Comparing Season Win Totals

[Ganchrow]

The attached spreadsheet may be used to compare different season totals between books.

For example:
Baltimore
Let's say that your handicapping tells you that you expect Baltimore on average to win 73 games next season. Pinnacle and Greek are offering the following:

WSEX
o73½ -125
u73½ -105

Greek
o70½ -240
u70½ +200

Which is better?

Enter Expected Season Wins (73), First Total (73.5), First Over Line (-125), First Under Line (-105), Second Total (70.5), Second Over Line (-240), and Second Under Line (+200) into the spreadsheet.

You see that the WSEX line implies a 15.8% expected loss on the over and a 4.0% expected win on the under.

You see that the Greek line implies a 7.5% expected loss on the over and a 4.2% expected win on the under.

Therefore, the Greek under would yield slightly greater expected profit than the WSEX under, and the Greek over would yield considerably lower expected loss than the WSEX over. This makes sense when we consider that in this particular market WSEX is charging 6.35% vig, market while The Greek is only charging 3.77%.^*

Lastly the equivalent over and under lines at the bottom of the spreadsheet show what the lines would need to be at the second book to match the expected profit or loss at the first book. In this case we see that in terms of expected profit, o70½ -345.4 is equivalent to o73½ -125 and u70½ +199.3 is equivalent to u73½ -105.
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^* -105/-125 => overround of 105/(100+105) + 125/(100+125) ≈ 1.067751; so vig = 1 - 1/1.067751 ≈ 6.35%.
^* +200/-240 => overround of 100/(100+200) + 240/(100+240) ≈ 1.039216; so vig = 1 - 1/1.039216 ≈ 3.77%.

[JoshW]

Pretty cool Ganchrow, thanks for the spreadsheet.

[Ganchrow]

Pretty cool Ganchrow, thanks for the spreadsheet.

No problem.

Someone asked me what distribution this is using and whether or not I believe it 100% appropriate.

The answers to teach of these question respectively are:

1. Binomial
2. No

The major problems with using the binomial in this regard are:

1. There are almost certainly different probabilities associated with winning home and away games. It's not the case that in general that the sum of two independent binomial random variables is distributed binomially, so as such it isn't appropriate to just look at the average win rate across home and road designations.
   
   Nevertheless, for home home/away win rates that are fairly close, results should be rather similar. For example, for a home win rate of 60% and an away win rate of 80%, the true probability of winning 121 or fewer game would be 49.407%; while using the binomial and the average win rate of 75%, results are 49.395%, a difference of 0.012%.
   
   This objection could further be extended with the realization that win rates aren't constant between opponents either. However, because each team plays a fairly large number of opponents, this issue will in most cases carries even less weight than the home/away issue.
   
   
2. Results between games are probably not independent. I'm not talking in reference to unexplained, generic streaks, which most statistical literature tends to discount, but in reference to specific events such as injuries, trades, and play-off contention. In other words, mathematically speaking, the data probably possesses some auto-correlation, meaning that the deviation from expected results of a given game is correlated with the deviation from expected results results of other games.
   
   In this manner autocorrelation will tend to widen the tails of a distribution, increasing the likelihood of extreme results. For example, given a team with a roughly 62% win prob (the probability associated with a team that you expect to win 100 games in the regular season), the probability of that team winning 75 games or fewer, is according to the binomial only about 0.00482%. In reality the true probability is likely several orders of magnitude higher.