# to maggie -

#### mirage

##### Member
Hi maggie
Thanks for that reference. Yes, that link to those filters are all things that I might find potentially useful as well.

#### mirage

##### Member
to johnph77 -

Hi johnph77
By now its probably becoming apparent that i'm a mathematical ignoramus. ?What do you mean by "positional analysis"?
Thanks!

#### mirage

##### Member
- to Beaker

Thanks Beaker for the info about Lotwin.
Ok, I'm totally dumb when it comes to math. Question: a "normal distribution" is a bell curve? Let me try to speak even more plainly. In my previous post, the "other list" is not a statistic, (as such?) - you are using statistic in a formal sense and I was being unclear(?). The 2nd list I referred to wanting is simply a list of 6 number combinations, not a statistic. What i've seemed to find is that there is also a tendency to a bell curve - of the sums of 6 number combos - which is the sum of when each number in the winning numbers this time - won the last time(!) (I.e., was it a repeater? Did it last win 2 draws ago, 3, 5, 16, another number 5 draws ago, etc?). Ok, so lets say in the last draw, one of the winning 6 numbers won in the draw before that, (a repeater), 2 numbers won 5 draws ago, one won in the last 2, one won in the last 3, and one won 16 draws ago. Therefore the sum in this case would be 1 + 2+ 3 + 5 + 5+ 16 = 32. Although the bell curve is not clear there is a tendency. And, certain sums - actual numbers - seem to reoccur - again and again - over the long time - of many many draws. Why this happens, I have no idea. It's interesting to see and may be just "coincidence" but the mind thinks it sees a pattern. However, the tendency to a bell curve makes sense, since over the long term, we are dealing with a finite set of possibilities. Therefore, what i would want is: give me a list of all the 6 number combos of which the sums of - the last time the number was drawn - equals, say, "50". Is there any program that produces a list of 6 number combos that does this? All the 6 number combos (it would naturally have to change after every draw) in the list would equal a number which is the sum of when each of the 6 numbers hit the last time? Hopefully that clarifies it a bit more?...

#### Beaker

##### Member
mirage thanks for that response - yes essentially the bell curve you refer to is a Normal Distribution - that is a formal statistical distribution - which, for this discussion is not important I think we understand what you mean.

What you are doing is adding the skips for each of the numbers and getting a sum. I, and others, have this data but I've never looked at sum of skips so let's see what it gives us.

#### Snides

##### Member
The last few draws (#2050 to 2106) for the Canadian 6/49 gives you these results.

First 6 columns are the regular 6 numbers, 7th column is the bonus, 8th column is the sum of the first 6, and the last column is the sum of the first 7

If you graph this it looks like the teeth on an old worn out saw blade, i didn't try bell curving it..

