Jobs & Education3 mins ago
Lottery Odds
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How much would I have to stake to guarantee winning £25 on the National Lottery (at £2 a line)?
I say this because I recall a competition in one of the Red Tops where the winner was allocated £20,000 I think to spend on the lottery. The newspaper divided the stake among 3 blocks of the lottery numbers (as in the old football perms). And guess what. Each block contained 2 of the numbers selected in the draw, meaning that the 'lucky' recipient received nothing!
I say this because I recall a competition in one of the Red Tops where the winner was allocated £20,000 I think to spend on the lottery. The newspaper divided the stake among 3 blocks of the lottery numbers (as in the old football perms). And guess what. Each block contained 2 of the numbers selected in the draw, meaning that the 'lucky' recipient received nothing!
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To be sure of having the first number drawn you'd need to buy 49 tickets. Then to be sure of having the first 2 numbers you'd have to buy 48 tickets for each of the 49. (49x48=2352) Then you'd need to buy 47 tickets for each of them. (2352x 47=110544)
I'll stick with a lucky dip when there's a big jackpot
I'll stick with a lucky dip when there's a big jackpot
Jim360. Thanks. It was how much would I have to stake to guarantee at least one £25 prize, irrespective of how much I was foolish enough to blow in one go. As I indicated above, if £20,000 wasn't enough to win £10 (as it was then), how much to be sure of £25. OK, you'd probably win more than £25 by covering all the odds but I was trying to get a handle on winning the smallest prize as opposed to the big one. (Sorry ABers, bit bored this morning!)
I think with a great day of care you can probably guarantee yourself a £25 prize with 5,528 different tickets, but actually achieving that minimum might turn out to be impossible.
The reason goes as follows: there are 48 x 48 x 47 = 110,544 distinct three-ball combinations that can occur in the Lotto draw, of which 20 appear in a single lotto draw. Equally, each ticket you buy can include 20 possible three-digit combinations, so if we divide the above figure by 20 we get 5,528 (actually 5527 and a bit, but you'd have to round up).
So in principle, 5,528 tickets can be bought that between them include all possible 3-digit combinations, which would guarantee 20 prizes of £25. This figure might turn out to be too small in practice, owing to the clash between the odd number of balls available and the even number of balls you can select in any individual ticket, so some repetition may be unavoidable, pushing the number of tickets you would need up, but the only way to check this would be to construct an actual set of such tickets (you could start by, e.g., buying the tickets "1 2 3 4 5 6", "1 2 7 8 9 10", "1 3 7 9 11 13"... and already you can see that it will be tough to avoid repeating any combination at all, and ensuring that each ticket covers twenty previously-unused 3-number sets.)
To guarantee winning just one of the available £25 prizes doesn't seem to change the picture much. In a worst-case scenario you can imagine covering 110,524 of the available 3-number sets and the actual draw includes the remaining 20, so your tickets need to cover at least 110,525 possible sets of three numbers, which implies only one ticket less than the scenario above!
Because of the relatively large numbers involved I can't really devote any more time to checking this, but I suspect that it should be possible to 100% guarantee winning a £25 prize for an initial stake of a little over £11,000.
The reason goes as follows: there are 48 x 48 x 47 = 110,544 distinct three-ball combinations that can occur in the Lotto draw, of which 20 appear in a single lotto draw. Equally, each ticket you buy can include 20 possible three-digit combinations, so if we divide the above figure by 20 we get 5,528 (actually 5527 and a bit, but you'd have to round up).
So in principle, 5,528 tickets can be bought that between them include all possible 3-digit combinations, which would guarantee 20 prizes of £25. This figure might turn out to be too small in practice, owing to the clash between the odd number of balls available and the even number of balls you can select in any individual ticket, so some repetition may be unavoidable, pushing the number of tickets you would need up, but the only way to check this would be to construct an actual set of such tickets (you could start by, e.g., buying the tickets "1 2 3 4 5 6", "1 2 7 8 9 10", "1 3 7 9 11 13"... and already you can see that it will be tough to avoid repeating any combination at all, and ensuring that each ticket covers twenty previously-unused 3-number sets.)
To guarantee winning just one of the available £25 prizes doesn't seem to change the picture much. In a worst-case scenario you can imagine covering 110,524 of the available 3-number sets and the actual draw includes the remaining 20, so your tickets need to cover at least 110,525 possible sets of three numbers, which implies only one ticket less than the scenario above!
Because of the relatively large numbers involved I can't really devote any more time to checking this, but I suspect that it should be possible to 100% guarantee winning a £25 prize for an initial stake of a little over £11,000.
That's not even close to right. The odds of getting a head in a fair coin toss are 1-1 (even), but tossing two coins in a row will not guarantee you a head (75% chance of at least one head, not 100%). Same applies here. The odds may be 57-1 but 58 tickets will not be enough. You would expect one ticket in 58 to win something... but you can't guarantee it. To guarantee a prize (with 100% certainty) you need to ensure that all possible three-number sets are covered at once (except possibly 19).
The following is not my words, I've copied and pasted it:
"Believe it or not, this is an open problem. Up to the present time, despite advanced mathematics and supercomputers, the precise minimum is unknown. It is known that the minimum value (called L(49,6,6,3) in mathematical terminology) lies between 87 and 163 . There is more information on this topic:
http://
... and here is a list of 163 tickets that will guarantee at least £25:
http://
"Believe it or not, this is an open problem. Up to the present time, despite advanced mathematics and supercomputers, the precise minimum is unknown. It is known that the minimum value (called L(49,6,6,3) in mathematical terminology) lies between 87 and 163 . There is more information on this topic:
http://
... and here is a list of 163 tickets that will guarantee at least £25:
http://
-- answer removed --
Goodness me! That's smaller than I expected. But there was one error in my post I should have noticed anyway, which was that I was counting permutations rather than combinations (ie my 110,544 cares about the order in which the three balls emerge, which isn't correct). I should have divided that by six in the first place, which knocks the figure down to closer to 900. And then it appears that I've oversimplified the problem anyway, in ways that I don't really understand because any paper entitled "combinatorics" could just as well be in Chinese for all the sense I can make of it.
Oh well. Sorry for misleading you, although it was an honest mistake. Still, on the plus side, at least there isn't a known answer!
Oh well. Sorry for misleading you, although it was an honest mistake. Still, on the plus side, at least there isn't a known answer!
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