5!

Reply #18

Fenix — Wed, 20 Oct 2004 13:10:26 GMT

Quote: Originally posted by JKING on October 20, 2004 You would not want to play a combination that is totally made up of numbers from column 1 only,(1,6,11,16,21,26,31,36). Because the percent of all combinations is only .04%.5x8 layout distribution by column gives:COL TOT %1 245 0.042 37240 6.473 258720 44.944 250880 43.575 28672 4.98= 575757 100This is only one reduction scheme, to help narrow down the amount of combinations and leave only the most likely one... [ More ]

Reply #17

JKING — Wed, 20 Oct 2004 12:46:36 GMT

Fenix,What were saying is:123456789101112131415161718192021222324252627282930313233343536373839You would not want to play a combination that is totally made up of numbers from column 1 only,(1,6,11,16,21,26,31,36). Because the percent of all combinations is only .04%.5x8 layout distribution by column gives:COL TOT %1 245 0.042 37240 6.473 258720 44.944 250880 43.575 28672 4.98= 575757 100This is only one reduction scheme, to help narrow down the amount of combin... [ More ]

Reply #16

Fenix — Wed, 20 Oct 2004 05:38:50 GMT

All this info and results open another question- which lines to play. These that show more often but there are too many of them, or wait for the ones that show once in a while but with significantly reduced odds..

Reply #15

JKING — Tue, 19 Oct 2004 23:24:12 GMT

Apagogeas, My solution is generic. I inferred that there would be variances in the percentages earlier. For those who don't go into the detail, the generic is fine. Your examples like mine show that a pick that includes 3 or 4 columns is the best bet. But, you are right when the concept is detailed out to each individual game. The same logic for the generic is applied to the programming loops for all numbers in any pick 5 game to get more precise results. Tanks for your interest.

Reply #14

lottoarchitect — Tue, 19 Oct 2004 13:40:17 GMT

JKING, I don't think these results are correct. The distribution is equivalent to find how many possible 5 number combinations fall in each category. Also, the result depends on the total numbers included in your lottery. For your 5of39 lottery(575757 combinations),4x10 layout distribution by column gives:COL TOT %1 882 0.152 78750 13.683 338625 58.814 157500 27.36= 575757 1005x8 layout distribution by column gives:COL TOT %1 245 0.042 37240 6.473... [ More ]

Reply #13

JKING — Tue, 19 Oct 2004 13:14:47 GMT

The distribution of 4 column in a pick 5 is:COLTOT142180360042401024COL%10.390625217.57813358.59375423.4375100The distribution of 5 columns in a pick 5 is:COLTOT152300315004120051203125COL%10.1629.6348438.453.84100Note-The above would be a 4 by N layout and 5 by N layout. It is applicable to ALL pick 5 games.

Reply #12

lottoarchitect — Wed, 13 Oct 2004 17:58:55 GMT

Quote

Reply #11

johnph77 — Tue, 12 Oct 2004 15:53:52 GMT

Fenix -Your second program with use of the +1 limiter to avoid skips will save a significant amount of execution time - it's what I use, and the larger the lottery being analyzed, the larger percentage of time it will save - talking on the order of 50%-90% here. When I saw the first part of your post that's what immediately came to mind. But you have it right now.gljohn

Reply #10

Fenix — Tue, 12 Oct 2004 15:00:36 GMT

Quote: Originally posted by JKING on October 11, 2004 FOR A=1 TO 35FOR B=2 TO 36FOR C=3 TO 37FOR D=4 TO 38FOR E=5 TO 39IF B<A OR B=A THEN GOTO [SKIP]IF C<B OR C=B THEN GOTO [SKIP]IF D<C OR D=C THEN GOTO [SKIP]IF E<D OR E=D THEN GOTO [SKIP]Then start taking counts of how the numbers are distributed across the columns for each combination[SKIP]NEXT ENEXT DNEXT CNEXT BNEXT Aprint resultsAlso, this might make your program works faster. FOR A=1 TO 35FOR B= A + 1 TO 36FOR C= B + 1 TO 37FOR D= C + 1 T... [ More ]

Reply #9

Fenix — Tue, 12 Oct 2004 03:03:42 GMT

I believe that already there are some software with these kinds of filters - using a certain amount of numbers from the certain columns Havent tried it, but probably apagogeass software might have something like that, how I concluded from some topics.

Reply #8

JKING — Tue, 12 Oct 2004 02:40:49 GMT

Hi FINIX, As you can see it is a work in progress,step 1 as you have seen is to setup a correspondence table123456789101112131415161718192021222324252627282930313233343536373839 So they have corresponding values of:123412341234123412341234123412341234123Step 2 is to get the results of the combinations to see if the data is worth pursuing.4 VALUES FOR PICK 4ONEVAL-4TWOVAL-84THREEVAL-144FOURVAL-24256COL%11.5625232.8125356.2549.375100 As you can see for the pick 4 people combinations that all fall

Reply #7

Fenix — Tue, 12 Oct 2004 01:39:49 GMT

Quote: Originally posted by JKING on October 11, 2004 Any comments?Sorry, cant understand what you are trying to do so far you are making combinations of 4 groups of 4 numbers ( 1024 combinations, or 3125 for 5 groups of 5 numbers) where these 4 numbers can be any from corresponding columns, which at the end will give all possible combinations with no any reduction.

Reply #6

JKING — Mon, 11 Oct 2004 15:39:30 GMT

I'm not really sure what to call it. The last place I saw it was for odd/even distribution:Pick 5 Odd/Even Distribution All Odd100000All Even1111111-Odd 4-Even4-Odd 1-Even1000011111102000102111013001003110114010004101115100005011112-Odd 3 Even3-Odd 2-Even10001111110020010121101030100131011041000140111050011051100160101061010171001070110180110081001191010090101110110001000111 Combinations By PercentageAll Odd13.125All Even13.1251-Odd 4-Even515.6254-Odd 1-Even515.6252-Odd 3 Even... [ More ]

Reply #5

dragon — Mon, 11 Oct 2004 13:46:18 GMT

Ok -- I had completely misunderstood. 5! means 1 * 2 * 3 * 4 * 5 = 120. That's all. You are talking about how many permutations you can have with 5 numbers where duplicates are permissible, right

Reply #4

johnph77 — Mon, 11 Oct 2004 11:15:36 GMT

Quick eye - there has to be a mistake somewhere - column distribution totals have to be 4, 84, 144 and 24.

Reply #3

JKING — Mon, 11 Oct 2004 10:32:57 GMT

I think I've answered my own question. What I looking at is distribution schemes. If you take the fantasy five numbers and arrange them if 4 columns:123456789101112131415161718192021222324252627282930313233343536373839 So they have corresponding values of:123412341234123412341234123412341234123You end up with the thirty nine numbers reduced to four. I did this, because when I'm filling out my lotto ticket I wondered what it should look like. Anyway, you find all the combinations of 1-4:4X4COLUMN... [ More ]

Reply #2

JKING — Mon, 11 Oct 2004 08:35:46 GMT

Yes, I believe 5 factorial is what I'm looking for. It would be like the following: 1,1,1,1,11,1,1,1,21,1,1,1,3etc. to5,5,5,5,5

Reply #1

dragon — Mon, 11 Oct 2004 04:13:23 GMT

Hi JKING,I think the reason nobody is responding is because we don't understand the question. With 5!, do you mean 5 factorial? If yes, what do you mean by all combinations? 5! is a calculation, it does not have combinations. Perhaps, however, 5! is another notation that I don't understand.

5!

JKING — Sun, 10 Oct 2004 14:02:59 GMT

Does anyone have a list of all the combinations for 5