers 
4.5 
110 over / 130 under 
4.8 
Wisconsin 
9.5 
135 over / 105 under 
9.1 
.
is in a downward spiral.
Yeah..I think your expected wins in Table 1 look very suspect as there are large inconsistent jumps as you move from number to number. In particular, the expected wins in the under scenario for 9.5 and 10.5 look strange. You have 7.32 as expected wins if under 9.5, while you have 9.5 wins as expected wins if under 10.5. A jump of over 2 wins expected in the under scenario when you move only point does not make sense.
The result of this is you overestimate the wins when using the 10.5 line relative to a team using the 9.5 line. The following makes it obvious:
MSU O/U 9.5 (155/+115): you have expected wins 9.2
OSU O/U 10.5 (140/+100): you have expected wins 10.4
So if we look at those two, we see that 1) MSU is more likely to hit the over than OSU is and 2) OSU O/U number is higher number so cant beat that number by as much since can only win 11 or 12 games. Despite these two points, the MSU expected wins is 0.3 less than its O/U line (so 9.2 vs 9.5) while the OSU expected wins is 0.1 less (10.4 vs 10.5). It is not logical for MSU expected wins to be at bigger discount to its O/U number than OSU's discount.
My comments only reflect on the relative discount of expected wins to O/U line between the two as it shows the inconsistencies in your expected wins. In my next post, I will give a simple model of what the expected wins could be.
Given the post above, I stated that your expected wins in various scenarios seemed very inconsistent. In this one, I create a simple model (with many flaws itself but at least it is consistent) to do the same exercise you did.
Let's assume that for a single team (say Michigan), each game can be modeled as a binomial distribution using the same win probability for each game. Clearly this is not true as there is a lower prob of beating OSU than beating Miami (OH) but those probabilities are not observed so this method will be used. Let's solve for Michigan's win probability in each game so that it matches exactly the probabilities implied in the O/U lines.
As an example for Michigan, the example says that if Michigan has a 63.46% probability of winning each game, then you can verify that the probability of Michigan winning more than 7.5 games is exactly 53.8% (which is the number you extracted from the O/U moneylines. We can do the same for each team where we follow same procedure: 1) Find individual game probability that exactly fits the O/U moneyline probability and 2) use those individual game probabilities to solve for the expected wins. Doing so yields the following table:
Single  Expected  
Game Prob  Wins  
Ind  0.419  5.03 
Mary  0.608  7.30 
Mich  0.635  7.62 
MSU  0.802  9.62 
Nebraska  0.628  7.54 
ND  0.766  9.19 
OSU  0.873  10.48 
PSU  0.690  8.28 
Rutgers  0.372  4.46 
Wisconsin  0.792  9.50 
If you notice this fixes the weird OSU/MSU discount issue as 1) MSU now at +0.12 premium to O/U line while 2) OSU is at tiny discount of 0.02.
Again, this isnt by any means the "right" answer but this model is at least consistent across all win totals where the original post had numbers that were inconsistent and "biased" towards certain O/U lines.
I like the proposed solution. For those playing at home with an excel sheet, pealw is proposing that you find q such that (1p) = BINOMDIST(O/U0.5,12,q,TRUE). (O/U is assumed to end with 1/2 win.) The expected number of wins is then 12*q. It's not that hard to calculate, and it has the benefit of implicitly incorporating the moneyline in the calculation of the conditional expected wins.
The downside of the method is that it implies a strict distribution of wins, which doesn't have to be the case even if Vegas in the long run makes money. For example, suppose teams with an over/under of 9.5, win 10 games half the time and win none the other half of the time. The appropriate moneyline is even, say 110/110 and 9.5 is a good O/U. Pearlw would calculate the expected number of wins as 9.5, but we know that the average number of wins is actually 5.0.
By empirically estimating the conditional expected wins, I was hoping to account for the possibility of a skewed distribution.
The problem of course is that with a small number of observations, you might just get a bad draw, which may be the case of the data I'm looking at. Of the 29 teams between 2010 and 2013 with O/U's that I have equal to 9.5, only 11 came over. Even accounting for the moneylines, if you bet $100 everytime, you would return an extra $510 for the $2900 bet.
Yeah..agree with your main points. Empirical data as you use is theoretically good but problems arise when the amount of data you have is limited. The issue becomes if in your limited amount of data that "9.5 O/U" teams have done bad. You would then assume 9.5 teams would always do bad even though the cause is limited data and might result in inconstencies when compared to "8.5 o/u" and "10.5 o/u" teams.
Also, in youre example about the 9.5 @ 110/110, I would actually calculate an expected wins of 9.40 not 9.5 as you state. In this case, a q of 0.7833 in order for the over on 9.5 to occur 50% of the time. This would result in 9.4 wins then (12 * 0.7833).
One problem is Im assuming that each of the 12 games are identically distributed variables. As I mention before, the fact that one of those games is Miami (OH) while another is OSU shows these arent identically distributed. This does have an effect on the distribution in total but no way to know what those probabilities for each game unless wer had moneylines on each of the individual games in addition to the O/U wins moneyline.
Thanks for posting this diary topic as it was very relevant topic given the coverage this past week. It did seem most were focusing only on the O/U number without even considering whether the moneylines were telling something different than that "somewhat arbitrary" initial O/U number.
I like the proposed solution. For those playing at home with an excel sheet, pealw is proposing that you find q such that (1p) = BINOMDIST(O/U0.5,12,q,TRUE). (O/U is assumed to end with 1/2 win.) The expected number of wins is then 12*q. It's not that hard to calculate, and it has the benefit of implicitly incorporating the moneyline in the calculation of the conditional expected wins.
The downside of the method is that it implies a strict distribution of wins, which doesn't have to be the case even if Vegas in the long run makes money. For example, suppose teams with an over/under of 9.5, win 10 games half the time and win none the other half of the time. The appropriate moneyline is even, say 110/110 and 9.5 is a good O/U. Pearlw would calculate the expected number of wins as 9.5, but we know that the average number of wins is actually 5.0.
By empirically estimating the conditional expected wins, I was hoping to account for the possibility of a skewed distribution.
The problem of course is that with a small number of observations, you might just get a bad draw, which may be the case of the data I'm looking at. Of the 29 teams between 2010 and 2013 with O/U's that I have equal to 9.5, only 11 came over. Even accounting for the moneylines, if you bet $100 everytime, you would return an extra $510 for the $2900 bet.
what the hell you are talking about, but kudos to the OP for acknowledging that a fellow poster could improve his method. OP, let me know if you ever run for office  you have my vote.