Re: Sorry not related to poker just a question thats bugging me. Need
[ QUOTE ]
Um, why not 49C5 = 1,906,844
To guarantee at least 5 numbers in any N size set drawn from the set of 49, all you need to have is every possible 5 number combination of 49.
[/ QUOTE ]
Since each 6-tuple contains 6 5-tuples, you can hope to cover all 5-tuples with 49C5/6 tickets. I'd guess this is possible. Usually, when there is no number theoretic obstruction, such designs exist.
However, covering all 5-tuples is not necessary. You can omit some, just not all 6 5-tuples contained in some 6-tuple.
The total number of tickets sharing at least 5 numbers in common with a given ticket is 1+6*43 = 259. Ideally, you would like the 259 neighbors of each ticket in your wheel to be distinct, in which case you would need only 49C6/259 tickets. That's a lower bound for the number of tickets needed. Unfortunately, 49C6/259 isn't an integer, so you need some overlap. The minimal wheel could be very messy.
|