Sounds more likely
To fully cover eight numbers you need to buy 28 tickets.

I'm not doing all that maths again though!
(Vice Admiral Sir Charles Cockbrush KCB MBE DSO, who is finding it increasingly difficult to log into b3ta.com wants to duck your aunt and kick your puppy, Tue 12 May 2009, 14:18, 1 reply)

I wonder
what the minimum number of tickets is to win at least one prize (i.e. get 3 of the numbers). I could probably work it out but don't care enough :)
(Richy_T hundreds if not thousands of dollars, Tue 12 May 2009, 17:08, closed)

This is a pretty hard problem
I've been thinking about it for ten minutes now and still can't work it out. My feeling is that it's a very, very high number. I'll think about it some more.
( flatfrog Staring vacantly into space, Tue 12 May 2009, 21:16, closed)

I think I got it
You'd have to guarantee 3 numbers out of 6 leaving 3 that don't matter so I reckon every combination of 3 from 46 numbers.

46 * 45 * 44 = £91,080.
( stopmeandslapme Cometh the hour, cometh your mum, Wed 13 May 2009, 8:10, closed)