[byronbb]

Lets assume NHL scoring is poisson. Say pinny has the TT for a team at O/U 2 at -120/110. I go to the poisson calc and via trial and error discover that at least 2 goals scored at -120 implies an expected goals scored of 1.8 goals. Is this correct? Is there a way to do this without the trial and error in the calc.


there is an error in my numbers but is this process okay?

[donjuan]

"At least 2" is the same thing as o1.5. For o2 you need to get rid of the % of the time the 2 pushes.

[FourLengthsClear]

Notwithstanding donjuan's (correct) observation, you would need to:

a) Familiarise yourself with the =Poisson function in MS Excel (it does the same as Ganchrow's Calculator).
b) Be able to "invert" the Poisson equation to give you the expected mean

The following should do b) if you enter it as a new function in MS Excel via VBA (google it if unsure, it is quite easy to do).

Function PoissonInv(dVal As Double, dMean As Double) As Variant
' flc

' For a Poisson process with mean dMean, _
' returns the largest integer such that the CDF <= dVal

' E.g., =POISSON(5, 10, TRUE) returns 0.0670859629
' PoissonInv(0.0670859629, 10) returns 5

Dim iX As Long
Dim dCDF As Double

' these variables are used to simplify this summation:
' dCDF = dCDF + Exp(-dMean) * dMean ^ iX / .Fact(iX)
Dim dExpMean As Double ' =Exp(-dMean)
Dim dFact As Double ' incremental factorial
Dim dPowr As Double ' incremental power

If dVal < 0 Or dVal >= 1 Then
PoissonInv = CVErr(xlErrValue)

ElseIf dVal > 0 Then
dExpMean = Exp(-dMean)
dFact = 1
dPowr = 1

Do While dCDF < dVal
dCDF = dCDF + dExpMean * dPowr / dFact
iX = iX + 1
dFact = dFact * iX
dPowr = dPowr * dMean
Loop

PoissonInv = iX - IIf(dCDF / dVal > 1.000000000001, 2, 1)
End If
End Function

[byronbb]

Thanks for the effort 4lengths.