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

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

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?...

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.

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

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.

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.

This is good GillesD but not what she is looking for

Do you have the stats for sum of skips?

No SW I know of does what you want It would have to be developed as a filter

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).

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.

Thanks GillesD,

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

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..

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!

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