Number drawn x times in last y drawsPrev Topic Next Topic

johnph77

CA
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December 10, 2003
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Aug 13, 2004, 3:25 pm

Nicely done! ty
gl
john

Blessed Saint Leibowitz, keep 'em dreamin' down there.....

Next week's convention for Psychics and Prognosticators has been cancelled due to unforeseen circumstances.

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JKING

United States
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Aug 14, 2004, 1:35 am

I tried something very simular to this. It didn't track very well when I compared it to history. Maybe it's just California. I like the calcutions though...I guess it deserves a second look. Nice work.
Datenda

Pretoria
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Aug 16, 2004, 6:03 am

apaqoqeas,
This is very interesting.
Will it be possible to provide similar equations to determine the chance to have a pair or a 3 No combination appear X times in the last Y draws of a lottery game
lottoarchitect

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Aug 22, 2004, 11:37 am

Datenda, if you could describe in more details what you mean by a pair & 3 No combination, I'd glady have a look at it.

If you have something to do, at least do it well...

Greece
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Aug 23, 2004, 1:22 pm

Well, about pairs and 3 number combinations, the idea is exactly the same. We have to determine the chance p to have a specific pair / 3 number combination to appear in a draw. Then the Bernoulli formula is applied as at the previous posts.

a and b parameters describe our lottery game (eg 5 of 45, a=5, b=45)

Pairs

the chance p to have a specific pair appear in a draw is equivalent to find how many combinations contain our pair among the whole possible combinations.
The combinations that contain a particular pair are C(b-2,a-2). In our example of a=5, b=45 we have exactly 12341 combinations from a total of C(45,5)=1221759 combinations. So, p=C(43,3)/C(45,5)=12341/1221759=0,010101

The relevant table for the above game is

	0	1	2	3	4	5
1	98,99%	1,01%
2	97,99%	2,00%	0,01%
3	97,00%	2,97%	0,03%	0,00%
4	96,02%	3,92%	0,06%	0,00%	0,00%
5	95,05%	4,85%	0,10%	0,00%	0,00%	0,00%
6	94,09%	5,76%	0,15%	0,00%	0,00%	0,00%
7	93,14%	6,65%	0,20%	0,00%	0,00%	0,00%
8	92,20%	7,53%	0,27%	0,01%	0,00%	0,00%
9	91,27%	8,38%	0,34%	0,01%	0,00%	0,00%
10	90,35%	9,22%	0,42%	0,01%	0,00%	0,00%

3 number combinations

the chance p to have a specific 3 number combination appear in a draw is equivalent to find how many combinations contain our 3 number combination among the whole possible combinations.
The combinations that contain a particular 3 number combination are C(b-3,a-3). In our example of a=5, b=45 we have exactly 861 combinations from a total of C(45,5)=1221759 combinations. So, p=C(42,2)/C(45,5)=861/1221759=0,0007047

The relevant table for the above game is

	0	1	2	3	4	5
1	99,93%	0,07%
2	99,86%	0,14%	0,00%
3	99,79%	0,21%	0,00%	0,00%
4	99,72%	0,28%	0,00%	0,00%	0,00%
5	99,65%	0,35%	0,00%	0,00%	0,00%	0,00%
6	99,58%	0,42%	0,00%	0,00%	0,00%	0,00%
7	99,51%	0,49%	0,00%	0,00%	0,00%	0,00%
8	99,44%	0,56%	0,00%	0,00%	0,00%	0,00%
9	99,37%	0,63%	0,00%	0,00%	0,00%	0,00%
10	99,30%	0,70%	0,00%	0,00%	0,00%	0,00%

As you can see, on both cases the rates are rather low to expect a pair or a 3 number combination to appear again in the next 10 draws. At least for the pair situation, we expect a specific pair to appear after the 10th draw with an error rate of about 10%. This means 90% of the time, a pair will appear again beyond the 10th draw after the draw it was appeared in.

If you have something to do, at least do it well...

Fenix

Canada
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August 21, 2004
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Aug 24, 2004, 7:56 am

The data has fine appearance. But, in a 6/49 game the repeating pairs from a last draw or last few draws act in a very different way.

lottoarchitect

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Aug 24, 2004, 8:07 am

In case of pairs in 6/49 game, we have

p=0,012755

The relevant table is

	0	1	2	3	4	5
1	98,72%	1,28%
2	97,47%	2,52%	0,02%
3	96,22%	3,73%	0,05%	0,00%
4	94,99%	4,91%	0,10%	0,00%	0,00%
5	93,78%	6,06%	0,16%	0,00%	0,00%	0,00%
6	92,59%	7,18%	0,23%	0,00%	0,00%	0,00%
7	91,41%	8,27%	0,32%	0,01%	0,00%	0,00%
8	90,24%	9,33%	0,42%	0,01%	0,00%	0,00%
9	89,09%	10,36%	0,54%	0,02%	0,00%	0,00%
10	87,95%	11,36%	0,66%	0,02%	0,00%	0,00%

This table means that the chance to have the same pair appear at the next draw is 1,28%. If it doesn't appear, then the chance to appear 2 draws after its first appearance is 2,52% etc. Anyway, the chances do not greatly differ from an 5/45 game. What have you observed?

