Moneyline Bayes

A possibly-novel formulation of Bayes' Theorem based on moneyline odds , where the updating of priors is based on lines rather than explicit probabilities (difficult, pre-loaded with meanings that do not necessarily correspond to their mathematical value) or odds (less well-understood by American non-horse gamblers.)

As a note, I will refer to fractional probability (e.g., 25% = 1/4) using forward slashes, and fractional odds (2:1, 3:2) using colons for clarity.

To begin, we need three lines

"Your line", i.e. the line that you think best represents to odds of some event (call it "A") occurring. For example, we will use -250.

Your line for your observation (call it "B") occurring given A occurs. For example, we will use +300.

Your line for B occurring given A doesn't occur. For example, we will use -200.

We will be adapting the odds form of Bayes' Theorem, which can be written as follows:

O(A|B) = O(A) * (P(B|A) / P(B|~A))

Our first step will be to get our second two lines set up in a manner conductive to calculation. A nice property of moneyline odds is that, when converting to fractional probability form (as opposed to fractional odds), your denominator will always be 100 + the line without a + or -, so your transformed version of the lines (with I will refer to as L generally, and Lx and Ly specically, for convenience) will look like this:

If L is a + line, 100/(100 + L)

If L is a - line, L/(100 + L)

So,

Lx = 100/400

Ly = 200/300

We will then divide them, representing the intermediate value as Ll

Ll = ((100/400) / (200/300)) = ((100/400) * (300/200)) [= 0.375]

Now, convert "your line" just to odds form using this rule:

If L is a + line, 100:L

If L is a - line, L:100

O(A) = 250:100

Then multiply for the fractional form of O(A|B), which we will refer to as provisional O(A|B), or pO(A|B).

pO(A|B) = O(A) * Ll = 15:16 [= 0.9375]

Now, we transform this back to moneyline odds using the following rule where X is the larger number in the odds and Y is the smaller:

+-O(A|B) = 100(X/Y)

So the example would be

O(A|B) = -16(100/15) = ~106.7

Then assign the valence (+ or -) this rule: if the result was X/Y, -O(A|B); else, +O(A|B).

So our final answer would be +106.7 (rounded)

Voila, you now have an updated line on the event based on your observation.

Notes:

Fractional odds here are "actual odds" (1:9 implies 10%) rather than gambling fractional odds (1:9 implies 90%)

100(X/Y) converts to the highest confidence case (i.e. > 50%) and uses the symmetry inherent to the moneyline representation to get our final answer. Another example: for an output odds of 1:9 or 9:1, 100(9/1) = 900; the former is +900, the latter is -900.

Yes, this is just a special case of the odds form. Probably a bit less wieldy, but it allows you to stay in moneyline odds mentally.