pick 3 separet, repit digit

[jack]

hello someone has the total possible combinations of the following criteria =
the last draw of pick3 the condition is to repeat a digit, but not in the same position next example 159 481 repeated the digit 1, need all the CIM conbinaçoes digit 1 in the 2nd and 3rd positions may not have the 1st position and so with digit 5:09 the conbinaçoes with the digit 5 can not be the 2nd position
and the total of all the digit 9 not pose be the 3rd position, anyone know how to calculate in excel? please

 

[Frank]

I'm not sure this has a rational solution Jack. If I understand your rules correctly, then starting with 000 then there would be 88 possible combinations of digits which could follow 000 whilst allowing a the first zero to repeat in a different position. Similarly for each of the other zeroes. So we have three possible streams of 88 combinations which could follow 000. Each of those possibilities could, in turn have three streams of 88 possibilities of digits to form the next 3 digit number, and so on ...... Making an infinite list within multiple chains of possibilities. Yes there would be repeated combinations along the way, just as the lottery can go on forever whilst containing repeated combinations. I don't believe this is solvable. Sorry.

 

[jack]

Hello, Frank, thanks for help, the problem is language, very bad vezestraduçao google
But this is how we have the standard in pick3 to repeat one of the three digits in 68%
* MASN not in position
* example
1st 2nd 3rd
4 5 2 last
2 9 6 = repeated the digit 2 but in 1st place
then we have
1x9x9 equal 81 for 1st place
9x1x9 equal 81 for 2nd place
Total = 162
The 3rd place will not because temdigito repeating and the rule is not repeated in verticamente position
81x3 equal 243
* You can repeat what one of three digits
* We 162x3 = 482
* Then hit condition
* Just give the pattern repeat of the digits that will profit,
* Or you can also pick up the other condition of the remaining 7 digits
* And make pairs (double digit) 7 x2 = 21
* The 3rd digit is repeating
As are three digits = 63, the setting is 100%
* But it has to be like = 163,631,316 ... 6 times
Frank as it can in the 1st condition of 482 formations decrease

 

[jack]

Hits

RED: repeat
GREEN: leapfrog
BLUE: hit
DATE DT 0 1 2 3 4 5 6 7 8 9
2015-07-25 1 0 1 2 3 4 5 6 7 8 9
2015-07-25 2 0 1 2 3 4 5 6 7 8 9
2015-07-26 1 0 1 2 3 4 5 6 7 8 9
2015-07-26 2 0 1 2 3 4 5 6 7 8 9
2015-07-27 1 0 1 2 3 4 5 6 7 8 9
2015-07-27 2 0 1 2 3 4 5 6 7 8 9
2015-07-28 1 0 1 2 3 4 5 6 7 8 9
2015-07-28 2 0 1 2 3 4 5 6 7 8 9
2015-07-29 1 0 1 2 3 4 5 6 7 8 9
2015-07-29 2 0 1 2 3 4 5 6 7 8 9

 

[jack]

Hello, Frank, it's ok to pick 3 Be Or the position repeats the digit can not be split, as we have We have 3 digits that anyone can repeat
It is 3 times = Example last draw
Next 459 gave 197, not repeated in the 3rd position

 

[jack]

Frank the question how to filter the positions where vain 9 digits intend to decrease
The value of 81, and 162, that is not all play, how to filter

 

[jack]

Hello, Frank, the only condition vain work with the repetition of a digit of the last to roximo of pick3 !! If repeat 2 or zero the last, will not work The condition to work at 100% is to repeat a digit in any of the three digits, why? as well as the remaining 7 for the pair can only repeat a last digit to the next to make it work, you know?

 

[jack]

Hello, Frank, can create a macro
Following the criteria =
starting from the last draw of pick3 (to be entered)
to couple the remaining 7 digits and add one digit of the last draw of the pick 3
Then we
21 = 63 = 3 formations

 

[Frank]

Jack, I answered your original question. My interest ended there. I'm not sufficiently interested in whatever it is to persue this.

Good luck.

 

[TheConcept]

hello someone has the total possible combinations of the following criteria =
the last draw of pick3 the condition is to repeat a digit, but not in the same position next example 159 481 repeated the digit 1, need all the CIM conbinaçoes digit 1 in the 2nd and 3rd positions may not have the 1st position and so with digit 5:09 the conbinaçoes with the digit 5 can not be the 2nd position
and the total of all the digit 9 not pose be the 3rd position, anyone know how to calculate in excel? please

Hi Jack.

