Positional Analysis for Odds & Evens
I thought a Nice Little Exercise to do would be to Produce the Data for the Odds & Evens Distribution Per Position.
Using Revised Data from my Positional Number Distribution Table ( in Excel ) I have Created the Following. This Table Shows the Total Combinations Achievable if the Ball Number Drawn in EACH Position is an Odd OR Even ( After Numerical Sorting ) Number.
Position 1 Odd = 7,443,260 Position 1 Even = 6,540,556
Position 2 Odd = 6,967,620 Position 2 Even = 7,016,196
Position 3 Odd = 6,992,920 Position 3 Even = 6,990,896
Position 4 Odd = 6,993,910 Position 4 Even = 6,990,896
Position 5 Odd = 6,967,620 Position 5 Even = 7,016,196
Position 6 Odd = 7,443,260 Position 6 Even = 6,540,556
I then Applied this Concept to the Canadian 649 Lotto as at Draw 2,217 ( the 20th of April 2005 ) and came up with the following Table.
The Table is Set Out with the Position Number ( Odd ), the Expected Drawn to Date, the Actual Drawn to Date & the Difference +/- Between the Two. Then EXACTLY the Same for Even.
Canadian 649 Lotto
Position 1 Odd = 1180.06 1195 +14.94 Position 1 Even = 1036.94 1022 -14.94
Position 2 Odd = 1104.65 1105 +0.35 Position 2 Even = 1112.35 1112 -0.35
Position 3 Odd = 1108.66 1106 -2.66 Position 3 Even = 1108.34 1111 +2.66
Position 4 Odd = 1108.82 1093 -15.82 Position 4 Even = 1108.34 1124 +15.66
Position 5 Odd = 1104.65 1096 -8.65 Position 5 Even = 1112.35 1121 +8.65
Position 6 Odd = 1180.06 1222 +41.94 Position 6 Even = 1036.94 995 -41.94
I then Applied EXACTLY the Same to the UK Main 649 Lotto as at Draw 973 ( the 20th of April 2005 ) and came up with the following Table.
UK Main 649 Lotto
Position 1 Odd = 517.91 512 -5.91 Position 1 Even = 455.09 461 +5.91
Position 2 Odd = 484.81 484 -0.81 Position 2 Even = 488.19 489 +0.81
Position 3 Odd = 486.57 509 +22.43 Position 3 Even = 486.43 464 -22.43
Position 4 Odd = 486.64 493 +6.36 Position 4 Even = 486.43 480 -6.43
Position 5 Odd = 484.81 476 -8.81 Position 5 Even = 488.19 497 +8.81
Position 6 Odd = 517.91 506 -11.91 Position 6 Even = 455.09 467 +11.91
Now if we Only Knew ( in the Next Draw for Example ) if the Number in Position One ( After Numerical Sorting ) was Going to be Odd OR Even, we would then in Theory ONLY be Playing with 7,443,260 Combinations if it is Odd, Or 6,540,556 Combinations if it is Even, a LOT Less than the FULL 13,983,816 Combinations.
I think that this is Probably a High Profile Filter that could be Used Front End ( at the Top Or Near the Top of Filters Being Applied ), Especially with it Being Based on Almost a 50/50 Chance of the First Number ( After Numerical Sorting ) Either Being Odd Or Even. This will Reduce the Combinations by 46.77% if the First Number is Odd, and 53.22% if the First Number is Even.
Obviously I have just Used Position One as an Example, it is Quite Easy to Associate the Data for the Other Five Positions.
The Bottom Line is, you Need to have a Really Good Feeling for Odd Or Even in Order to Even Consider this as a Viable Proposition. But Reducing the Combinations to this Degree, and then Applying Any Other Filters that you Use, you should be Left with a More Manageable List of Combinations, Hopefully Still Including the JACKPOT Combination.
I just thought I would Throw this out for Discussion.
I Double Checked my Figure for Position One ( if it was Even ) with the Macro Below. It takes About 15 Seconds to Run.
Option Explicit
Option Base 1
Dim A As Integer
Dim B As Integer
Dim C As Integer
Dim D As Integer
Dim E As Integer
Dim F As Integer
Sub PositionOne()
Dim nCount As Long
Application.ScreenUpdating = False
nCount = 0
For A = 1 To 44
For B = A + 1 To 45
For C = B + 1 To 46
For D = C + 1 To 47
For E = D + 1 To 48
For F = E + 1 To 49
If PositionOneIsEven Then
nCount = nCount + 1
End If
Next F
Next E
Next D
Next C
Next B
Next A
ActiveCell.Value = nCount
Application.ScreenUpdating = True
End Sub
Private Function PositionOneIsEven() As Long
PositionOneIsEven = False
Select Case A
Case 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, _
26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48
PositionOneIsEven = True
Exit Function
End Select
End Function
Good Luck Everyone.
