hi David and Mike,
under the heading:
"The distribution of the number of appearances of each number"

You appear to assume that the number of balls is m (6 for UK lottery) and on this basis you are proceeding to prove that the distribution is binomial. The implicit assumption here is that one is drawing a sample of 6 balls, whereas in reality "one is drawing one ball at a time without substitution, up to the number of m (6 for UK lottery)". The implication is that whereas the former assumption leads to a BINOMIAL DISTN, the latter reality would lead to some other distribution ( HYPERGEOMETRIC ???).

Although the first assumption may be justified for the purpose of the study at hand, I would appreciate your comments... of course I realise that my remarks come a few years later; nevertheless lottery seems to remain a very lively topic

hi David and Mike,

under the heading:

"The distribution of the number of appearances of each number"

You appear to assume that the number of balls is m (6 for UK lottery) and on this basis you are proceeding to prove that the distribution is binomial. The implicit assumption here is that one is drawing a sample of 6 balls, whereas in reality "one is drawing one ball at a time without substitution, up to the number of m (6 for UK lottery)". The implication is that whereas the former assumption leads to a BINOMIAL DISTN, the latter reality would lead to some other distribution ( HYPERGEOMETRIC ???).

Although the first assumption may be justified for the purpose of the study at hand, I would appreciate your comments... of course I realise that my remarks come a few years later; nevertheless lottery seems to remain a very lively topic

Regards

Tassos Chryssafis (tchryssafis@hotmail.com and tassos@talktalk.net)