Final Digit sequence combos

Rebeckah

Member
According to my notes there are 531,441 possible different sequences of final digits for a 6 ball lotto. Can anybody give me a correct number of sequences possible if you removed all the doubles and triples?

Example:
So wining combo: 6-13-23-26-30-31
would have FD seq of: 633601 or 013366 sorted.
Reduced, this would be: 0136

I'm thinking it'll be less data intensive without dbls & trpls. & easier to predict. All comments welcome. :) Thanks. & hey, if anybody just happens to have a listing of all possible sequences, posting it in the free section would be awesome! Or you could just email it to me. Thanks you lotto Gods. :agree2: :agree: :kiss:
 

GillesD

Member
Sequences of last digits

Rebeckah

Yes, a very good question. But before I start working on reducing the number of possible different sequences of last digits, I would like to know where or how you obtained the value 531,441 for possible different sequences of last digits for a 6 ball lotto. I have tried to get this but I always end up with 991,998 possible different sequences of last digits.

In my way of looking at your question, the sequence 123456 (coming from numbers 1, 22, 23, 34, 45 and 46) is different from the sequence 213456 (coming from numbers 12, 21, 23, 34, 45 and 46). But eventually after sorting and eliminating possible doubles and triples, they would identical (123456) and be only one of the possible different sequence of last digits.

Beaker

What is the reason you say "The number must depend on the last ball number."?
 

hot4

Member
Re: Sequences of last digits

GillesD said:
Rebeckah

Yes, a very good question. But before I start working on reducing the number of possible different sequences of last digits, I would like to know where or how you obtained the value 531,441 for possible different sequences of last digits for a 6 ball lotto. I have tried to get this but I always end up with 991,998 possible different sequences of last digits.

In my way of looking at your question, the sequence 123456 (coming from numbers 1, 22, 23, 34, 45 and 46) is different from the sequence 213456 (coming from numbers 12, 21, 23, 34, 45 and 46). But eventually after sorting and eliminating possible doubles and triples, they would identical (123456) and be only one of the possible different sequence of last digits.

Beaker

What is the reason you say "The number must depend on the last ball number."?
GillesD,

yes you are right. There's 991,998 possible figures of last digits. Sometime ago, trying to understand Mr Kashani claiming he cracked UK lottery, I calculate that in a lotto 6/49 there are:

Maximum of theoretical figures : 10*10*10*10*10*10 = 1 000 000
Not possible figures: 8002
Total of real figures: 991998

Figures with 1 combinations : 174195 * 1 = 174195 ( 1,2456% of 13983816 lines)
Figures with 7 combinations : 504315 * 7 = 3530205 (25,2449%)
Figures with 28 combinations : 286860 * 28 = 8032080 (57,4383%)
Figures with 84 combinations : 26544 * 84 = 2229696 (15,9448%)
Figures with 210 combinations : 84 *210 = 17640 ( 0,1261%)

TOTAL =13983816 (99,9997%) (because of not rounded at 4th digit)

Hoping to help ...

Frank
 

PAB

Member
Hi Everybody,

Just for Information, the UK 649 Lotto upto and including Draw 822 ( Saturday the 8th of November 2003 ) has the Total Final Digits Distribution as follows :-

Description Combs Percent
1-1-1-1-1-1 153 18.61%
2-1-1-1-1-0 406 49.39%
2-2-1-1-0-0 179 21.78%
2-2-2-0-0-0 8 0.97%
3-1-1-1-0-0 57 6.93%
3-2-1-0-0-0 18 2.19%
3-3-0-0-0-0 0 0.00%
4-1-1-0-0-0 1 0.12%
4-2-0-0-0-0 0 0.00%
5-1-0-0-0-0 0 0.00%
-----------------------------------------------
Totals > 822 100%
-----------------------------------------------

Best Regards
PAB
:wavey:

I copied this from Excel 2002, does anybody know why this Table shows like this when it looked OK when I was replying to this thread.
 

GillesD

Member
Last digit distribution

The table gives the distribution for the last digita in draws for a lottery. The first 6 numbers gives the distribution itself with 1-1-1-1-1-1 meaning all digits are unique and 3-2-1-0-0-0 meaning one digit is repeated 3 times, another digit 2 times and another digit once. Some 0 are added to get an uniform 6-digit output. After that, you have the number of occurences for draws analyzed and finally the percentage it gives.

