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I think the most important thing to remember is that we're all sat in the middle of our offices, in front of other people, maths-debating. Disgusting, really.

I completely agree; they are very, very close to identical. But not quite - to see the difference: a) Buying all 13,983,816 tickets on any one week would guarantee you a win b) Buying one ticket for 13,983,816 draws would not guarantee you a win.

This is exactly what I said. The odds of each ticket winning is 1/13,983,816. The odds of one of the two tickets winning is 2/13,983,816. But then that's what you're saying. So... what? I'm not arguing with that. What are you talking about?

Sorry, adding a bit more just to make clear what I'm saying and not saying. 20 x 20 does nothing in terms of odds directly, the only purpose for multiplying 20x20 is to work out how many items are on the list. That is it. Nothing more. Step 1: Do you agree that if I pick 3 from 6 on Ticket A...

If both tickets are for the same draw, and both tickets are different (i.e. you're not just buying lucky dips where technically you could end up with the same line twice) then yes the odds do half. If you bought all 13,983,816 tickets in the same week, you'd be guaranteed to win. Your logic...

Well I'm not going to like it because you've completely misunderstood my point. I am saying nothing about the chances of winning with 3 numbers on just one line. All I am doing is listing the number of possible "winning" combinations, where in this case "winning" is defined as having exactly 3...

Edited to make this bit clearer: I don't mean 3 out of the 6 winning numbers will satisfy (b); I mean any 3 of the 6 numbers written on Ticket B. We know there are no duplicates, so any 3 of those 6 will do. The 400 combinations that I'm referring to are: Pick any 3 out of the 6 written on...

By my workings, assuming no duplicate numbers on the lines you buy: A winning line (as in the real numbers that are drawn) would leave you in this position if it had (a) ANY three from Ticket A, and (b) ANY three from Ticket B. There are 20 ways of picking 3 numbers out of 6, so 20 possible...