Calculating Wager Variance (Received via PM)

[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]

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]

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

[SBR Lou]

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

[Thremp]

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]

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]

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:

<style> .GANCH { font-size: 10px; font-family: Verdana, Arial, Helvetica, sans-serif; }</style> <table border=0 cellpadding=2 cellspacing=2> <tr> <td align=left class=GANCH> n ^&nbsp; </td> <td align=left class=GANCH> d^&nbsp; </td> <td align=left class=GANCH> σ²[n] </td> <td align=left class=GANCH> σ[n]^&nbsp; </td> </tr> <tr> <td align=left class=GANCH> 3 </td> <td align=right class=GANCH> 6 </td> <td align=right class=GANCH> 4.99 </td> <td align=right class=GANCH> 223.5% </td> </tr> <tr> <td align=left class=GANCH> 4 </td> <td align=right class=GANCH> 11 </td> <td align=right class=GANCH> 10.06 </td> <td align=right class=GANCH> 317.2% </td> </tr> <tr> <td align=left class=GANCH> 5 </td> <td align=right class=GANCH> 21 </td> <td align=right class=GANCH> 23.13 </td> <td align=right class=GANCH> 481.0% </td> </tr> </table>

[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.

[Ganchrow]

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

Actually as pointed out by Data those figures were off.

Nevertheless, it's still possible for larger parlays to have lower variance than smaller parlays if the payout odds are on the larger parlay sufficiently low.

If the 5-team payout odds were only +1200, for example, then the std. dev. would drop from 481.0% to 284.2%.

Conversely if the 5-team payout odds were +2500, then the std. dev. would rise to 509.9%.

To determine necessary and sufficient conditions for std. dev. increasing when moving up in parlay size:

Given a single leg win probability of q, we have that

σ²[n] = qⁿ * (1 - qⁿ) * d²[n]

differencing then gives us:

^Δσ2[n]/_σ²[n] = ( qⁿ⁺¹ * (1-qⁿ⁺¹) * d²[n+1] ) / ( qⁿ * (1-qⁿ) * d²[n])
= q * (1-qⁿ⁺¹) / (1-qⁿ) * d²[n+1] / d²[n]

Which will be ≥ 1 iff

(d[n]/d[n+1])² ≤ q * (1-qⁿ⁺¹) / (1-qⁿ)

So in other words, variance (and hence std. dev.) will increase when moving up in parlay size if and only if the the square of the ratio of the of the smaller parlay odds to the larger parlay is less than the quantity q * (1-qⁿ⁺¹) / (1-qⁿ).

So to put it in concrete terms, if the payout on a 6-team parlay were lower than about +2,831.7, the 6-team would have lower variance than the 5-team.

[Data]

σ²[n] = 55%ⁿ * (1 - 55%)ⁿ * d²[n]

The way I see it:

σ[n]² = 55%ⁿ * (1 - 55%ⁿ) * d[n]²

So to summarize:
σ[3]² = 4.99
σ[4]² = 10.06
σ[5]² = 23.13

[Ganchrow]

The way I see it:

σ[n]² = 55%ⁿ * (1 - 55%ⁿ) * d[n]²

So to summarize:
σ[3]² = 4.99
σ[4]² = 10.06
σ[5]² = 23.13

You are of course completely correct. I had that down in my notebook, but it got all screwy when I transcribed to BBCode. I'll fix my earlier post. Thanks to you and my apologies to Mike.

I think I'm losing my composure here.

[mikevegas]

I got confused because I thought you use decimals
when dealing with percents.

.55 instead of 55.

Now it makes sense.

Thank you.

edit: it still does not make sense. sorry.

[mikevegas]

I am still not following this.
1-55 is a negative number.

[Data]

55%=0.55

[mikevegas]

55%=0.55

I must be doing something wrong

variance for 3 teamer = (.55^3)*(1-.55)^3*(6^2)
= .166*.091*36 = .54

[Ganchrow]

I must be doing something wrong

variance for 3 teamer = (.55³)*(1-.55)^3*(6^2)
= .166*.091*36 = .54

It's actually (.55^3)*(1-.55^3)*(6^2) ≈ 4.99.

That corresponds to the transcription error in this post that Data identified. My apologies again.

[mikevegas]

I am with you so far and thank you for the correction.

The concept I don't understand is why does increasing the pay out
in a 3 team parlay increase your variance?

I would think that if the pay out increased the parlay would
be "better." Also, the payout will not affect the odds of the parlay winning or losing. It will affect your expectation, though.

It may be that I am misusing the word variance when I try to
analyze parlays. But my simple way of looking at parlays is the higher the variance the worse the parlay. But the higher the pay out the better the parlay.

I think I need a drink.

[Ganchrow]

I am with you so far and thank you for the correction.

The concept I don't understand is why does increasing the pay out
in a 3 team parlay increase your variance?

I would think that if the pay out increased the parlay would
be "better." Also, the payout will not affect the odds of the parlay winning or losing. It will affect your expectation, though.

It may be that I am misusing the word variance when I try to
analyze parlays. But my simple way of looking at parlays is the higher the variance the worse the parlay. But the higher the pay out the better the parlay.

I think I need a drink.

By increasing the odds on a wager the bet is made better because it has a higher expectation, but that effect is slightly offset by the increased risk.

We can make the axiomatic claim that all bettors should always prefer higher payout odds to lower on a given bet.

The following thought experiment might help clear up some of the confusion:

Imagine you have coin-flip bet paying out at even odds. This corresponds to standard deviation of sqrt(2-1) = 100%.

Now imagine that the payout odds on the bet were increased from +100 to let's say +1,000,000.

This will increase standard deviation to sqrt(50%*(1-50%))*10,001 = 500,050%.

In the former case the outcomes are both quite similar in magnitude. On a $1 bet you either make a dollar of you lose a $1.

In the latter case, there's much more variability in your results. You either lose $1 or you make $10,000. That's some pretty significant variance.

What this goes to show is that as expectation grows because it becomes increasingly important to consider factors besides just expectation and variance in order to assess the value of a particular bet. This a bet management system such as Kelly comes in.