01 17 03 04 10 02 21 | 37 | 58
05 03 04 09 12 01 02 | 34 | 36
03 01 07 08 09 10 05 | 38 | 43
07 12 02 02 07 07 03 | 37 | 40
09 17 07 03 04 13 05 | 53 | 58
03 02 13 07 22 07 04 | 54 | 58
03 43 13 11 02 01 03 | 73 | 76
01 02 05 01 01 02 04 | 12 | 16
06 11 08 05 10 02 03 | 42 | 45
04 19 14 05 01 07 11 | 50 | 61
01 01 10 07 01 10 03 | 30 | 33
04 06 01 10 12 07 04 | 40 | 44
08 04 03 18 15 02 01 | 50 | 51
01 05 15 07 03 01 01 | 32 | 33
01 07 01 03 07 10 03 | 29 | 32
08 07 03 01 13 01 01 | 33 | 34
22 14 10 01 13 03 02 | 63 | 65
10 28 03 06 11 18 02 | 76 | 78
01 10 05 03 01 02 13 | 22 | 35
06 09 04 03 03 03 01 | 28 | 29
04 08 10 02 07 03 01 | 34 | 35
04 06 17 01 10 03 03 | 41 | 44
02 11 10 04 03 05 07 | 35 | 42
02 12 01 09 04 08 14 | 36 | 50
03 16 01 14 03 04 15 | 41 | 56
05 03 01 26 03 01 03 | 39 | 42
05 02 07 25 06 03 03 | 48 | 51
01 02 02 06 01 09 04 | 21 | 25
11 10 33 04 19 04 09 | 81 | 90
02 29 05 10 03 07 05 | 56 | 61
01 02 03 02 10 03 07 | 21 | 28
06 04 01 03 03 10 12 | 27 | 39
02 01 07 30 02 03 03 | 45 | 48
08 11 01 02 08 05 03 | 35 | 38
01 14 06 05 26 11 12 | 63 | 75
09 09 06 03 03 19 05 | 49 | 54
03 01 09 05 19 15 05 | 52 | 57
02 02 03 01 21 05 27 | 34 | 61
08 24 07 05 01 02 09 | 47 | 56
01 03 01 08 03 02 03 | 18 | 21
03 01 01 08 02 05 03 | 20 | 23
03 28 08 01 02 03 04 | 45 | 49
03 03 04 19 16 07 13 | 52 | 65
15 17 01 31 09 01 04 | 74 | 78
04 09 01 06 14 02 04 | 36 | 40
02 04 03 03 08 04 03 | 24 | 27
03 02 01 12 05 02 06 | 25 | 31
03 15 08 02 06 02 06 | 36 | 42
03 44 04 02 01 11 02 | 65 | 67
02 13 09 15 08 03 09 | 50 | 59
06 02 03 02 01 01 01 | 15 | 16
01 01 09 15 03 18 05 | 47 | 52
05 04 18 04 21 06 01 | 58 | 59
02 13 03 07 26 02 05 | 53 | 58
02 20 09 01 02 01 01 | 35 | 36
04 07 12 20 05 03 05 | 51 | 56
07 01 12 01 09 04 12 | 34 | 46

#### GillesD

##### Member
Distribution - Normal or not

Snides

With only 57 samples, you can not expect a very nice and clearly identified distribution. Repeat this for all draws and I would expect a nice normal distribution centered around a sum of 50 and decreasing slowly as you go down or up. This I assume as I do not have the actual data to calculate it. But the limited data you provided seems to prove this.

For the delta values calculated on ordered numbers (not considering the bonus number), the distribution of the sum of deltas is exponential, with an average of 43.2, a minimum of 13 (once), a maximum of 49 (260 times) and nearly 80% of the values with a sum of 40 and over.

#### mirage

##### Member
Normal distribution yes?

Hello everyone,
So, does anyone know of a software that will produce a list of 6 number combos that will equal a sum? All you programmers are at an advantage here as I suspect one may have to be developed.

#### Beaker

##### Member
Re: Distribution - Normal or not

GillesD said:
Snides

With only 57 samples, you can not expect a very nice and clearly identified distribution. Repeat this for all draws and I would expect a nice normal distribution centered around a sum of 50 and decreasing slowly as you go down or up. This I assume as I do not have the actual data to calculate it. But the limited data you provided seems to prove this.

For the delta values calculated on ordered numbers (not considering the bonus number), the distribution of the sum of deltas is exponential, with an average of 43.2, a minimum of 13 (once), a maximum of 49 (260 times) and nearly 80% of the values with a sum of 40 and over.
This is good GillesD but not what she is looking for

Do you have the stats for sum of skips?

#### Beaker

##### Member
Re: Normal distribution yes?

mirage said:
Hello everyone,
So, does anyone know of a software that will produce a list of 6 number combos that will equal a sum? All you programmers are at an advantage here as I suspect one may have to be developed.
No SW I know of does what you want It would have to be developed as a filter

#### GillesD

##### Member
Distribution of sums of skips

Beaker, I never worked with "skips" (or how many numbers between two successive numbers) but I have data for deltas (or the difference between two successive numbers). By substracting 6 to all sum of deltas mentionned, you should get the sum of skips you want.