I tried doing the Maths side and it was getting complicated.

I found it a lot easier using Visual Basic for Applications in an Excel spreadsheet. It gave me an answer of 729, which makes me think that the Maths side may be not to bad after all. 729 is 9 x 9 x 9!

Here's my code. I used 159 as an example. It cycles through all possible three digit permutations and only counts it if there is not a 1 in position 1, 5 in position 2 or 9 in position 3.

Sub Pick3PermutationsCount()
Dim b1 As Byte
Dim b2 As Byte
Dim b3 As Byte
Dim intTotal As Integer

intTotal = 0
For b1 = 0 To 9
For b2 = 0 To 9
For b3 = 0 To 9
If Not ((b1 = 1) Or (b2 = 5) Or (b3 = 9)) Then
intTotal = intTotal + 1
End If
Next b3
Next b2
Next b1

MsgBox "Total = " & intTotal
End Sub

 

[jack]

Hello, the concept, thank you to see the macro, but there is an error in the construction
Because it is always used the last draw as the base (reference) the reference is the Last draw, and follow the standard to take advantage of repeating the last one digit to the next of pick3, but can not repeat in position. example =
1st 2nd 3rd 4 5 8 = base last draw 8 6 7
1x9x9 equal 81
9x1x9 equal 81
Total 162
Equal 162x3 = 486 729 = error
486 = The concept corret without filters you have to review the macro 729 error, the corret is 486 but has the account of macro remaining 7 didgitos making pairs (doubles)
7 2 = 21 times per each of three digits Total = 63, these being very promising paresis
Theconcept think !! It will only hit 100% if the pattern to repeat a digit will work if, repeat double-digit or none at all will not funionas, will only work if repeat a digit Repetition of a digit is 68%, since the right confidence

 

[jack]

The concept, purpose enjoy the repeat pattern of a digit, the last draw for the next draw, but not possible, put in position
You must create the macro on the following criteria
*Type in a field in Excel 2010 or higher
And the macro creates combinations, watching not to repeat the positions below the last draw of pick3 are 462 formations.

 

[TheConcept]

Hello, the concept, thank you to see the macro, but there is an error in the construction
Because it is always used the last draw as the base (reference) the reference is the Last draw, and follow the standard to take advantage of repeating the last one digit to the next of pick3, but can not repeat in position. example =
1st 2nd 3rd 4 5 8 = base last draw 8 6 7
1x9x9 equal 81
9x1x9 equal 81
Total 162
Equal 162x3 = 486 729 = error
486 = The concept corret without filters you have to review the macro 729 error, the corret is 486 but has the account of macro remaining 7 didgitos making pairs (doubles)
7 2 = 21 times per each of three digits Total = 63, these being very promising paresis

Hi Jack. Just so I'm sure...

Let's say you had 4 5 8 in the previous draw. You want to generate all arrangements with the permutations
n 4 n (0 4 0, 0 4 1, 0 4 2, 0 4 3, 0 4 5, 0 4 6, ... , 9 4 8, 9 4 9)
n n 4 (0 0 4, 0 1 4, etc, and for the others below)
5 n n
n n 5
8 n n
n 8 n
where n is any digit 0...9 except for the one indicated?

Would you want all of the arrangements output to cells on a spreadsheet, or into a text file?

 

[jack]

Hello, the , I would in a spreadsheet in Excel 2010 or higher on window serveth
I need you believed in field or button where I type the last draw all the time Ai macro makes automatic qye every time I type in a field a new draw for pick3
Reminder = the formations can not repeat positions, as we do not know is three times

 

[TheConcept]

Hi Jack. See how you get on with the code below, which can be pasted into the code section of Excel to run as a macro. Hit [ALT]+[F11], then on the left hand side under 'Project - VBAProject', double-click on 'This Workbook'. You can then paste all of the code below, and can be run from the spreadsheet by pressing [ALT]+[F8] to bring up the macros combo box.

I've limited the digits in the draw to only occur only once, whereas other digits can occur twice.

Any comments/feedback welcome.

