"joeu2004" wrote in message
...
Errata.... I wrote:
Given the condition for full wheels ("all 6 of the drawn numbers are in
your subset"), the probability that the condition is met is COMBIN(18,6)
/ COMBIN(36,6) for a subset is 18 numbers.

Similarly, for abbreviated wheels, the probability that the condition
("at least 4 of the drawn numbers are in your subset") is met is
COMBIN(18,4) / COMBIN(36/6).

While I believe the full-wheel conditional probability is correct, I have
my doubts about the formula for the abbreviated-wheel conditional
probability. I need to give that more thought.

=COMBIN(36,6)

That returns 1947792 which is indeed the number of 6-number combinations my
routine churns out before 'filtering'. Now if you can derive some formula
that returns 16431 for the filtered '4 out of 6' that would validate my
routine!

Regards,
Peter T