Again I agree the number but your reasoning seems all wrong. The chance of a correct line is 1/45C6 and so the chances of an incorrect line is 1-1/45C6. Agreed and this is NCr/(LCs)^r*(1-1/LCs)^(N-r) where:
N is the number of lines submitted by punters (2M)
r is the number of correct entries (6)
L is the panel of Lotto numbers (45), and
s are the number of selections in a line (6)