-----------

'
' Macro assumes three different digits
' Cells A1, B1 and C1 must contain the three digits from the draw
' The arrangements are output in columns A, B, C, starting from the second row
' - The first digit is shown in column B, then column C
' - The second digit is shown in column A, then column C
' - The third digit is shown in column A, then column B
'
Sub Pick3Perm()
Const CODED_PERMS = "121313122123231231233213"
Const RNG_DRAW_CELL = "A1"
Const RNG_OUTPUT_CELL = "A2"
'
Dim b As Byte
Dim b1 As Byte
Dim b2 As Byte
Dim blnOutput As Boolean
Dim bytOtherDigitPos1 As Byte
Dim bytOtherDigitPos2 As Byte
Dim bytRptDigitIPPos As Byte
Dim bytRptDigitOPPos As Byte
Dim i As Integer
Dim rngDraw As Range
Dim rngOutput As Range
'
' Setup
Set rngDraw = Range(RNG_DRAW_CELL)
Set rngOutput = Range(RNG_OUTPUT_CELL)
'
' Cycle through the different options
i = 0
For b = 0 To 5
bytRptDigitIPPos = Val(Mid$(CODED_PERMS, 4 * b + 1, 1)) - 1
bytRptDigitOPPos = Val(Mid$(CODED_PERMS, 4 * b + 2, 1)) - 1
bytOtherDigitPos1 = Val(Mid$(CODED_PERMS, 4 * b + 3, 1)) - 1
bytOtherDigitPos2 = Val(Mid$(CODED_PERMS, 4 * b + 4, 1)) - 1
'
' Output the arrangements
For b1 = 0 To 9
For b2 = 0 To 9
blnOutput = True
If b1 = rngDraw.Offset(0, 0).Value Then
blnOutput = False
ElseIf b1 = rngDraw.Offset(0, 1).Value Then
blnOutput = False
ElseIf b1 = rngDraw.Offset(0, 2).Value Then
blnOutput = False
ElseIf b2 = rngDraw.Offset(0, 0).Value Then
blnOutput = False
ElseIf b2 = rngDraw.Offset(0, 1).Value Then
blnOutput = False
ElseIf b2 = rngDraw.Offset(0, 2).Value Then
blnOutput = False
End If
'
If blnOutput Then
rngOutput.Offset(i, bytOtherDigitPos1).Value = b1
rngOutput.Offset(i, bytOtherDigitPos2).Value = b2
rngOutput.Offset(i, bytRptDigitOPPos).Value = rngDraw.Offset(0, bytRptDigitIPPos).Value
i = i + 1
End If
Next b2
Next b1
Next b
'
' Tidy up
Set rngDraw = Nothing
Set rngOutput = Nothing
End Sub

 

[jack]

Hello, perfect, congratulations, very good job, thank you
So just give the pattern repeat at least one digit (67%) that the setting is 100%, thecomcept Now, how do we filter it?

 

[jack]

Hello, theconcept
* In my pick3 official draw, gave 974
* Next gave this = 137
Okay hit the condition to repeat a digit (the digit 3) and out of position
Ok, So hit, !!!! Explendido !! so that the 486 formations are many,, we need to filter, now that we have every opportunity
* How can we filter to reduce the 486 formations?

 

[jack]

hello, we have a problem, the digit zero (0) yes ok are 9 digits, perfect macro quantity, but it is the digit zero (0)
* in order to use the filtering ???

 

[TheConcept]

Hello, theconcept
* In my pick3 official draw, gave 974
* Next gave this = 137
Okay hit the condition to repeat a digit (the digit 3) and out of position
Ok, So hit, !!!! Explendido !! so that the 486 formations are many,, we need to filter, now that we have every opportunity
* How can we filter to reduce the 486 formations?

Hi. As things stand, there are 294 combinations, not 486.

In the coding, I have not allowed more than one digit to be present from the previous draw, even if it is in a different position.
So e.g. starting with 137, all combinations with the '1' in the middle (n1n) do not contain 3 or 7 either, so that has filtered out quite a few formations.

In terms of further filtering, one possibility would be not to allow repeating digits (e.g. 010).

I've not done any analysis with Pick3 lotteries before so would need to investigate further as to what other filter conditions could apply. Perhaps something like not allowing all three digits to be even or all three odd?

 

[TheConcept]

hello, we have a problem, the digit zero (0) yes ok are 9 digits, perfect macro quantity, but it is the digit zero (0)
* in order to use the filtering ???

Hi. Not sure what you mean here? Are you saying that only the digits 1...9 can be used?