using the 4 patterns even/odd/low/high in pick3.

[jorel]

using the 4 patterns even/odd/low/high in pick3, how many permutations do we have at 100%
even = 1 0,2,4,6,8
odd = 2 1,3,5,7,9
low = 3 0.1,2,3,4
high = 4 5,6,7,8,9
For example: say the last draw was 158. That's (Odd & Low) & (Odd & High) & (Even & High), or 23 & 24 & 14.
so using 4 patterns how many combinations two by two is possible?
23 24 14
?
how many in 100%

[jorel]

https://www.excelforum.com/attachmen...1&d=1638127655

[jorel]

hello please!

[PrizeGotti]

Hi Jorel,

I don't think you question is in the right part of the forum. This is for help with VBA in Excel.

[jorel]

ok,prezi = I'll put it on VBA thank you

[jorel]

up! but it's on the vba forum!!!

[MrShorty]

How much of this is actually Excel programming related, and how much is more generic number theory/permutation/combination?

With each digit fitting into four overlapping categories, I see only 4 permutations -- even&low=13, even&high=14, odd&low=23, odd&high=24. Am I misunderstanding something?

Computing the "score" for each digit looks like it should be as simple as =ISEVEN(digit)*10+ISODD(digit)*20+(digit<=4)*3+(digit>=5)*4. Am I misunderstanding something? If you prefer VBA, VBA has similar functions, so it should be straightforward to translate that worksheet formula into VBA.

[jorel]

so see step by step, we have 4 patterns for each of the three pick3 positions ok
each position is possible two patterns, the question is =
how many permutations are there in the total possible
two by two
xx xx xx?
example
333= equals 2.3 2.3 2.3
other ex=
334= 2.3 2.3 1.4

[MrShorty]

So, this becomes a basic permutation question (here's a math tutorial on permutations and combinations: https://www.mathsisfun.com/combinato...mutations.html ). If you read the tutorial, the applicable section seems to be section 1. permutations with repetition, where it explains that the number of permutations when you have n (4 in your scenario) possibilities and r (3 in your scenario) repetitions, the number of permutations is n^4 (4^3=64). If desired, you can use the PERMUTATIONA() function for this in Excel: https://support.microsoft.com/en-us/...9-2857b25cae4e

[jorel]

hello on the link I posted is not doing the desired permutations ex = 12 12 12
12 12 13 .............. 34 34 34

[MrShorty]

I'm not sure I understand. Are you saying that you want to actually generate the permutations, not just count them? I know in past projects where you have wanted to generate permutations, people have supplied various procedures using nested For..Next loops to generate permutations. Do you have a specific question about how to use For..Next loops (or other programming strategies) for generating permutations?

If it were me and I were having any trouble coming up with a way to programmatically generate the permutations, I would probably simply hand enter the permutations. There are only 64 of them in this case. While it would be a few minutes of tedium, it would probably be easier and faster than trying to come up with a program for the task -- especially since the program probably only needs to run once.

[jorel]

hello, mrshoty, almost there, yes! there are 64 4 to 3 patterns, but look mrshoty, there are three doubles each line and not 3 digits
example = 1,2,3...... 4,4,4 64 patterns
but it's the doubles I need, ex=12,13,14
maybe first you have to do
how many pairs do each of the 3 positions have
12
13
14
23
...
34
you swap you doubles

[MrShorty]

I'm lost. I don't know what you mean by "doubles" or what you want to do with them.

[jorel]

ok each position can have two of the 4 patterns, you have to make the list of the 1st position after the two remaining positions are equal
look at an example pick3
963
9= is odd and high = or 2.4
mrshoty you have to consult the table of the 4 patterns to assemble the list of the three pairs

[MrShorty]

I still don't understand, isn't that basically the same thing I proposed in post #7?

[jorel]

hello mrshort
1ª posiçao da pick3
temos
1,2
1,3
1,4
2,1
2,3
2,4
3,1
3,2
3,4
4,1
4,2
4,3
entao 12 x 12 x12 = 1728

[jorel]

1278 no error= correct 64

[MrShorty]

Now I'm even more confused.