I can never quite understand you. In part, it is due to your difficulty with English. Too bad I don't speak Portuguese (or whatever your native language is). But in part, I suspect you are using terms which have special meaning to people who are more familiar with the so-called "systems" that are discussed in these forums.

In Excel, the number of __combinations__ (no repeats, and ignoring order) of 0 to 9 taken 3 at a time is computed with the formula =COMBIN(10,3). That returns 120.

And the number of __permutations__ (no repeats, but order matters) of 0 to 9 taken 3 at a time is computed with the formula =PERMUT(10,3). That returns 720 because each "combination" of 3 can be ordered 6 different ways.

In order to generate all 120 combinations or 720 permutations, I would write an algorithm using VBA. I can provide implementations. But it is unclear to me that those are the algorithms that you want.

Obviously, neither count is close to the 30 that you expect. I don't know if you are talking about something else, or if your expectations are simply completely wrong.

I don't know what you mean by "bma", "mma" and "maa". I don't know what you mean by your example, to wit: "478 maa".

I don't know what you mean by "help excel in the next". Presumably "help __with__ Excel". But what does "in the next" mean in that context?

Considering your difficulty with communicating clearly in English (and my inability to communicate in Portuguese), it would help if you gave __complete__ examples of your expectations.

For example, instead of simply "478 maa", which is cryptic to me (again, perhaps because I don't understand the terminology that this forum is using), it would help __me__ if you provided the complete 30(?) "permutations" (combinations?) that you expect. Or at least some representative number of __concrete__ "permutations" using the numbers 0 to 9 that make your expectations clear; not some cryptic non-numeric representation.

bgonÃ§alves Brasil Member #92564 June 9, 2010 2196 Posts Offline

Posted: September 11, 2012, 7:05 pm - IP Logged

Quote: Originally posted by mathhead on September 11, 2012

I can never quite understand you. In part, it is due to your difficulty with English. Too bad I don't speak Portuguese (or whatever your native language is). But in part, I suspect you are using terms which have special meaning to people who are more familiar with the so-called "systems" that are discussed in these forums.

In Excel, the number of __combinations__ (no repeats, and ignoring order) of 0 to 9 taken 3 at a time is computed with the formula =COMBIN(10,3). That returns 120.

And the number of __permutations__ (no repeats, but order matters) of 0 to 9 taken 3 at a time is computed with the formula =PERMUT(10,3). That returns 720 because each "combination" of 3 can be ordered 6 different ways.

In order to generate all 120 combinations or 720 permutations, I would write an algorithm using VBA. I can provide implementations. But it is unclear to me that those are the algorithms that you want.

Obviously, neither count is close to the 30 that you expect. I don't know if you are talking about something else, or if your expectations are simply completely wrong.

I don't know what you mean by "bma", "mma" and "maa". I don't know what you mean by your example, to wit: "478 maa".

I don't know what you mean by "help excel in the next". Presumably "help __with__ Excel". But what does "in the next" mean in that context?

Considering your difficulty with communicating clearly in English (and my inability to communicate in Portuguese), it would help if you gave __complete__ examples of your expectations.

For example, instead of simply "478 maa", which is cryptic to me (again, perhaps because I don't understand the terminology that this forum is using), it would help __me__ if you provided the complete 30(?) "permutations" (combinations?) that you expect. Or at least some representative number of __concrete__ "permutations" using the numbers 0 to 9 that make your expectations clear; not some cryptic non-numeric representation.

hello mathheard It is for apick3Lotto that draws one ball from numbers 0-9, replaces it, draws another, replaces it and draws one final ball. The code splits the numbers 0-9 for each group of A,B & C into three groups for low, medium and high numbers.

bgonÃ§alves Brasil Member #92564 June 9, 2010 2196 Posts Offline

Posted: September 11, 2012, 7:10 pm - IP Logged

Quote: Originally posted by dr san on September 11, 2012

hello mathheard It is for apick3Lotto that draws one ball from numbers 0-9, replaces it, draws another, replaces it and draws one final ball. The code splits the numbers 0-9 for each group of A,B & C into three groups for low, medium and high numbers.

