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5/39: Consecutive Drawing Simulations

cottoneyedjoe's avatar - cuonvFT

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 distinct numbers: 12.1178%
25 distinct numbers: 18.1189%
26 distinct numbers: 20.9162%
27 distinct numbers: 18.2370%
28 distinct numbers: 12.1156%
29 distinct numbers:  6.0332%
30 distinct numbers:  2.2738%
31 distinct numbers:  0.6316%
32 distinct numbers:  0.1232%
33 distinct numbers:  0.0167%
34 distinct numbers:  0.0016%

The random simulations never resulted in a set of 8 consecutive drawings with fewer than 17 nor more than 34 distinct numbers. The distribution has a bell curve shape, but the underlying true distribution isn't Gaussian. More than half the time you should see 25-27 distinct numbers over 8 consecutive drawings. I checked it against the actual draw data and it hews pretty close. Not sure how you can apply it to picking winning numbers, but maybe someone will find it of use.

I also ran some sims for other k/N games:

number of distinct numbers in 7 consecutive draws of 5/35
1 million simulations
15 distinct numbers:  0.0013%
16 distinct numbers:  0.0083%
17 distinct numbers:  0.0678%
18 distinct numbers:  0.4022%
19 distinct numbers:  1.6290%
20 distinct numbers:  4.9901%
21 distinct numbers: 11.1579%
22 distinct numbers: 18.3341%
23 distinct numbers: 22.1149%
24 distinct numbers: 19.6929%
25 distinct numbers: 12.8480%
26 distinct numbers:  6.1063%
27 distinct numbers:  2.0623%
28 distinct numbers:  0.4931%
29 distinct numbers:  0.0822%
30 distinct numbers:  0.0089%
31 distinct numbers:  0.0007%

number of distinct numbers in 7 consecutive draws of 5/36
1 million simulations
14 distinct numbers:  0.0001%
15 distinct numbers:  0.0001%
16 distinct numbers:  0.0059%
17 distinct numbers:  0.0463%
18 distinct numbers:  0.2715%
19 distinct numbers:  1.2275%
20 distinct numbers:  3.9813%
21 distinct numbers:  9.3899%
22 distinct numbers: 16.4923%
23 distinct numbers: 21.5070%
24 distinct numbers: 20.7106%
25 distinct numbers: 14.8097%
26 distinct numbers:  7.7064%
27 distinct numbers:  2.9005%
28 distinct numbers:  0.7789%
29 distinct numbers:  0.1519%
30 distinct numbers:  0.0193%
31 distinct numbers:  0.0006%
32 distinct numbers:  0.0002%


number of distinct numbers in 8 consecutive draws of 5/36
1 million simulations
16 distinct numbers:  0.0002%
17 distinct numbers:  0.0017%
18 distinct numbers:  0.0140%
19 distinct numbers:  0.0917%
20 distinct numbers:  0.4958%
21 distinct numbers:  1.8420%
22 distinct numbers:  5.2649%
23 distinct numbers: 11.1472%
24 distinct numbers: 17.8356%
25 distinct numbers: 21.4305%
26 distinct numbers: 19.3556%
27 distinct numbers: 12.9918%
28 distinct numbers:  6.4208%
29 distinct numbers:  2.3664%
30 distinct numbers:  0.6154%
31 distinct numbers:  0.1118%
32 distinct numbers:  0.0129%
33 distinct numbers:  0.0017%


number of distinct numbers in 10 consecutive draws of 5/47 (CA SLP white balls)
1 million simulations
22 distinct numbers:  0.0006%
23 distinct numbers:  0.0049%
24 distinct numbers:  0.0220%
25 distinct numbers:  0.1239%
26 distinct numbers:  0.4837%
27 distinct numbers:  1.5579%
28 distinct numbers:  3.9742%
29 distinct numbers:  8.1618%
30 distinct numbers: 13.3472%
31 distinct numbers: 17.6032%
32 distinct numbers: 18.7832%
33 distinct numbers: 15.9078%
34 distinct numbers: 10.7101%
35 distinct numbers:  5.7900%
36 distinct numbers:  2.4449%
37 distinct numbers:  0.8233%
38 distinct numbers:  0.2142%
39 distinct numbers:  0.0397%
40 distinct numbers:  0.0065%
41 distinct numbers:  0.0008%
42 distinct numbers:  0.0001%

I can do other games by request.

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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 first count, ( last three drawings), was for ex 6 and your second count, (last two drawings), was 4 then, to pass this filter, any combination you are considering must contain 1,2, or 3 digits that are not contained in you second count.

this does not eliminate a lot of combinations but you can have a high level of confidence that you have not eliminated the next winner.

one caveat if the indicated digit count is less than 5 then ignore it and use 6  or nothing.

cottoneyedjoe's avatar - cuonvFT
In response to phileight

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%
9 distinct numbers: 0.3657%

number of distinct numbers in 3 consecutive draws of pick 4
1 million simulations
3 distinct numbers: 0.0057%
4 distinct numbers: 0.2932%
5 distinct numbers: 4.1653%
6 distinct numbers: 19.9344%
7 distinct numbers: 38.0076%
8 distinct numbers: 28.8614%
9 distinct numbers: 8.1036%
10 distinct numbers: 0.6288%

Do you use this filter to play box or straight in p3/p4? I was thinking of using it as a filter in the same way for 5-distinct-numbers type games, working it into a random ticket generator and comparing the results to an unfiltered random ticket generator.

db101's avatar - RB55Ms1

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.

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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 favorite casino games) many times where in 38 spins, 12 or 13 numbers are missing.

I haven't checked it out myself, but someone might check the last 1000 draws in the pick 3 for their state, and see if there are approximately 333  three digit numbers that are missing.

