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Don't feel too bad, it took me a while to get my head round DTES's approach (to start with I thought he was wrong) and I used to be an A-Level Maths teacher! Also, once we've thrashed Birmingham you'll have forgotten all about it.

This where the two different approaches comes in. You are thinking of the winning numbers being known and then calculating the probability of picking them. This is not DTES's approach. Consider buying two tickets on a Wednesday, i.e. we don't know what the winning numbers are yet. The first...

The confusion lies around the fact that there are two ways to approach the problem. One approach, Triggaaar's, is to think of the winning numbers as fixed and calculate the probability of picking 3 of these fixed numbers on one ticket and three of them on a second ticket. The other approach...

You haven't understood the method that DTES is using. DTES's method is: - Buy two tickets and assume no number appears on both tickets. - There are 20 different sets of 3 numbers possible on the first ticket. - There are 20 different sets of 3 numbers possible on the second ticket. - Therefore...

And buying all 13,983,816 tickets in any one week means it's impossible to win 2 or more times, while buying one ticket for 13,983,816 draws would give you a very good chance of winning more than once.

2 in 13,983,816 and 1 in 6,991,908 are the same odds.