--A--- --B--- --C--- ---D---

1 H L M Total

2 3/10 3/10 4/10 1

3

4 Ball1 Ball2 Ball3 Prob

5 L L L 2.7%

6 L L M 3.6%

7 L L H 2.7%

8 L M L 3.6%

9 L M M 4.8%

10 L M H 3.6%

11 L H L 2.7%

12 L H M 3.6%

13 L H H 2.7%

14 M L L 3.6%

15 M L M 4.8%

16 M L H 3.6%

17 M M L 4.8%

18 M M M 6.4%

19 M M H 4.8%

20 M H L 3.6%

21 M H M 4.8%

22 M H H 3.6%

23 H L L 2.7%

24 H L M 3.6%

25 H L H 2.7%

26 H M L 3.6%

27 H M M 4.8%

28 H M H 3.6%

29 H H L 2.7%

30 H H M 3.6%

31 H H H 2.7%

32

33 Total 100.0%

Hello,mathheard,changedcoursein the previous postthe lettersrepresenting thethreegroups Must havea system thatlowof 120to 30to 27 (reduction)isthesepermutationswithoutrepetitionsI need,iepatternsaremore orless likely tocome outin the draws, so we haveset thestandardof3 letters,the digitsareor notin, Objectiveandsee patternsreduced tothreegroupseachnumber,

bgonÃ§alves Brasil Member #92564 June 9, 2010 2196 Posts Offline

Posted: September 11, 2012, 7:26 pm - IP Logged

Quote: Originally posted by dr san on September 11, 2012

Hello,mathheard,changedcoursein the previous postthe lettersrepresenting thethreegroups Must havea system thatlowof 120to 30to 27 (reduction)isthesepermutationswithoutrepetitionsI need,iepatternsaremore orless likely tocome outin the draws, so we haveset thestandardof3 letters,the digitsareor notin, Objectiveandsee patternsreduced tothreegroupseachnumber,

Helloexampleflorida951 =turning intoletters(standards)=MLH Converting951 inMLH(groups)last post,the advantageand can seepatterns inverticalpositionsinthe list ofsweepstakes,this listof lettersyou can seethehot,mediumand coldandlate(groups of letters)then justconvertthedigits of eachgroup

bgonÃ§alves Brasil Member #92564 June 9, 2010 2196 Posts Offline

Posted: September 11, 2012, 7:44 pm - IP Logged

Quote: Originally posted by dr san on September 11, 2012

Helloexampleflorida951 =turning intoletters(standards)=MLH Converting951 inMLH(groups)last post,the advantageand can seepatterns inverticalpositionsinthe list ofsweepstakes,this listof lettersyou can seethehot,mediumand coldandlate(groups of letters)then justconvertthedigits of eachgroup

Hello,mathheardthesepermutations27-30,covering100%in any lottery? Sothe advantage is thatyou can seethe defaultfontforposition,eg Ifthe 1stposition ofthe predictionforpick 3L(namingthe lastpost) Thenwe havethe 1stposition3,4,5,because thegroups weredividedsoH(0,1,2)L(3,4,5) M(6,7,8,9)

United States Member #130795 July 25, 2012 80 Posts Offline

Posted: September 11, 2012, 11:50 pm - IP Logged

Quote: Originally posted by dr san on September 11, 2012

hello mathheard It is for apick3Lotto that draws one ball from numbers 0-9, replaces it, draws another, replaces it and draws one final ball. The code splits the numbers 0-9 for each group of A,B & C into three groups for low, medium and high numbers.

--A--- --B--- --C--- ---D---

1 H L M Total

2 3/10 3/10 4/10 1

3

4 Ball1 Ball2 Ball3 Prob

5 L L L 2.7%

6 L L M 3.6%

7 L L H 2.7%

8 L M L 3.6%

9 L M M 4.8%

10 L M H 3.6%

11 L H L 2.7%

12 L H M 3.6%

13 L H H 2.7%

14 M L L 3.6%

15 M L M 4.8%

16 M L H 3.6%

17 M M L 4.8%

18 M M M 6.4%

19 M M H 4.8%

20 M H L 3.6%

21 M H M 4.8%

22 M H H 3.6%

23 H L L 2.7%

24 H L M 3.6%

25 H L H 2.7%

26 H M L 3.6%

27 H M M 4.8%

28 H M H 3.6%

29 H H L 2.7%

30 H H M 3.6%

31 H H H 2.7%

32

33 Total 100.0%

With this and your follow-up postings, you did a job of clarifying what you are doing. Unfortunately, there is still one very important detail that is unclear to me.

Your original question was how to do some things in Excel. So at this point, I would like to refer you to an uploaded Excel file that might clarify my comments below. But I am still just a "new member", and the draconian rules of this forum do not permit me to post a URL or even an email address to contact me directly.

I will do my best to represent the Excel solutions here. Alternatively, Google "Microsoft Answers" and post your Excel questions there. We should not discuss lottery strategies there. But at least I would be able to point you to Excel files and demonstrate Excel usage graphically.

Summarizing your several postings, "dr san" wrote:

``The code splits the numbers 0-9 [...] into three groups for low, medium and high numbers.

