Question received via PM:
Is it possible to calculate the variance on teasers and parlays?
e.g., Your standard 3 team +500 parlay card requires 55% legs to breakeven.
You must win 1/6th or 16.7% to breakeven.
How do you calculate the variance of that parlay?
I am not even sure if what I am asking is logical.
I am not sure that standard deviation and variance applies to a single parlay or teaser wager.
Technically speaking, it doesn't make sense to calculate the variance on a binary bet (be it a parlay, teaser, or straight bet) without first specifying:
1. the win probability or expectation and
2. the amount wagered
So let's assume that on a bet paying out at decimal odds of d with win probability p, we were to wager X (and X can be in whatever units we like: dollars, percent of bankroll, drachmas, hamburgers, or just generic "units").
This implies an expectation on that parlay of
E(X|p,d) = p * (d-1) * X - (1-p) * X
E(X|p,d) = (p*d - 1) * X
The variance is just the expected sum of the squared deviations from the mean above, which is:
V = p*((d-1)*X - (p*d -1)*X )2 + (1-p) * (-1*X - (p*d -1)*X )2
V = p(1-p)*(Xd)2
The variance is in the same units of measurement as X2. We can then normalize V, by looking at V/X2 (henceforth, we'll refer to the dimensionless quantity V/X2 simply as σ2) so:
V/X2 = σ2 = p(1-p)*d2
The standard deviation, σ, is just the square root of the variance so for a 1-unit bet we have the following formula for standard deviation:
σ = sqrt(p(1-p))*d
From above, expectation E(X|p,d) = (p*d - 1) * X, and so normalized expectation E(X|p,d) / X = (p*d - 1), which we'll call E. So standard deviation given purely in terms of normalized expectation E and decimal odds d would be:
σ = sqrt((d - E - 1) * (E + 1))
If we assume the bet is being offered at "fair" odds, then d = 1/p and the standard deviation is:
σ = sqrt(1/d * (1-1/d))*d
σ = sqrt(d-1)
This last equation serves as a convenient approximation for standard deviation provided that edge is sufficiently close zero. The accuracy of the approximation increases the closer payout odds get to even.
So given your example of a breakeven bet, σ2 = d - 1 = 5, and σ ≈ 224%. If we assume that a player has identified a 5% edge on that bet, standard deviation would be σ = sqrt((d - E - 1) * (E + 1)) = sqrt((6-.05-1) (1.05) ≈ 228%, which depending upon your purposes is fairly close to the estimated zero-edge standard dev. of 224%.