I don't know if this information cold possibly help anyone, and it could fall under the category of "voodoo math" for all I know.

Stay safe, everyone.

Kynge

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In response to cottoneyedjoe

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

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In response to KyngeRycharde

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.

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hello cottoney, my lottery is from 01 to 25
where 15 numbers are drawn (lotofacil brasil)
25/15. can you run this lottery? please

cottoneyedjoe's avatar - cuonvFT
In response to db101

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 distinct numbers: 17.8863%
36 distinct numbers: 16.3473%
37 distinct numbers: 12.2242%
38 distinct numbers:  7.3941%
39 distinct numbers:  3.6348%
40 distinct numbers:  1.4381%
41 distinct numbers:  0.4583%
42 distinct numbers:  0.1154%
43 distinct numbers:  0.0225%
44 distinct numbers:  0.0038%
45 distinct numbers:  0.0003%
46 distinct numbers:  0.0001%

 

number of distinct numbers in 10 consecutive draws of 6/59
1 million simulations
27 distinct numbers:  0.0001%
28 distinct numbers:  0.0003%
29 distinct numbers:  0.0028%
30 distinct numbers:  0.0151%
31 distinct numbers:  0.0713%
32 distinct numbers:  0.2594%
33 distinct numbers:  0.8017%
34 distinct numbers:  2.0996%
35 distinct numbers:  4.5406%
36 distinct numbers:  8.2407%
37 distinct numbers: 12.5586%
38 distinct numbers: 15.9349%
39 distinct numbers: 16.9399%
40 distinct numbers: 14.9425%
41 distinct numbers: 11.1066%
42 distinct numbers:  6.8198%
43 distinct numbers:  3.4715%
44 distinct numbers:  1.4989%
45 distinct numbers:  0.5176%
46 distinct numbers:  0.1439%
47 distinct numbers:  0.0279%
48 distinct numbers:  0.0058%
49 distinct numbers:  0.0005%

 

number of distinct numbers in 2 consecutive draws of 15/25
1 million simulations
16 distinct numbers:  0.0049%
17 distinct numbers:  0.1420%
18 distinct numbers:  1.6621%
19 distinct numbers:  8.7723%
20 distinct numbers: 23.1675%
21 distinct numbers: 32.1073%
22 distinct numbers: 23.6671%
23 distinct numbers:  8.8598%
24 distinct numbers:  1.5192%
25 distinct numbers:  0.0978%

cottoneyedjoe's avatar - cuonvFT
In response to KyngeRycharde

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 little more than a third, so that's where the rule of the third comes from.

Unfortunately, the distribution is a little more complicated for draw lotteries where you drawing 5 or 6 distinct balls at a time, but in all the simulations I've run, it does seem that on average a little more than a third of the numbers do not come out.

cottoneyedjoe's avatar - cuonvFT

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 distinct pairs: 0.0004%
170 distinct pairs: 0.0016%
171 distinct pairs: 0.0036%
172 distinct pairs: 0.0076%
173 distinct pairs: 0.0192%
174 distinct pairs: 0.0384%
175 distinct pairs: 0.0624%
176 distinct pairs: 0.1360%
177 distinct pairs: 0.2384%
178 distinct pairs: 0.4228%
179 distinct pairs: 0.7420%
180 distinct pairs: 1.2468%
181 distinct pairs: 2.0108%
182 distinct pairs: 3.0396%
183 distinct pairs: 4.2592%
184 distinct pairs: 5.9032%
185 distinct pairs: 7.5420%
186 distinct pairs: 9.4320%
187 distinct pairs: 10.7200%
188 distinct pairs: 11.3444%
189 distinct pairs: 11.0996%
190 distinct pairs: 9.9740%
191 distinct pairs: 8.1188%
192 distinct pairs: 5.9168%
193 distinct pairs: 3.8672%
194 distinct pairs: 2.2208%
195 distinct pairs: 1.0608%
196 distinct pairs: 0.4104%
197 distinct pairs: 0.1256%
198 distinct pairs: 0.0296%
199 distinct pairs: 0.0044%
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hello , cottoney thank you!

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hello cottoney, please

 lottterys

 7/31=

5/80

cottoneyedjoe's avatar - cuonvFT
In response to dr san

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 numbers:  0.4694%
28 distinct numbers:  0.0610%
29 distinct numbers:  0.0044%
30 distinct numbers:  0.0002%

number of distinct numbers in 16 consecutive draws of 5/80
1 million simulations
37 distinct numbers:  0.0001%
38 distinct numbers:  0.0001%
39 distinct numbers:  0.0008%
40 distinct numbers:  0.0029%
41 distinct numbers:  0.0120%
42 distinct numbers:  0.0367%
43 distinct numbers:  0.1231%
44 distinct numbers:  0.3616%
45 distinct numbers:  0.9095%
46 distinct numbers:  1.9600%
47 distinct numbers:  3.7909%
48 distinct numbers:  6.4133%
49 distinct numbers:  9.4766%
50 distinct numbers: 12.3855%
51 distinct numbers: 14.1713%
52 distinct numbers: 14.2195%
53 distinct numbers: 12.5695%
54 distinct numbers:  9.7374%
55 distinct numbers:  6.5644%
56 distinct numbers:  3.8795%
57 distinct numbers:  1.9758%
58 distinct numbers:  0.8930%
59 distinct numbers:  0.3468%
60 distinct numbers:  0.1229%
61 distinct numbers:  0.0349%
62 distinct numbers:  0.0087%
63 distinct numbers:  0.0027%
64 distinct numbers:  0.0005%

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hello cottoneyedjoe, ok corret thank you!!

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