4 Ball1 Ball2 Ball3 Prob 5 L L L 2.7% 6 L L M 3.6% 7 L L H 2.7% [....] 29 H H L 2.7% 30 H H M 3.6% 31 H H H 2.7% [....] Then we have the 1st position 3,4,5, because the groups were divided so H (0,1,2) L (3,4,5) M (6,7,8,9) [....] these permutations 27-30, covering 100% in any lottery?``

It is unclear to me how you determined those probabilities. It is unclear whether you simply counted wrong; or if understanding your computations might give us some insight into the answer to the point that is still unclear to me (below).

For the L L L case, is there only one combination (3 4 5), ignoring order; or are there six combinations, namely all permutations of 3 4 5?

To be clear, there are only 27 permutations (with replacement) of the LMH categories. Why do you keep saying 30?! (Rhetorical question.)

And those 27 categories would cover all 720 Pick3 permutations (without replacement) only if L L L covers all permutations of 3 4 5, and likewise for the other categories.

Suppose "ball1", "ball2" and "ball3" are in columns A, B and C starting in row 1.

If L L L represents all six permutations of 3 4 5, the number of Pick3 combinations for each category can be computed by the following formula in D1 and copied down through D27:

In that case, SUM(D1:D27) is indeed 720. But the number of Pick3 combinations represented by L L L is 6; so the probability is 6/720, which is about 0.83%.

On the other hand, if L L L represents just one combination (3 4 5), the number Pick3 combinations for each category is computed by the following formula in F1 and copied down through F27:

In that case, SUM(F1:F27) is 456(!). And the probability of L L L is 1/456, which is about 0.22%.

Note that SUM(F1:F27) is 456, not 120. This is because if we only consider __combinations__ in each category, your methodology is a __mix__ of combinations and permutations. For example, L L M and M L L are simply permutations of each other.

For that reason, the latter case does not make sense to do, IMHO.

The following macro generates a table similar to yours with all the permutations in each category. The total represents all 720 Pick3 permutations without replacement (i.e. no repeats).

-----

Option Explicit

Sub genPermut() Dim v(1 To 27, 1 To 3 + 2 + 36) As Variant Dim b1 As Variant, b2 As Variant, b3 As Variant Dim n1 As Variant, n2 As Variant, n3 As Variant Dim n1Array As Variant, n2Array As Variant, n3Array As Variant Dim r As Long, c As Long, totl As Long, i As Long ActiveSheet.Cells.Clear r = 0: totl = 0 For Each b1 In Array("L", "M", "H") n1Array = IIf(b1 = "H", Array(0, 1, 2), _ IIf(b1 = "L", Array(3, 4, 5), Array(6, 7, 8, 9))) For Each b2 In Array("L", "M", "H") n2Array = IIf(b2 = "H", Array(0, 1, 2), _ IIf(b2 = "L", Array(3, 4, 5), Array(6, 7, 8, 9))) For Each b3 In Array("L", "M", "H") n3Array = IIf(b3 = "H", Array(0, 1, 2), _ IIf(b3 = "L", Array(3, 4, 5), Array(6, 7, 8, 9))) r = r + 1 v(r, 1) = b1: v(r, 2) = b2: v(r, 3) = b3 ' skip 2 columns for count in category and for probability c = 5 For Each n1 In n1Array For Each n2 In n2Array For Each n3 In n3Array If n1 <> n2 And n1 <> n3 And n2 <> n3 Then c = c + 1 v(r, c) = --(n1 & n2 & n3) End If Next n3, n2, n1 v(r, 4) = c - 5 ' count in category totl = totl + v(r, 4) Next b3, b2, b1 For i = 1 To r v(i, 5) = v(i, 4) / totl ' probability Next Range("a1").Resize(r, UBound(v, 2)).Value = v Range("e1").Resize(r).NumberFormat = "0.00%" Range("f1").Resize(r, 36).NumberFormat = "000" End Sub

bgonÃ§alves Brasil Member #92564 June 9, 2010 2196 Posts Offline

Posted: September 12, 2012, 12:12 am - IP Logged

Hellomath,fanstastico,very goodjob,goodcalculationof the probabilitiesof the 27groupsItiswheremostcombinations,thesamplegroup(6,7,8,9)hascombinedwith anotherprobabidadesmore,because of thegreaternumber ofcombinations, very goodyourmacro,Nowyou can doso withinthe 27groupssee10.20inthe 30pairings(smallintervalssweepstakesworks best)thenamong the 27groupsthegroupsaveragehotand cold,andlate,eachoftheverticalposition,since the goalsee patternsisthatifthe 1000wasinpick3groups27,10 or20forexamplein the verticalpositionof the 1stViewstatistics on delayson 10raffles, these patternswill work wellforverticalpositionof each