The data given below is valid for the first 2106 draws and does not take into consideration the bonus number. All calculations are made on ordered numbers in a draw. The data is first the sum of the deltas (6 to 49), then the number of times this sum occured and the relative percentage for 2106 draws.

- 06: never (0.0%)
- 07: never (0.0%)
- 08: never (0.0%)
- 09: never (0.0%)
- 10: never (0.0%)
- 11: never (0.0%)
- 12: never (0.0%)
- 13: 1 time (0.0%)
- 14: never (0.0%)
- 15: never (0.0%)
- 16: never (0.0%)
- 17: never (0.0%)
- 18: 1 time (0.0%)
- 19: 1 time (0.0%)
- 20: 2 times (0.1%)
- 21: 2 times (0.1%)
- 22: 1 time (0.0%)
- 23: 3 times (0.1%)
- 24: 2 times (0.1%)
- 25: 4 times (0.2%)
- 26: 6 times (0.3%)
- 27: 8 times (0.4%)
- 28: 10 times (0.5%)
- 29: 13 times (0.6%)
- 30: 16 times (0.8%)
- 31: 22 times (1.0%)
- 32: 20 times (0.9%)
- 33: 29 times (1.4%)
- 34: 29 times (1.4%)
- 35: 47 times (2.2%)
- 36: 44 times (2.1%)
- 37: 50 times (2.4%)
- 38: 56 times (2.7%)
- 39: 70 times (3.3%)
- 40: 89 times (4.2%)
- 41: 94 times (4.5%)
- 42: 121 times (5.7%)
- 43: 138 times (6.6%)
- 44: 157 times (7.5%)
- 45: 172 times (8.2%)
- 46: 175 times (8.3%)
- 47: 250 times (11.9%)
- 48: 213 times (10.1%)
- 49: 260 times (12.3%)

The sum of deltas can go from 6 (for 6 consecutive numbers) to 49 (many combinations such as 7, 19, 31, 43, 45 and 49).

One thing that bothers me is the values for the sums 47 and 48. If inverted (213 for 47 and 250 for 48), it would give an almost perfect exponential distribution. I have not found anything in my data to indicate my calculations are wrong (yet).

#### GillesD

##### Member
Re: Normal distribution yes?

mirage said:
So, does anyone know of a software that will produce a list of 6 number combos that will equal a sum? All you programmers are at an advantage here as I suspect one may have to be developed.

mirage, if you use Excel, check my post on another thread that indicates how to do this. If necessary, I will post the macro again with the modification to list only combinations meeting a certain sum.

#### Beaker

##### Member
Thanks GillesD,

I'll take a look for myself. That distributution will definitely skew right

#### Snides

##### Member
Re: Distribution - Normal or not

GillesD said:
Snides

With only 57 samples, you can not expect a very nice and clearly identified distribution. Repeat this for all draws and I would expect a nice normal distribution centered around a sum of 50 and decreasing slowly as you go down or up.

For the delta values calculated on ordered numbers (not considering the bonus number), the distribution of the sum of deltas is exponential, with an average of 43.2, a minimum of 13 (once), a maximum of 49 (260 times) and nearly 80% of the values with a sum of 40 and over.

I just did a small snippit to show some sample data without making everyone scroll though tons of data.. I just did a graph to try out the bell curve idea on this and it looks like it peaks a bit earlier than you expected, highest numbers from 30 to 43. Here's the info I've compiled, first column sum, second column how many times..

