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My "algorithm" is brute force - totally independent of IMAX. I gave up on a looping "step 3" that was limited to 36×36, but instead looked at the whole 45×45 at once, using a scheme to try to find maximum number of new pairs for each ticket. I started by accepting 12 or more, then when those dried up accepted tickets with fewer new pairs.
After looping by 1's and by 7's to get the first 15 tickets with 15 pairs each, the procedure is:This method runs runs into trouble (that is, infinite loop) when the remaining number of pairs is less than 60 or so, so at that point I switch to finding pairs that haven't yet been used, and trying to find some with common values. About the last 10 tickets are found that way.Code:1. Find a value M that has been used in fewest pairs so far. 2. Find a value I which has not yet been paired with M. 3. Look for J that has not been paired with M or with I: if none found, try another I with same M if no more (M,I) pairs can be found, try a different M 4. For all K, count number of new pairs in set (M,I,J,K) and use K that is max 5. For all L, count number of new pairs in set (M,I,J,K,L) and use L that is max 6. for all N, count number of new pairs in set (M,I,J,K,L,N) and use N that is max 7. If number of new pairs [U]>[/U] current criterion, keep the ticket else try again from step 1. 8. If 10 in a row fail, decrease criterion by 1.
Every possible pair occurs in some ticket with 4 other "random" values. That makes the probability of having any specific set of 3 values P(3) = 4/43. Is that enough to skew the odds in your favor? Is there a payout for getting 3 hits?
So essentially, there is no true algorithm, but it's more a trial-and-error method?
There's no payout with getting 3-hits unfortunately. I've been playing this method for a few weeks now, more out of interest than to expect any significant return. I've actually won a smaller prize every week with my 94 games. Albeit, the system is not as simple as the math problem I proposed as there is actually 8 numbers drawn with 2 of those being considered as supplementary. I wasn't going to complicate things, as I had already withheld the second part of my question which was whether or not an algorithm could be found where, instead of pairs, groups of n>2 could also be found (ie: How many subsets required to cover all quadruples using subsets of 9, from a field of 200). This had nothing to do with lottery anymore, but was more just out of interest for me.