United States Member #130795 July 25, 2012 80 Posts Offline

Posted: September 12, 2012, 3:20 am - IP Logged

Quote: Originally posted by dr san on September 12, 2012

Hellomath,fanstastico,very goodjob,goodcalculationof the probabilitiesof the 27groupsItiswheremostcombinations,thesamplegroup(6,7,8,9)hascombinedwith anotherprobabidadesmore,because of thegreaternumber ofcombinations, very goodyourmacro,Nowyou can doso withinthe 27groupssee10.20inthe 30pairings(smallintervalssweepstakesworks best)thenamong the 27groupsthegroupsaveragehotand cold,andlate,eachoftheverticalposition,since the goalsee patternsisthatifthe 1000wasinpick3groups27,10 or20forexamplein the verticalpositionof the 1stViewstatistics on delayson 10raffles, these patternswill work wellforverticalpositionof each

I would like to help, but I really cannot understand you. What does "see 10.20 in the 30 pairings" mean? What do you mean by "vertical position"? And what "30 pairings" are you referring to?

Remember: concrete examples breach the language barrier.

For Pick3 of the numbers 0 to 9, there are 45 pairings altogether -- COMBIN(10,2).

But I do notice that among the 27 LHM groups, 12 have more permutations than the others -- specifically 36 permutations. Those 432 triads (12*36) comprise 60% of the total 720.

So perhaps your strategy is to focus on the 60% most common triads.

And I did discover that among the 432 triads, only 39 pairings are represented. And there are 30 pairings that appear most often -- 36 times instead 24 times for the others.

But I don't know if that is the information you are looking for, and how you would know that.

And even if it is, you started by asking how to write something in "Excel" -- VBA -- to discover them. Right? Or do you just what the answer; and is that the answer you were looking for?

New Mexico United States Member #86099 January 29, 2010 11341 Posts Offline

Posted: September 12, 2012, 11:17 am - IP Logged

DR San may be refering to vertical positions as :

4 5 6

2 3 4

2 2 2 <<<<<< This would be the vertical difference or absolute value between the last draw example 456 and the previous 234 draw. This is the way I use the vertical difference in some of my systems . As far as the groups 20,30 etc., he may be grouping them in groups of ten. I'm a little lost in translation myself. The end result or usage is a mystery to me.

bgonÃ§alves Brasil Member #92564 June 9, 2010 2196 Posts Offline

Posted: September 12, 2012, 12:07 pm - IP Logged

Quote: Originally posted by lakerben on September 12, 2012

DR San may be refering to vertical positions as :

4 5 6

2 3 4

2 2 2 <<<<<< This would be the vertical difference or absolute value between the last draw example 456 and the previous 234 draw. This is the way I use the vertical difference in some of my systems . As far as the groups 20,30 etc., he may be grouping them in groups of ten. I'm a little lost in translation myself. The end result or usage is a mystery to me.

Hellolakerandmathhiseasyvertical,eg= 465 234 222 1st position422,islookingverticallyeach position,as totherangeof 10lotterydraw, it seems besttoanalyzeandgroupsinlate1000 or more Example10inthissweepstakeslatelastgroupMLHunderstand,thenyou can Cruisingwithgraphicsumsthathavethis forum withthe chartof the 27groups

New Mexico United States Member #86099 January 29, 2010 11341 Posts Offline

Posted: September 12, 2012, 12:27 pm - IP Logged

Quote: Originally posted by dr san on September 12, 2012

Hellolakerandmathhiseasyvertical,eg= 465 234 222 1st position422,islookingverticallyeach position,as totherangeof 10lotterydraw, it seems besttoanalyzeandgroupsinlate1000 or more Example10inthissweepstakeslatelastgroupMLHunderstand,thenyou can Cruisingwithgraphicsumsthathavethis forum withthe chartof the 27groups

In Nm the p3 has repeat pairs that hits consistantly over the last 10 draws. So, 1000 draws and graphs would be not usefull.

bgonÃ§alves Brasil Member #92564 June 9, 2010 2196 Posts Offline

Posted: September 12, 2012, 12:40 pm - IP Logged

Quote: Originally posted by lakerben on September 12, 2012

In Nm the p3 has repeat pairs that hits consistantly over the last 10 draws. So, 1000 draws and graphs would be not usefull.

Hello,lakerthis systemis similar tothe graphof the sums,you can take thedigitor pairrepeatsand look forthisgroupwherethe digitor pairrepeats, understand!!