6 1
7 0
8 0
9 1
10 1
11 2
12 1
13 4
14 7
15 3
16 13
17 16
18 26
19 15
20 27
21 29
22 26
23 31
24 28
25 40
26 43
27 47
28 47
29 47
30 67
31 59
32 58
33 51
34 57
35 51
36 62
37 52
38 60
39 66
40 48
41 66
42 62
43 65
44 39
45 38
46 53
47 47
48 36
49 38
50 32
51 37
52 41
53 40
54 29
55 32
56 34
57 25
58 28
59 19
60 24
61 20
62 17
63 21
64 17
65 13
66 9
67 7
68 7
69 7
70 6
71 17
72 7
73 11
74 8
75 5
76 7
77 2
78 7
79 8
80 2
81 2
82 2
83 3
84 2
85 1
86 3
87 3
88 3
89 0
90 1
91 2
92 3
93 1
94 0
95 0
96 0
97 2
98 0
99 0
100 1
101 0
102 0
103 0
104 2
105 1
106 1
107 0
108 0
109 0
110 0
111 0
112 0
113 1
114 0
115 0
116 0
117 0
118 0
119 1
120 1

the low numbers on this data are slightly inflated, since I started with the beginning of the draws, the first time each number comes up it calculates it as how many skips since the first draw, as it does not have any history prior to that..

#### mirage

##### Member
-to GillesD - re Normal distribution again - where's the macro?

GillesD said:
mirage, if you use Excel, check my post on another thread that indicates how to do this. If necessary, I will post the macro again with the modification to list only combinations meeting a certain sum.

Hi GillesD,
Sorry if I'm being a pest. I've looked through many threads and cannot find seem to find the Excel macro that you are saying indicates how to do the list of 6 number combos that equals a sum. Can you post with modification?
Thanks!

#### GillesD

##### Member
Combinations meeting a sum

The macro Comb_Sum below lists all combinations that meet specific sums (between a minimum and a maximum). The combinations are listed in the form N1-N2-N3-N4-N5-N6 each in a single cell, a format which allows to list all 13,983,816 combinations in an unique Excel sheet.

The column width should be set to allow at least 20 characters in a cell. The program will list 65,00 combinations in a column and then move one column to the right and continue. If necessary, the command Convert of Excel can be used to split the combinations into 6 adjacent columns (using "-" as a delimiter).

Adjust the values for nSumMin and nSumMax in the macro, place the cursor in the sheet you want the combinations to appear and start the macro. If you want the combinations to meet a single sum, set both nSumMin and nSumMax to that value.

Sub Comb_Sum()
Dim A As Integer, B As Integer, C As Integer, D As Integer, E As Integer, F As Integer
Dim N As Long, nSum As Integer, nSumMin As Integer, nSumMax As Integer
' Set nSumMin and nSumMax to the values (minimum and maximum) you want the sum to cover
' if you want to meet a single value, set both nSumMin and nSumMax to that value
nSumMin = 145
nSumMax = 155
Range("A1").Select
Application.ScreenUpdating = False
N = 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
nSum = A + B + C + D + E + F
If N = 65000 Then
N = 1
ActiveCell.Offset(-65000, 1).Select
Application.ScreenUpdating = True
Application.ScreenUpdating = False
End If
If nSum >= nSumMin And nSum <= nSumMax Then
N = N + 1
ActiveCell.Value = Application.WorksheetFunction.Text(A, "00") & "-" & Application.WorksheetFunction.Text(B, "00") _
& "-" & Application.WorksheetFunction.Text(C, "00") & "-" & Application.WorksheetFunction.Text(D, "00") _
& "-" & Application.WorksheetFunction.Text(E, "00") & "-" & Application.WorksheetFunction.Text(F, "00")
ActiveCell.Offset(1, 0).Select
End If
Next F
Next E
Next D
Next C
Next B
Next A
Application.ScreenUpdating = True
End Sub

#### Nick Koutras

##### Member
The following sub will do the
count of sums far faster,
but it will not display the actual sets,
but only the count of the sets complying to
the sum of your interest, for any 6/xx Lotto.

1. Create a new XLS sheet.
1a On sheet1

2. In cell A1 enter "Max NB"
3. In cell B1 enter "Sum"
4. In cell C1 enter "Sets"
5. In cell D1 enter "Time"

6. Press Alt+F11 to open VBA
8. Copy the code below to that module

10. Create a Button and link to the Sub you just copied.

Now you are ready to enter the values and find the result!