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Fenix

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Aug 24, 2004, 8:14 am

As I said, the table has a nice appearance. Unfortunately, the pairs act in a very different manner than the formula shows. If you check a number of repeating pairs in a last couple of draws in any 6/49 game, you will see a little different percentage than 1 or 2 %&.No need to speak about some more nasty occurrences like repeating triplets or quads&.
Fenix

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Aug 24, 2004, 8:23 am

Or to simplify, by the formula once in 78 draws (100/1.28) a player may expect a repeating pair. That is quite far from the reality.
lottoarchitect

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Aug 24, 2004, 8:35 am

So, you understand that this 6/49 game does not behave as a random lottery. I haven't analysed this on real lotteries.
Keep in mind that we talk about a specific pair to appear again and not any two numbers common with a previous draw (which effectively means a pair as well). The chance at this case is 13.24% to appear at the next draw.
For example, if we have the draw 1 13 15 22 33 40, then the chance to have each pair
1-13, 1-15, 1-22, 1-33, 1-40, 13-15, 13-22,13-33, 13-40, 15-22, 15-33, 15-40, 22-33, 22-40, 33-40 individually to appear at the following draw is 1.28%.
On the other hand, the chance to have any two common numbers appear at the next draw (and therefore have one of the above pairs included) is 13.24%. Which case have you observed?

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Fenix

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Aug 24, 2004, 8:44 am

________________________________________________________

QUOTE:

On the other hand, the chance to have any two common numbers appear at the next draw (and therefore have one of the above pairs included) is 13.24%. Which case have you observed?

________________________________________________________

I wonder how did you get 13.24%? That percentage looks quite OK.
lottoarchitect

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Aug 24, 2004, 9:26 am

I got this result from my program. A full mathematical solution is possible of course but it requires lengthy analysis because there are overlapping conditions which makes the problem more complex. If I have some free time, I'll provide the mathematical solution to this.

If you have something to do, at least do it well...
Fenix

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Aug 24, 2004, 9:35 am

________________________________________________________

QUOTE:

A full mathematical solution is possible of course but it requires lengthy analysis because there are overlapping conditions which makes the problem more complex. If I have some free time, I'll provide the mathematical solution to this.

________________________________________________________

Thanks,that would be wonderful.
lottoarchitect

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Aug 24, 2004, 10:05 am

Here is the mathematical solution to the above problem.
The game in consideration is 6/49. Some explanation is needed to understand the following. As already said, each draw of the 6/49 game produces C(6,2)=15 different pairs. We have to find out how many different combinations produced by each pair, excluding combinations generated by that pair which include any of the other 14 pairs (this is the overlapping condition). The solution is hopefully simpler than I first thought.
Now, each draw will have 2 fixed numbers (the numbers of the pair). We want to find out how many combinations generated with that pair without including any other pair. This means we have to remove from our number selection all numbers drawn at the previous draw. This leave us with 49-6=43 numbers to fill the remaining 4 positions of our combinations. These can be arranged in C(43,4)=123410 different ways. The conclusion is that each pair can generate 123410 different combinations where no other pair from our previous draw included (and therefore we do not count the same combinations multiple times-overlapping condition).
Finally, the total combinations generated by all our 15 pairs are 15*C(43,4)=15*123410=1851150 and these are the total combinations that have 2 common numbers with our previous draw because we have included all possible 2 common numbers (our 15 pairs provide this).
Finally, all possible combinations of 6/49 are C(49,6)=13983816 and therefore
%=1851150/13983816=13,2378%

Using the above, we can construct a general solution to this problem.
Game eg 6/49 (a=6, b=49). c=common numbers with the previous draw (eg c=2).
p=C(a,c)*C(b-c,a-c)/C(b,a)
For the above problem, a=6, b=49, c=2
C(6,2)*C(43,4)/C(49,6)=15*123410/13983816=0.132378 or 13,2378%

If you have something to do, at least do it well...
Fenix

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August 21, 2004
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Aug 24, 2004, 11:35 am

At the first sight, it seems that any two numbers from any preceding draws have the same chance to be drawn in the next draw? Exactly two numbers from the five draws ago for example, have the same probability as two numbers from the previous draw?

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