Your data is right for the UK lottery and for the Canadian Lotto 6/49 lottery, the values for 2066 draws are:

1-1-1-1-1-1 -- 444 -- 21.49% -- 20.65%
2-1-1-1-1-0 -- 1031 -- 49.90% -- 49.56%
2-2-1-1-0-0 -- 407 -- 19.70% -- 19.82%
2-2-2-0-0-0 -- 15 -- 0.73% -- 0.76%
3-1-1-1-0-0 -- 122 -- 5.91% -- 6.61%
3-2-1-0-0-0 -- 42 -- 2.03% -- 2.27%
3-3-0-0-0-0 -- 1 -- 0.05% -- 0.03%
4-1-1-0-0-0 -- 4 -- 0.19% -- 0.28%
4-2-0-0-0-0 -- 0 -- 0.00% -- 0.03%
5-1-0-0-0-0 -- 0 -- 0.00% -- 0.00%

In this table, I added at the end the theorical value each distribution should come out, based on the 13,983,816 possible combinations. These percentage would also apply to the UK 6/49 lottery.
 

Beaker

Member
Re: Sequences of last digits

GillesD said:
Rebeckah

Beaker

What is the reason you say "The number must depend on the last ball number."?
Consider this: 49 balls you could have 5 numbers with 9 like 9 - 10-19-29-39-49

If you have 6/40, say, you can't have 5 9's because they don't exist.

You need to know the last number because that defines the potential solution set.
 

hot4

Member
Re: Re: Sequences of last digits

Beaker said:
Consider this: 49 balls you could have 5 numbers with 9 like 9 - 10-19-29-39-49

If you have 6/40, say, you can't have 5 9's because they don't exist.

You need to know the last number because that defines the potential solution set.
And there are 8002 not possible figures in that circunstance, Beaker:

Maximum of theoretical figures : 10*10*10*10*10*10 = 1 000 000
Not possible figures: 8002
Total of real figures: 991998

Frank
 

GillesD

Member
Rebeckah's question

Actually I think the solution to your question is quite simple, since the numbers have to be sorted and no duplicates are allowed.

It can be calculated from the number of combinations that can be made with 6, 5, 4, 3 and 2 numbers out of 10 numbers (from 0 to 9).

The number of combinations are:
with 6 digits: COMBIN(10,6) or 210 combinations from 012345 to 456789
with 5 digits: COMBIN(10,5) or 252 from 01234 to 56789
with 4 digits: COMBIN(10,4) or 210 from 0123 to 6789
with 3 digits: COMBIN(10,3) or 120 from 012 to 789
with 2 digits: COMBIN(10,2) or 45 from 01 to 89
with 1 digit: impossible

For a total of 837 combinations.

And I do not see how the last number can have any effect on the numbers of combinations. But the first one has an effect. If the first number is 44, then the only possibility for the sequence of last digits is 456789.
 
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Beaker

Member
If you only have 40 balls, given your example, the first ball can be 35 for 35-36-37-38-39-40 or 567890

How can a 6/49 lotto have as many as a 6/40 lotto? :confused:
 
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GillesD

Member
6/40 or 6/46 last digit sequences

Actually, the type of lottery has little effect whether it is a 6/40, a 6/47 or a 6/49 lottery. Only the last digits (obviously from 0 to 9) and the number of balls chosen (6) have an effect. But I do not say it applies to all lotteries but I believe it would apply to any lottery with at least 40 balls.

You have to remember Rebeckah's requirements:
- first, identify the last digit of all numbers;
- then, place them in order starting with the lowest one to the highest one;
- finally, remove any double, triple or quadruple numbers.

Let's take 2 examples that could happen in both a 6/40 and 6/49 lottery.

1) Numbers 01, 10, 14, 16, 25 and 39 (draw # 2059 for Can. 6/49): this would give 104659, then 014569 and that's it.

2) Numbers 06, 13, 23, 26, 33 and 34 (draw #2038 for Can. 6/49): this would give 633634, then 333466 and finally 346.

You can see that since only the last digits are considered and some can be eliminated (double, triple, quadruple numbers), then the same formulas apply.

So by Rebeckah's rules, the 13,983,816 combinations of a 6/49 lottery or the 3,838,380 combinations of a 6/40 lottery can both be reduced to 837 possible different sequences of last digits.
 