All the Best.
PAB
I thought a Nice Little Exercise to do would be to Produce the Data for the Odds & Evens Distribution Per Position.
Using Revised Data from my Positional Number Distribution Table ( in Excel ) I have Created the Following. This Table Shows the Total Combinations Achievable if the Ball Number Drawn in EACH Position is an Odd OR Even ( After Numerical Sorting ) Number.
Position 1 Odd = 7,443,260 Position 1 Even = 6,540,556
Position 2 Odd = 6,967,620 Position 2 Even = 7,016,196
Position 3 Odd = 6,992,920 Position 3 Even = 6,990,896
Position 4 Odd = 6,993,910 Position 4 Even = 6,990,896
Position 5 Odd = 6,967,620 Position 5 Even = 7,016,196
Position 6 Odd = 7,443,260 Position 6 Even = 6,540,556
I then Applied this Concept to the Canadian 649 Lotto as at Draw 2,217 ( the 20th of April 2005 ) and came up with the following Table.
The Table is Set Out with the Position Number ( Odd ), the Expected Drawn to Date, the Actual Drawn to Date & the Difference +/- Between the Two. Then EXACTLY the Same for Even.
Canadian 649 Lotto
Position 1 Odd = 1180.06 1195 +14.94 Position 1 Even = 1036.94 1022 -14.94
Position 2 Odd = 1104.65 1105 +0.35 Position 2 Even = 1112.35 1112 -0.35
Position 3 Odd = 1108.66 1106 -2.66 Position 3 Even = 1108.34 1111 +2.66
Position 4 Odd = 1108.82 1093 -15.82 Position 4 Even = 1108.34 1124 +15.66
Position 5 Odd = 1104.65 1096 -8.65 Position 5 Even = 1112.35 1121 +8.65
Position 6 Odd = 1180.06 1222 +41.94 Position 6 Even = 1036.94 995 -41.94
I then Applied EXACTLY the Same to the UK Main 649 Lotto as at Draw 973 ( the 20th of April 2005 ) and came up with the following Table.
UK Main 649 Lotto
Position 1 Odd = 517.91 512 -5.91 Position 1 Even = 455.09 461 +5.91
Position 2 Odd = 484.81 484 -0.81 Position 2 Even = 488.19 489 +0.81
Position 3 Odd = 486.57 509 +22.43 Position 3 Even = 486.43 464 -22.43
Position 4 Odd = 486.64 493 +6.36 Position 4 Even = 486.43 480 -6.43
Position 5 Odd = 484.81 476 -8.81 Position 5 Even = 488.19 497 +8.81
Position 6 Odd = 517.91 506 -11.91 Position 6 Even = 455.09 467 +11.91
Now if we Only Knew ( in the Next Draw for Example ) if the Number in Position One ( After Numerical Sorting ) was Going to be Odd OR Even, we would then in Theory ONLY be Playing with 7,443,260 Combinations if it is Odd, Or 6,540,556 Combinations if it is Even, a LOT Less than the FULL 13,983,816 Combinations.
I think that this is Probably a High Profile Filter that could be Used Front End ( at the Top Or Near the Top of Filters Being Applied ), Especially with it Being Based on Almost a 50/50 Chance of the First Number ( After Numerical Sorting ) Either Being Odd Or Even. This will Reduce the Combinations by 46.77% if the First Number is Odd, and 53.22% if the First Number is Even.
Obviously I have just Used Position One as an Example, it is Quite Easy to Associate the Data for the Other Five Positions.
The Bottom Line is, you Need to have a Really Good Feeling for Odd Or Even in Order to Even Consider this as a Viable Proposition. But Reducing the Combinations to this Degree, and then Applying Any Other Filters that you Use, you should be Left with a More Manageable List of Combinations, Hopefully Still Including the JACKPOT Combination.
I just thought I would Throw this out for Discussion.
I Double Checked my Figure for Position One ( if it was Even ) with the Macro Below. It takes About 15 Seconds to Run.
Option Explicit
Option Base 1
Dim A As Integer
Dim B As Integer
Dim C As Integer
Dim D As Integer
Dim E As Integer
Dim F As Integer
Sub PositionOne()
Dim nCount As Long
Application.ScreenUpdating = False
nCount = 0
For A = 1 To 44
For B = A + 1 To 45
For C = B + 1 To 46
For D = C + 1 To 47
For E = D + 1 To 48
For F = E + 1 To 49
If PositionOneIsEven Then
nCount = nCount + 1
End If
Next F
Next E
Next D
Next C
Next B
Next A
ActiveCell.Value = nCount
Application.ScreenUpdating = True
End Sub
Private Function PositionOneIsEven() As Long
PositionOneIsEven = False
Select Case A
Case 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, _
26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48
PositionOneIsEven = True
Exit Function
End Select
End Function
Good Luck Everyone.
All the Best.
PAB