In Cells A2 Enter the Maximum Numbers
In Cell B2 Enter the Sum of your interest.

Click the Button to run the Macro.

The result of such sets with that Sum
and Time taken will be displayed

==========Code ======
Option Explicit

Sub Sum_Of_Sets()

Dim Sum&, MinVal%, MaxVal%, Target&
Dim I&, J&, K&, L&, M&, N&, Cnt&
Dim xTime

MinVal=1
MaxVal = Range("a2")
Target = Range("b2")
Sum = 0

xTime = Timer
For I = MinVal To (Target + 5) / 6 - 3
Sum = Sum + I
For J = I + 1 To (Target + 4 - Sum) / 5 - 2
Sum = Sum + J
For K = J + 1 To (Target + 3 - Sum) / 4 - 2
Sum = Sum + K
For L = K + 1 To (Target + 2 - Sum) / 3 - 1
Sum = Sum + L
For M = L + 1 To (Target + 1 - Sum) / 2 - 1
N = Target - Sum - M
If (N <= MaxVal) Then
Cnt = Cnt + 1
If Cnt Mod 1000 = 0 Then Cells(4, 1) = Cnt
End If
Next
Sum = Sum - L
Next
Sum = Sum - K
Next
Sum = Sum - J
Next
Sum = Sum - I
Next

Cells(2, 3) = Cnt
Cells(2, 4) = Timer - xTime
Cells(4, 1) = ""
Cells(1, 1).Select

End Sub

=========End of Code

#### mirage

##### Member
Thank you GillesD!

GillesD -
Thanks a million (that would be nice) for your help! Going out today to get manuals on Excel macros and/or VB. May have to modify slightly to get what I want, but will let you know how it works out!

#### mirage

##### Member
- Nick Koutras - Thank you!

Thank you Nick! This macro could certainly be useful too...

#### GillesD

##### Member
Sets for a specific sum

Nick, your macro is certainly very fast and your coding is far better than mine.

I ran your macro for a 6/49 lottery and tested it initially for three sums, the lowest possible values: 21, 22 and 23.

Placing successively those values in cell B2, I get respectively the values 1, 2 and 3 in cell C2 indicating the number of sets in each case.

For a sum of 21, there is only one set (1, 2, 3, 4, 5, 6) as the macro indicates. But for a sum of 22, the first set is 1, 2, 3, 4, 5 and 7 but I can not find any other set. The same thing for a sum of 23, I know 2 sets (1, 2, 3, 4, 5, 8 and 1, 2, 3, 4, 6, 7) but can not see any other one to make it to 3.

#### Nick Koutras

##### Member
This is a funny error.

Do the following:

Remark the following:

Option explicit
and all the Dim statments.

This shall work!

====Code ========
Sub Sum_Of_Digits()

Sum = 0
MinVal = 1
MaxVal = Range("a2")
Target = Range("b2")
Cnt = 0

xTime = Timer
For I = MinVal To (Target + 5) / 6 - 3
Sum = Sum + I
For J = I + 1 To (Target + 4 - Sum) / 5 - 2
Sum = Sum + J
For K = J + 1 To (Target + 3 - Sum) / 4 - 2
Sum = Sum + K
For L = K + 1 To (Target + 2 - Sum) / 3 - 1
Sum = Sum + L
For M = L + 1 To (Target + 1 - Sum) / 2 - 1
N = Target - Sum - M
If (N <= MaxVal) Then

'If you wish to display them, do it below
'by printing i,j,k,l,m,n

Cnt = Cnt + 1
If Cnt Mod 1000 = 0 Then Cells(4, 1) = Cnt

End If
Next
Sum = Sum - L
Next
Sum = Sum - K
Next
Sum = Sum - J
Next
Sum = Sum - I
Next

Cells(2, 3) = Cnt
Cells(2, 4) = Timer - xTime
Cells(4, 1) = ""
Cells(1, 1).Select

End Sub

======end of code=======