[Ganchrow]

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%.

[jjgold]

98% of people have no clue what this stuff means including a player like me.

[Ganchrow]

Originally Posted by jjgold

98% of people have no clue what this stuff means including a player like me.

If so, then I guess my audience for this type of post is the remaining 2% of the population.

[Ganchrow]

Originally Posted by Ganchrow

If so, then I guess my audience for this type of post is the remaining 2% of the population.

But anyway, the executive summary is this:

If you place one 1-unit bet with an edge close to zero, at decimal odds d, the standard deviation of that bet would be

σ = sqrt(d-1)

If you place n 1-unit bets with edges close to zero, at decimal odds, d, the standard deviation of across those n bets would be

σ = sqrt( n * (d-1) )

So for example, if over the course of a season you were to place 200 $100 zero-edge bets at odds of , your standard deviation (in dollars) would be:

σ = $100 * sqrt(200 * 10/21) ≈ $975.90

This means that your 95% confidence interval would be approximately ±1.96 * $975.90 ≈ ±$1,912.73.

In plain English, this means that after all 200 $100 bets there's be a 95% probability that you would have won or lost less no more than $1,912.73.

[Scorpion]

i dont get it

[jjgold]

How does this help you win or increase your chances of winning?

[Santo]

It doesn't, it helps you manage your bankroll

[CrazyL]

Make some different spreadsheets Ganchrow, I'm playing with excel and it's pretty damn cool.

[Thremp]

Originally Posted by Santo

It doesn't, it helps you manage your bankroll

Which in the long term is what really ships the most $$$.

[Santo]

Agreed, depends how you define "chance of winning" I guess. I read it as "chance of picking winners", but could also be "chance of ending up in profit"

[Jamie_UK]

I got to "Technically speaking" and a headache came on.

[Data]

Originally Posted by Ganchrow

If so, then I guess my audience for this type of post is the remaining 2% of the population.

Being an admirer of your posts I tend to expect an excellent answer coming from you and an excellent answer must actually answer the question. This did NOT happen here even though you math is fine as always. The question did not make sense and as such could not have a good meaningful answer.

What you probably should have done is clarifying the question and asking to rephrase it without using the word "variance".

[mikevegas]

Ganchrow,

I am sorry but I phrased the question incorrectly.

If you wish to bet 3 team, 4 team or 5 team parlays or teasers, is there a formula or mathematical expression which measures your volatility or "variance?" Obviously, you win a 5 teamer less frequently than a 3 teamer but is it quantifiable?

For discussion purpose assume that a 3 team parlay pays +500 (6.0), a 4 team +1000 (11.0), & a 5 teamer +2100 (22.0).
These are typical payoffs on parlay cards around town.

Also assume that the parlay card is stale and that each leg has a 55% win rate verse the closing line.

[Ganchrow]

Originally Posted by mikevegas

Ganchrow,

I am sorry but I phrased the question incorrectly.

If you wish to bet 3 team, 4 team or 5 team parlays or teasers, is there a formula or mathematical expression which measures your volatility or "variance?" Obviously, you win a 5 teamer less frequently than a 3 teamer but is it quantifiable?

For discussion purpose assume that a 3 team parlay pays +500 (6.0), a 4 team +1000 (11.0), & a 5 teamer +2100 (22.0).
These are typical payoffs on parlay cards around town.

Also assume that the parlay card is stale and that each leg has a 55% win rate verse the closing line.

From the above:

variance = σ² = (p(1-p))*d²

where as above, p is win probability, and d is decimal odds. Extrapolating to an n-team parlay with each leg winning at 55% and paying out at decimal odds of d²[n] n-team parlay variance σ²[n] is given by:

σ²[n] = 55%ⁿ * (1 - 55%ⁿ) * d²[n]

So to summarize:

n	d	σ2[n]	σ[n]
3	6	4.99	223.5%
4	11	10.06	317.2%
5	21	23.13	481.0%

[mikevegas]

Ganchrow,

Thank you for the formula.
I am surprised that the variance is higher on a 3 team parlay
instead of a 5 team parlay.