Number drawn x times in last y draws

Reply #16

Datenda — Thu, 26 Aug 2004 15:07:26 GMT

apagogeas,I see that you have already answered the question you asked me.What yo have done is fantastic.Here is an analysis of the South African 6/49 Lottery up to Draw 384 (Bonus excluded) which you can use to compare with the theory. The % Chance calculated by your formula for y = 384 Draws * 49 yields the number of numbers that theoretically appear x times.For Single Numbers:1 Number has appeared 68 times. Theoretically at x = 68, 0.02 numbers should have appeared. Theoretically at x = 67, 0.... [ More ]

Reply #15

lottoarchitect — Tue, 24 Aug 2004 16:03:35 GMT

At first sight this is what the common sense says. Any two numbers from any past draw, have the same chance to appear at the next draw. However, what the bernoulli suggests is different.Here is the table for 6/49 game to have 2 common numbers appear:Hor. axis - occurencies of 2 common numbersVer. axis - future draws012345186,76%13,24%275,28%22,97%1,75%365,31%29,89%4,56%0,23%456,67%34,58%7,91%0,81%0,03%549,16%37,51%11,45%1,75%0,13%0,00%642,66%39,05%14,90%3,03%0,35%0,02%737,01%39,53%18,09%4,60%0,7... [ More ]

Reply #14

Fenix — Tue, 24 Aug 2004 15:35:10 GMT

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

Reply #13

lottoarchitect — Tue, 24 Aug 2004 14:05:35 GMT

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 h... [ More ]

Reply #12

Fenix — Tue, 24 Aug 2004 13:35:21 GMT

________________________________________________________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.

Reply #11

lottoarchitect — Tue, 24 Aug 2004 13:26:17 GMT

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.

Reply #10

Fenix — Tue, 24 Aug 2004 12:44:56 GMT

________________________________________________________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.

Reply #9

lottoarchitect — Tue, 24 Aug 2004 12:35:48 GMT

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,... [ More ]

Reply #8

Fenix — Tue, 24 Aug 2004 12:23:57 GMT

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.

Reply #7

Fenix — Tue, 24 Aug 2004 12:14:06 GMT

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 .

Reply #6

lottoarchitect — Tue, 24 Aug 2004 12:07:12 GMT

In case of pairs in 6/49 game, we havep=0,012755The relevant table is012345198,72%1,28%297,47%2,52%0,02%396,22%3,73%0,05%0,00%494,99%4,91%0,10%0,00%0,00%593,78%6,06%0,16%0,00%0,00%0,00%692,59%7,18%0,23%0,00%0,00%0,00%791,41%8,27%0,32%0,01%0,00%0,00%890,24%9,33%0,42%0,01%0,00%0,00%989,09%10,36%0,54%0,02%0,00%0,00%1087,95%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 i... [ More ]

Reply #5

Fenix — Tue, 24 Aug 2004 11:56:40 GMT

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.

Reply #4

lottoarchitect — Mon, 23 Aug 2004 17:22:00 GMT

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)Pairsthe 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 pa... [ More ]

Reply #3

lottoarchitect — Sun, 22 Aug 2004 15:37:29 GMT

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.

Reply #2

Datenda — Mon, 16 Aug 2004 10:03:32 GMT

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

Reply #1

JKING — Sat, 14 Aug 2004 05:35:16 GMT

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.

Number drawn x times in last y draws

johnph77 — Fri, 13 Aug 2004 19:25:36 GMT

Nicely done! tygljohn