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.