Rebeckah

Member
Beaker said:
If you only have 40 balls, given your example, the first ball can be 35 for 35-36-37-38-39-40 or 567890

How can a 6/49 lotto have as many as a 6/40 lotto? :confused:
Beaker I understand what you're saying.... but it's a reduced & sorted list. so the list lists only possible combinations of 0-9 in a pick6 format. The number of lottery balls in the game doesn't matter. it's based soley on reduced sorted combos of the final digits 0-9. And a 6/20 lotto will use the same *master* data list as a 6/99. The actual stats would differ of course.

Your example would actually be 056789. So ending lotto ball # of 40 doesn't matter, that counts as zero. Got it? It's hard to explain as we don't really see many *master* data lists in lottery.

& as for my # of *531,441 possible different sequences* That was in my notes that I've copied from here, RGL, & LP. Not surprised that it's wrong. Maybe that's for a pick5 game?
 

Rebeckah

Member
Re: Rebeckah's question

GillesD said:
The number of combinations are:
with 6 digits: COMBIN(10,6) or 210 combinations from 012345 to 456789
with 5 digits: COMBIN(10,5) or 252 from 01234 to 56789
with 4 digits: COMBIN(10,4) or 210 from 0123 to 6789
with 3 digits: COMBIN(10,3) or 120 from 012 to 789
with 2 digits: COMBIN(10,2) or 45 from 01 to 89
with 1 digit: impossible

For a total of 837 combinations.
YES! Thank you Gilles! I knew it was a simple small thing. & I was kinda hoping you would tackle it. I know you like to exercise your excel skills. Did you figure it out formula wise or actually tally up a list? I'm asking because I really want the list too. :D ;) :lol: Thank you again for the data.
 

GillesD

Member
Listing of all sequences

I tried to figure it out based on all combinations for a 6/49 lottery but then a light come on and it was so simple. The macro was developed to veriy the answer.

The following macro is not very elegant but it does the job.

Sub LD_Combo()
Dim A As Long, B As Long, C As Long, D As Long, E As Long, F As Long
Application.ScreenUpdating = False
' Set-up
Range("A1").Select
ActiveCell.Offset(0, 0).Value = " 6# comb."
ActiveCell.Offset(0, 1).Value = " 5# comb."
ActiveCell.Offset(0, 2).Value = " 4# comb."
ActiveCell.Offset(0, 3).Value = " 3# comb."
ActiveCell.Offset(0, 4).Value = " 2# comb."
Columns("A:A").Select
Selection.NumberFormat = "000000"
Columns("B:B").Select
Selection.NumberFormat = "00000"
Columns("C:C").Select
Selection.NumberFormat = "0000"
Columns("D:D").Select
Selection.NumberFormat = "000"
Columns("E:E").Select
Selection.NumberFormat = "00"
' 6-number combo
Range("A2").Select
For A = 0 To 4
For B = A + 1 To 5
For C = B + 1 To 6
For D = C + 1 To 7
For E = D + 1 To 8
For F = E + 1 To 9
ActiveCell.Value = 100000 * A + 10000 * B + 1000 * C + 100 * D + 10 * E + F
ActiveCell.Offset(1, 0).Select
Next F
Next E
Next D
Next C
Next B
Next A
' 5-number combo
Range("B2").Select
For A = 0 To 5
For B = A + 1 To 6
For C = B + 1 To 7
For D = C + 1 To 8
For E = D + 1 To 9
ActiveCell.Value = 10000 * A + 1000 * B + 100 * C + 10 * D + E
ActiveCell.Offset(1, 0).Select
Next E
Next D
Next C
Next B
Next A
' 4-number combo
Range("C2").Select
For A = 0 To 6
For B = A + 1 To 7
For C = B + 1 To 8
For D = C + 1 To 9
ActiveCell.Value = 1000 * A + 100 * B + 10 * C + D
ActiveCell.Offset(1, 0).Select
Next D
Next C
Next B
Next A
' 3-number combo
Range("D2").Select
For A = 0 To 7
For B = A + 1 To 8
For C = B + 1 To 9
ActiveCell.Value = 100 * A + 10 * B + C
ActiveCell.Offset(1, 0).Select
Next C
Next B
Next A
' 2-number combo
Range("E2").Select
For A = 0 To 8
For B = A + 1 To 9
ActiveCell.Value = 10 * A + B
ActiveCell.Offset(1, 0).Select
Next B
Next A
Application.ScreenUpdating = True
Range("A1").Select
End Sub

Take note that the line "Columns("D:D").Select"
should actually read "Columns("D : D").Select" but with no spaces on each side of the :
Smileys can be fun but here it is a problem.
 
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