Heres my take on it, for what its worth. First let me say that PAB's assessment of the situation is spot on. I agree with his mathematical assessment of what we are up against precisely. The problem is that even though we confine our macro to dealing with the data available, the number of required unique items of collected data we need to store still has to be available to cover the whole range of possible outcomes (some as yet unknown) for pairing a number n1 with number n2 where for any new result set, n1 can be as low as 1 or as high as 80. And thats just for pairs, the reasoning continues for triples quads etc right up the range of families of results as far as groupls of 10 from 20 per result. Without resorting to macros I looked at the problem from the point of view of what was the likelihood of finding any useful data at the end of such an excercise.
The Keno is new to me as well, since I too come from the UK so I've studied the various odds involved in the game.
I know that one can choose how many numbers from the 20 drawn you wish to match and have to then specify that number of balls in your bet. This probably old hat to our US based friends so forgive me for going through this again as its new to me, I'm sure you've seen all this.
To work out the odds on a spreadsheet for all the various choices of numbers of balls to match you can use the HYPOGEOMDIST function. =HYPOGEOMDIST(number succeeeded in matching, number you promised to match, number of balls drawn, highest ball number ).
So for The Keno if you want to match 10 balls the probability is =HYPOGEOMDIST(10,10,20,80) = 1.12212E-07 giving odds of 8,911,710.18 to 1.
If you are choosing to match 4 balls the same formula would be =HYPOGEOMDIST(4,4,20,80)= 0.003063392 giving odds of 325.4355 to 1.
If you are feeling keen you can (and probably already have) created a table in this way for 1,2,3,4,5,6,7,8,9,10 balls you hope to match.
I've uploaded mine for reference anyway. ..http://www.mediafire.com/view/?rdb9z6ctz736oou
Why is this relevant to the exercise we are trying to perform ?
Well each row of results in your 4.5k rows of results can be thought of as a 20 number entry into the draw. All the other results can be thought of as draws you tried to match that result with. The odds are exactly the same as if you were playing the game and are as calculated above. The only difference is each row of 20 numbers can be regarded as a full wheel of pairs or triples or quads or quins etc. to match against all the other rows of results. I cannot see simmcity's data spreadsheet as it appears to be no longer available (it doesn't matter) but you can imagine if you were counting all the triples of a specific combination say 12,35,77 then you are counting them up over your 4.5k lines of 20 numbers and each new result set gives you =COMBIN(20,3) =1140 chances of finding that combination. Over 4500 draws you have 4500 x 1140 = 5130000 chances of finding it . The computed odds in Keno if you were entering those three numbers would be 71.072 to 1 of finding it in a single draw. However we are comparing a 20 number wheel containing the required 3 so those odds are optimistic by a factor of =COMBIN(20,3) =1140. All this leads to a conclusion that I would expect to find 5130000/1140 = about (on average) 63 occurrences of that pair. This would be useful information when compared with other triples so would be worth doing.
Contrast that with quints. Doing the same exercise but this time taking the short cut of multiplying the probability of matching 5 (p= 0.000644925) by 4500 comes to an expected occurence (in 4500 results) of 2.9. Over 4500 lines of results I would expect to find 2.9 ocurrences of a given quint and many others wouldn't be found yet. Is this worthwhile information to compare ? NO.
Furthermore, the more numbers in the n-tuplet you want data for e.g a family of 10, the LOWER the expectation of finding that n-tuplet in your data. Youd need 10 MILLION draws to fine ONE named family of 10.

So its not worth the effort in my opinion, no useful information would be found anyway.
Thats my view, I hope I've got my sums right, no doubt someone will tell me if not.
