Pair numbers in historic drawing
- Thread starter febausa
- Start date
Frank
Member
Well I've looked at your example and I think I know whats happening from cell G3 to BM 13 , you signify where a number appears in each result. What I do not understand is what any of this has to do with Column J ? Column J is the fifth number in each result (Ball 5) so what did you mean by:-
determine repetition numbers pair >=2 (column J) ?
If you enter the formula =IF(COUNTIF($B3:$F3,G$2),G$2,"") in cell G3 and copy this formula along to column BM and also copy the formulas down to row 13 so as to fill in the are G3:BM13 with copies of the formula, I think that does what you illustrated.
Also if you enter =IF(COUNTIF(H3:H13,">1")>1,H2,"") in cell H16 and copy that along the row to cell BM16, then that flags up the identity of a repeated number over the 11 draws you show, which is what your example appeared to be doing ( a few differences here and there).
Just what you are doing with the dates, I'm not quite sure since there seems to be no consistent pattern in your illustration.
For example:- in column H the number 2 appears three times , two of them on 19/3/2014 and once on 15/03/2014. you have entered both these dates under date repetition.
In column O the number 9 appears twice, once on each date but you only enter 19/03/2014.
In column BE number 51 appears twice, once on each date, but you only enter 19/03/2014.
So what is your criterion for entering a repetition date ?
determine repetition numbers pair >=2 (column J) ?
If you enter the formula =IF(COUNTIF($B3:$F3,G$2),G$2,"") in cell G3 and copy this formula along to column BM and also copy the formulas down to row 13 so as to fill in the are G3:BM13 with copies of the formula, I think that does what you illustrated.
Also if you enter =IF(COUNTIF(H3:H13,">1")>1,H2,"") in cell H16 and copy that along the row to cell BM16, then that flags up the identity of a repeated number over the 11 draws you show, which is what your example appeared to be doing ( a few differences here and there).
Just what you are doing with the dates, I'm not quite sure since there seems to be no consistent pattern in your illustration.
For example:- in column H the number 2 appears three times , two of them on 19/3/2014 and once on 15/03/2014. you have entered both these dates under date repetition.
In column O the number 9 appears twice, once on each date but you only enter 19/03/2014.
In column BE number 51 appears twice, once on each date, but you only enter 19/03/2014.
So what is your criterion for entering a repetition date ?
Hello Frank:Frank said:
Excuse the information submitted was wrong.
I'm sending the correct information https://www.dropbox.com/s/z2m6au03yt69xxx/repetition numbers pair-2.xlsx?dl=0
Again I apologize
Frank
Member
Febusa this is an entirely different question! One that cannot be attempted because you are not revealing the full extent of what you want. In your example you have highlighted certain pairs in results and indicated that those specific highlighted pairs want counting over the range of the results you show. If you only EVER want pair 20 and 41 counting thats fine, or pair 13 and 20 from these specific results, but I suspect this is NOT what you really want!
For there to be any point to this, you would need a way of entering elsewhere which pair(s) you want counting and it looks like you want more than one set of pairs counting which would change whenever you changed your mind.
This is not a job for formulas, its a job for macros, and I'm not going to get involved any further with this, there are already in existence macros that count ALL pairs, and I can't understand why counting them all, and then looking at specific ones afterwards is not the way forward. Its what everyone else does.
You don't say what format of lottery you are working on (looks like 6-59) there may be a macro somewhere that counts pairs for that format. Im pretty sure I did a 6/60 macro, probably a whole spreadsheet recently that would do for 6/59 counting. Ah here it is, 72 people have downloaded it ...http://www.mediafire.com/view/h9bbuhgbo37h608/GillesD_Ntups_six_ball_60s.xlsm you only need run the pairs macro from the toolbar if you are only counting pairs.
I do have a specific pairfinder I wrote 10 years ago that counts only the pair you type in, its for 6/49 might be able to modify it ....
For there to be any point to this, you would need a way of entering elsewhere which pair(s) you want counting and it looks like you want more than one set of pairs counting which would change whenever you changed your mind.
This is not a job for formulas, its a job for macros, and I'm not going to get involved any further with this, there are already in existence macros that count ALL pairs, and I can't understand why counting them all, and then looking at specific ones afterwards is not the way forward. Its what everyone else does.
You don't say what format of lottery you are working on (looks like 6-59) there may be a macro somewhere that counts pairs for that format. Im pretty sure I did a 6/60 macro, probably a whole spreadsheet recently that would do for 6/59 counting. Ah here it is, 72 people have downloaded it ...http://www.mediafire.com/view/h9bbuhgbo37h608/GillesD_Ntups_six_ball_60s.xlsm you only need run the pairs macro from the toolbar if you are only counting pairs.
I do have a specific pairfinder I wrote 10 years ago that counts only the pair you type in, its for 6/49 might be able to modify it ....
febausa, would not the average range
* output of each pair, if you want the total number of pairs (double the numbers 01-59 = 1711 at 100% of the pairs hit
* Now if you divide into 4 groups 3 groups of 15 and a group
* 14 lottery 01-59, you are in 406 pair or pairs of numbers, hit at least a pair of 100% entaofoi taking 1300 double pairs and even the 100% because the standard of minimum condition of a quadrant have two numbers,
* you can ride in 406 pairs of digits without repetition
* 12.46 example we have 4 digits = 1,2,3,6
* output of each pair, if you want the total number of pairs (double the numbers 01-59 = 1711 at 100% of the pairs hit
* Now if you divide into 4 groups 3 groups of 15 and a group
* 14 lottery 01-59, you are in 406 pair or pairs of numbers, hit at least a pair of 100% entaofoi taking 1300 double pairs and even the 100% because the standard of minimum condition of a quadrant have two numbers,
* you can ride in 406 pairs of digits without repetition
* 12.46 example we have 4 digits = 1,2,3,6