lottery

Hello, someone knows about positional probabilities of the lottery numbers? That is, more likely to see certain numbers fall into certain positions?

dr san

The link below has two calculators one for digits and the other for numbers.  They both calculate the probabilities

by position.

https://www.box.com/s/y1y81bjobtg15rkkzcay

RL

Rl Hello, thanks for help, and give you data, Rl was seeing it from your lottery 39/5, combine pairs of extremes (lowest to highest number) of the limit from 01 to 11 (number of 1 oumenor) and 30 to 39) number of the 5 th or higher number = we have 210 pairs, giving the suit within the limits of each pair   Example 30 par 11 the trio will be central in 11 to 30 (12-29) are 816 = 100% of the suit in the central trio 02-38 but because the 01 can only be the 1st and 5th position 39da

dr sam

I forgot to mention how to use the calcs, first set the matrix at the very top of the program.

Set the number of numbers in each set then set the total numbers in the game.  L-Click counts

up and R-Click counts down.  Then just click on any digit or number and the results will be displayed.

RL

Hello rl randomlogic thanks good work, it can help me now? I plan to unite three probabalidades to form a suit positional Example of a lotto 49/6 we have 20 positions   Rl the best way to see the positional probability of a suit? the formula should count pascal triangles multiple, low-probability calculation spreadsheet, tab positional probability formula is correct.


review:

td = total scores
a = affixed by dzs
d1, d2, ..., dx = tens,
p1, p2, ..., px = positions,

where d1 is the least ten, d2 is the second lowest decade
where p1 is the lowest position, p2 is the second lowest position,
etc.


Duke = C (td, d2, the-P2) * C (d2-d1-1, P2-P1-1) * C (d1-1, P1-1) Suit = C (td-d3, the-p3) * C (d2-d3-1, P3-P2-1) * C (d2-d1-1, P2-P1-1) * C (d1-1, P1 -1) Block = C (td-d4; to-p4) * C (d4-d3-1, P4-P3-1) * C (d2-d3-1, P3-P2-1) * C (d2-d1-1 , p1-p2-1) * C (d1-1, p1-1)   Rl or the simple multiplication of the three probability calculator?   example =   The suit 12,25,56 = the best position among the 20 possible positions? The way to calculate pascal triangle with multiple (because it has three numbers) or a simple multiplication by 3? what is the best? For the probability of each suit? thank you

Hello, rl rando, it is possible to create this filter, for example umaloteria 49/6 I want to do a trio with a suit or fixed ok 07,23,46 The suit has six digits in their composition that do not repeat each other    Q = the system will use the suit with fixed, and the other three numbers missing   They have repeated the six digits, then suit = 07,23,46, three other 17 33 43 So the three numbers qie complete the meso one digit repeats, And so with peers also codiçao or warranty of accuracy, has hit the suit 07,23,46 Of course it would be nice to see what position should be, so the study of the positions   This pattern did not suit repeated digits is 72% Of course the wheel fic as well (07,23,46) 17,33,43 but you see what best position

7 23 46 from a pool of 49 numbers and you pick 6

Each ball appears a total of 1,712,304 times in the 13,983,816 possible combinations no more no less

Each 3 ball combination appears a total of 15,180 times in the 13,983,816 possible combinations no more no less

The possible positions the 3 numbers can occupy on a single ticket = 20

So we have 3 numbers from 6 choosen which produce the following 20 possible positions

07 23 46 ?? ?? ?? = 1 line contains the 3 numbers in these positions

07 23 ?? 46 ?? ?? = 66 lines contain the 3 numbers in these positions

07 23 ?? ?? 46 ?? = 693 lines contain the 3 numbers in these positions

07 23 ?? ?? ?? 46= 1540 liness contain the 3 numbers in these positions

07 ?? 23 46 ?? ?? = 45 lines contain the 3 numbers in these positions

07 ?? 23 ?? 46 ?? = 990 lines contain the 3 numbers in these positions

07 ?? 23 ?? ?? 46 = 3465 lines contain the 3 numbers in these positions (Best One)

07 ?? ?? 23 46 ?? = 315 lines contain the 3 numbers in these positions

07 ?? ?? 23 ?? 46= 2310 lines contain the 3 numbers in these positions

07 ?? ?? ?? 23 46= 455 lines contain the 3 numbers in these positions

?? 07 23 46 ?? ?? = 18 lines contain the 3 numbers in these positions

?? 07 23 ?? 46 ?? = 396 lines contain the 3 numbers in these positions

?? 07 23 ?? ?? 46= 1386 lines contain the 3 numbers in these positions

?? 07 ?? 23 46 ??= 270 lines contain the 3 numbers in these positions

?? 07 ?? 23 ?? 46= 1980 lines contain the 3 numbers in these positions

?? 07 ?? ?? 23 46= 630 lines contain the 3 numbers in these positions

?? ?? 07 23 46 ??= 45 lines contain the 3 numbers in these positions

?? ?? 07 23 ?? 46= 330 lines contain the 3 numbers in these positions

?? ?? 07 ?? 23 46= 225 lines contain the 3 numbers in these positions

?? ?? ?? 07 23 46 = 20 lines contain the 3 numbers in these positions

To prove the math right

1 + 66 + 693 + 1540 + 45 + 990 + 3465 + 315 + 2310 + 455 + 18  + 396 + 1386 +270 + 1980 + 630 + 45 + 330 + 225 + 20 = 15,180 as expected

I wrote a tool to calculate this for any combination in any draw some time ago.

My above examples prove that any 3 ball combination you choose has exactly the same chance to appear as any other 3 ball combination in the draw.

13,983,816 / 15180 = 921.2

So on average in a 6/49 each 3 ball combination will appear once every 921.2 draws.

So again to prove the math

921.2 x the 20 possible combinations =18424 which happens to be the number of 3 ball combinations in a 6/49 draw

Althought the 20 combinations vary from combination to combination the total will always be 15,180 for any combination choosen.

Hello, developer, OK 04 23 46 suit was good or more likely to position 1,4,5 perfect! Look, the big problem is finding the suit pivot, or the tender reference , Not to walk in circles, ah! Then hitting the suit based, in position, the rest is best for forecasting, thank you

A intresting fact for 6/49

13,983,816 / 15180 = 921.2 the average number of draws to pass before each 3 ball combination will appear

921.2 / 20 = 46.06 the average number of draws to pass before you hit a 3 ball combination.

To cover all 49 numbers each draw costs 9 tickets

01 02 03 04 05 06

07 08 09 10 11 12

13 14 15 16 17 18

19 20 21 22 23 24

25 26 27 28 29 30

31 32 33 34 35 36

37 38 39 40 41 42

43 44 45 46 47 48

?? ?? ?? ?? ?? 49

So 46.06 / 9 = 5.11 the average number of draws to pass before you hit a 3 ball combination betting a total of 46 lines over the period.

The odds are off a little as you have the 5 repeated numbers you would have to use but needless to say any system can't beat the lottery, the math shows us that.

In Belgium we have a full lotto, which is the a qp using all numbers once, twice or more often, to your choice.

Then the math in that draw becomes much more complex and the odds become massive, I personally would not play such a draw unless they offered prizes for match 2

I have read this 5 times and can't get my head around it, it is late so maybe I will get it in the morning.