dr san wrote: ``04 = (A)> (greater than) space between the draws. 05 = (B) <less than) space between the draws``
I am having trouble understanding what you want. But if it has something to do with determining the spacing between drawing the same number (or set of numbers), perhaps the following will be helpful.
Consider the "mega" number of the Mega Millions lottery. There are 46 "mega" numbers. The probability of the same "mega" number appearing k drawings later (and not before) is given by the binomial distribution. Specifically: (1/46)*(1 - 1/46)^(k-1), where the operator "^" means "to the power of". (For example, 4^3 = 4*4*4 = 64.)
For example, suppose the most-recently drawn "mega" number is 30 (21 Aug 2012). The probability that 30 will be the "mega" number in the next drawing is 1/46. The probability that 30 will be the "mega" number in 2 drawings later (and not before) is (1/46)*(1-1/46). In 3 drawings later (and not before): (1/46)*(1 - 1/46)^2.
If we compute that probability for k=1,2,3,..., we find that 50% of the time, 30 (really any particular "mega" number) will be appear again 13 to 63 drawings later (that is, the middle two quartiles: 25% to 75% cumulative probability).
I have analyzed the 742 Mega Million drawings from 24 June 2005 through 31 July 2012 (I'm a little behind). And in fact, the same "mega" number appears again 11 to 58 drawings later. Pretty close to expectations based on statistical theory.
The statistical theory is similar for the "distance" between drawing a particular number among the 5 "regular" numbers (non-megas). There are 56 "regular" numbers. So the probability of including a particular "regular" number in the next drawing is COMBIN(55,4) / COMBIN(56,5), or about 8.93%.
So I would expect a particular number from the last drawing to appear in the next drawing with a probability of 8.93%. In 2 drawings later (and not before) with probability 8.93%*(1 - 8.93%). In 3 drawings later (and not before) with probability 8.93%*(1 - 8.93%)^2.
So 50% of the time, I would expect a particular number from the last drawing to appear again 4 to 8 drawings later (again, the middle two quartiles: 25% to 75% probability). I have not vetted that against the actual historical drawings.
But that is just one particular number. I find it more useful to ask: given a selection of 5 "regular" numbers, how recently should we expect to have seen __any__ 1, 2 or 3 of our selected numbers in past drawings.
Based on statistical theory, I would expect any 1 of 5 selected "regular" numbers to appear 2 to 4 drawings earlier 50% of the time. Any 2 numbers to appear 6 to 13 drawings earlier. Any 3 numbers to appear 88 to 212 drawings earlier.
Analyzing the 742 historical Mega Millions drawings, I find that any 1 of 5 selected "regular" numbers appears 1 to 4 drawings earlier 50% of the time. Any 2 numbers appear 6 to 25 drawings earlier. Any 3 numbers appear 46 to 225 drawings earlier. It is very rare for any 4 or 5 numbers to appear in a previous drawing. Again, pretty close to expectations based on statistical theory.
So in choosing 5 numbers to play in the next drawing, we might reject any combination that matches 4 or 5 numbers of any previous drawings, or that matches 3 numbers within the last 45 drawings or no recently than 226 drawings earlier. Et cetera.
Of course, such a choice does not alter the probabililty of matching some or all numbers in the next drawing.
It is just a rationalization for choosing ("filtering") among the 3.8 million 5-of-56 combinations that we might play. It just makes us feel better about our choices.