Calculating Wager Variance (Received via PM)
Question received via PM:
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
The variance is just the expected sum of the squared deviations from the mean above, which is:
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:
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:
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:
If we assume the bet is being offered at "fair" odds, then d = 1/p and the standard deviation is:
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%.
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.
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
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
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)
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%.