Converting from a money-line to a point-spread

I've seen many people attempt to answer the question, "How do I convert a point-spread to a money-line?" (or conversely, "How do I convert a money-line to a point-spread?" The answers I've seen given have been unequivocally bad; but that’s because the correct answer is not simple, and people like simple answers.

To understand that a simple plug-and-play formula won’t do the trick, just check out the lines at your favorite sportsbook. I checked betus.com at 8:00 PM on the 16th of October and found the following lines:

St. Louis +9.5 -110 (points) or +400 (moneyline)
Jacksonville -9.5 -110 (points) or -500 (moneyline)

Buffalo +9.5 -110 (points) or +350 (moneyline)
NY Jets -9.5 -110 (points) or -450 (moneyline)

From this it is clear that there is more involved than simply knowing the point-spread to determine the moneyline (or more than just knowing the odds to determine the pointspread).

The function that is used to convert from a pointspread to a moneyline is the normal distribution function. This function must be integrated from negative infinity to the point-spread to determine the probability of an outcome. In all cases the mean is zero. If the pointspread is -9.5 that is the value used for x. The value of the integral from –∞ to -9.5 is also a function of the standard deviation; and this is what varies from game to game.

We can back-calculate the standard deviation the casino used by iterating (such as tools  goal-seek from MS Excel). In the game where Jacksonville is -500 you have to lay $500 to win $100. But remember the 10% juice. Without it you’d only have to lay $450 to win $100. This means that Jacksonville should win 4.5 times more often than it loses. Neglecting ties this means that Jacksonville should win 4.5 / 5.5 of the contests or 81.8%. Taking the inverse of the integrated normal distribution function (=norminv(0.818, 0, guess) in MS Excel) you can change your guess (using tools  goal-seek) to get a point-spread of 9.5. In the case of the Jacksonville vs St. Louis game you’ll find that the sportsbook used a standard deviation of 10.46. You can now verify this by using the same standard deviation but a negative spread; you’ll find St. Louis has a 18.2% chance of winning. One divided by 18.2% = 5.5. This means that the casino without juice would give you $450 winnings for every $100 bet (you get $550 off $100 18.2% of the time). Since they take their juice off your winnings, the line should be 90% of 450 or 405. Note that they’re actually offering +400. I have a feeling rounding errors tend to favor the house!

A similar exercise will reveal that the standard deviation used in the Bills vs Jets game is 11.2. I would have simplified to assume that the standard deviation tracks the over/under, but that is clearly not the case here. The casino most likely has team specific standard deviations (due to a more complex model than mine).

This brings up a separate, but related issue. If the casino’s model is more complex than mine, how can I expect to win in the long run? The key is that I don’t have to be smarter than the sportsbook. The book is balancing their desire for profit (holding their line) with risk mitigation (splitting the betting public evenly). If I have a decent model (I’m right greater than 52.4% of the time) I can win in the long run. Currently we’re running 58% season to date.