5/39: Consecutive Drawing Simulations

Reply #16

db101 — Thu, 07 May 2020 21:31:04 GMT

Thanks. It gives me something new to consider when I start playing the lottery again. I can see there is something to the rule of the third.

Reply #15

dr san — Tue, 05 May 2020 07:31:58 GMT

hello cottoney, can you make separate pairs or doubles numbers for the 25/15 lottery? please

Reply #14

dr san — Mon, 04 May 2020 03:02:21 GMT

hello cottoneyedjoe, ok corret thank you

Reply #13

cottoneyedjoe — Sun, 03 May 2020 22:13:11 GMT

Here you go

number of distinct numbers in 5 consecutive draws of 7/31

1 million simulations

15 distinct numbers: 0.0013%

16 distinct numbers: 0.0181%

17 distinct numbers: 0.1370%

18 distinct numbers: 0.8124%

19 distinct numbers: 3.1465%

20 distinct numbers: 8.7027%

21 distinct numbers: 16.8465%

22 distinct numbers: 23.0876%

23 distinct numbers: 22.2622%

24 distinct numbers: 15.0714%

25 distinct numbers: 7.1303%

26 distinct numbers: 2.2490%

27 distinct... [ More ]

Reply #12

dr san — Sun, 03 May 2020 20:53:06 GMT

hello cottoney, please

lottterys

7/31=

5/80

Reply #11

dr san — Sun, 03 May 2020 11:07:50 GMT

hello , cottoney thank you

Reply #10

cottoneyedjoe — Sat, 02 May 2020 22:01:02 GMT

I'm also looked at the number of distinct pairs that come out in consecutive drawings of 5/39. I decided to simulate sets of 20 consecutive draws. Nothing special about the number 20, just that it's manageable. Here's what I got for 250,000 simulations

number of distinct pairs in 20 consecutive draws of 5/39

250,000 simulations

164 distinct pairs: 0.0004%

165 distinct pairs: 0.0%

166 distinct pairs: 0.0%

167 distinct pairs: 0.0004%

168 distinct pairs: 0.0008%

169... [ More ]

Reply #9

cottoneyedjoe — Sat, 02 May 2020 17:08:56 GMT

The rule is true and can be verified mathematically. When you are drawing just one item with replacement from a set of N, the number of distinct items follows a known distribution:

If you replace A with N in the formula above, and compute the average for k = 1,...N, the average number of unique items is approximately (1 - 1/e)*N, where e is the mathematical constant 2.718281828... The larger the value of N the closer the approximation. The reciprocal, 1/e, is about 0.36788 or a... [ More ]

Reply #8

cottoneyedjoe — Sat, 02 May 2020 16:59:16 GMT

For db101 and dr san:

number of distinct numbers in 9 consecutive draws of 6/53

1 million simulations

23 distinct numbers: 0.0001%

24 distinct numbers: 0.0001%

25 distinct numbers: 0.0007%

26 distinct numbers: 0.0043%

27 distinct numbers: 0.0276%

28 distinct numbers: 0.1233%

29 distinct numbers: 0.4591%

30 distinct numbers: 1.3980%

31 distinct numbers: 3.5024%

32 distinct numbers: 7.1014%

33 distinct numbers: 11.8502%

34 distinct numbers: 16.0076%

35

Reply #7

dr san — Fri, 01 May 2020 11:25:02 GMT

hello cottoney, my lottery is from 01 to 25

where 15 numbers are drawn (lotofacil brasil)

25/15. can you run this lottery? please

Reply #6

phileight — Fri, 01 May 2020 10:26:03 GMT

i don't think i have ever seen this. I'm getting old and maybe just forgot. a quick look at pick3 shows me roughly 30% of digits, in various positions, are out 10 or more draws, i do keep mid and eve drawings separate.

something to look into a bit further.

Reply #5

phileight — Fri, 01 May 2020 10:13:28 GMT

Do you use this filter to play box or straight in p3/p4?

I play for straight almost exclusively, however i use this as a filter for what numbers may come up in the next game. I use other indicators to pick the order

Reply #4

KyngeRycharde — Fri, 01 May 2020 04:45:56 GMT

Roulette players put a lot of stock in the so-called Law of the Third , which, if I understand it correctly, states that given a set of n numbers, drawn n times with replacement, approximately one-third of the numbers will not show. Applied to the pick-3, that means that in 10 drawings, 3 or 4 digits will not show in each position. In a 5/45 lottery, 45 divided by 5 = 9 games. In those 9 games, approximately 15 of the 45 numbers will not show.

I have seen it happen in roulette (one of my favo... [ More ]

Reply #3

db101 — Fri, 01 May 2020 03:57:39 GMT

Can you do it for 6-ball draw games? Florida Lotto is 6/53 and NY Lotto is 6/59. I'd like to know how many different numbers should come out for 9 consecutive drawings in FL Lotto, and 10 consec drawings in NY. Thanks.

Reply #2

cottoneyedjoe — Thu, 30 Apr 2020 16:31:40 GMT

Pretty cool, Phileight. You're right about 6 and 7 distinct numbers being the rule for three draws of Pick 3. And it looks like 7 to 8 distinct numbers is the rule for three draws of Pick 4:

number of distinct numbers in 3 consecutive draws of pick 3

1 million simulations

2 distinct numbers: 0.0019%

3 distinct numbers: 0.2163%

4 distinct numbers: 3.8962%

5 distinct numbers: 20.9695%

6 distinct numbers: 40.1181%

7 distinct numbers: 27.9183%

8 distinct numbers: 6.5140... [ More ]

Reply #1

phileight — Thu, 30 Apr 2020 08:40:27 GMT

I use this basic concept in my pick3/4 selection. it is a very effective filter. the way i use it is count the number of different digits in the past three drawings. this will typically be 6 or 7 but can be higher or lower. now count the number of different digits for the past 2 games.

when the combination being evaluated is combined with the previous 2 drawings the resulting digit count has to be the same, plus or minus 1, as the earlier 3 day count, if not is is filtered out.

if your fi... [ More ]

5/39: Consecutive Drawing Simulations

cottoneyedjoe — Thu, 30 Apr 2020 04:21:47 GMT

I was looking at the distribution of the number of distinct numbers that come out in 8 consecutive drawings of 5/39 (CA Fantasy Five). This is what I got for 1 million simulations.

number of distinct numbers in 8 consecutive draws of 5/39

1 million simulations

17 distinct numbers: 0.0002%

18 distinct numbers: 0.0022%

19 distinct numbers: 0.0237%

20 distinct numbers: 0.1564%

21 distinct numbers: 0.7009%

22 distinct numbers: 2.3915%

23 distinct numbers: 6.1395%

24 dis... [ More ]