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potterfan3 | 14:09 Sat 01st Mar 2008 | How it Works
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Is there anything against the rules about buying all 13,983,816 different possible combinations of lottery numbers, so you'll definitely win?
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No, but it would take you forever to fill out the tickets.

You'd have to ensure the prize was more than the outlay, and that you weren't going to share it with a dozen other winners.

Big risk.
There have been times in other lotteries, when there has been a couple of rollovers and the prize is huge, where people have tried this.

Maybe not buying all the tickets, but a syndicate buying as many as they can.

As Munkeeluvva says, the problem is the logistics of actually buying the tickets.

First you need to get people to fill out thousands of lottery forms, making sure all the combinations of numbers are covered (not easy in itself).

Then you need loads of people to actually BUY millions of lottery tickets, again not easy to organize.

This may work if you are the only winner, but if the top prize is shared out by many winners you may find it has cost you a lot of money to win very little.
someone tried that here in Ireland a few years back using a syndicate of lotto machine operators.

The Lotto company instantly spotted the unusual activity and shut the machines down remotely.
You need many more than 14 million tickets. The number 49 factorial (49x48x47x46x etc) is far bigger than 14 million.
>You need many more than 14 million tickets

Not true.

The odds of winning main lottery are 13,983,816 to one.

So you need just under 14 million tickets.

I think you work it out like this - 49!/(6!*(49-6)!)
It would only be worth considering buying all the permutations if the value of the prize is worth more than the cost of the tickets. But suppose the prize is �20 million and you spend �14 million...and someone else has the same idea. Then the most you can get is �10 million as you'd have to share the winnings- and you'd get even less if others had the same idea or picked the winning combination
I think buildersmate is on the right lines but I think his calculation assumes you have to pick the six numbers in the right order. But you still win if you pick the right numbers but in the wrong order. I think the figure of circa 14 million looks right
13983816 to one is correct.

Also, if you were to do this, and get away with it, and be unlucky enough to do it on the same week as someone else, you could still potentially be on to a nice little earner as you would also have every combination of 5 matching nubers, 4 matching etc etc
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buildersmate - you're assuming the order matters, it doesn't.
49*48*47*46*45*44, will give you every possible combination in every possible order, and since every 6! combinations will be the same numbers just in a different order, you do (49*48*47*46*45*44)/6!, which equals 13,983,816
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also, it isn't 49!, because you don't pick 49 numbers, you pick 6.

49 different possible numbers * 48 remaining numbers * 47 remaining numbers * 46 remaining numbers * 45 remaining numbers * 44 remaining numbers

Then like I said before there is 6! (720) different ways of having the same numbers but in a different orders.
I remember hearing about someone doing this and it wasn't allowed. Apparently the lottery is a game of chance and this is taking away the "chance" element of it.
Not sure what the outcome of this case was though
Yup - you are right (it was obviously too early on a Sunday to be thinking about it).
ive. been studying the lottery for a while and toying with the idea of covering all but the improbable combinations. ie. all lines with 3 or more consequitive numbers, all lines with all even no,s, and all odd no. lines. all evenly spaced combinations. and a few others too numerous to mention. if i cut it down to about a million lines i can get a backer(s) anybody any idea how i go about working it out. soupy.

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