Can the €190m euromillions lottery be rigged?

[jay_mcfc]

There are around 76 million different permutations and a ticket costs 2e in Europe (people in the UK are ripped off). So to cover all permutations it would cost you 152m euros. Obviously there is a chance someone else will win it also, but there would also be millions in other prizes, which of course you would win a lot of.

Now, you could probably knock off about 30 million by not having 4 or 5 numbers consecutively, for example you wouldn't have 1,2,3,4,5 as your main 5 numbers. There are probably a ton of these you could save money on without reducing your chances of winning the lottery in real terms.

So if anyone has a spare 120 million euros and wants to possibly double their money within a few days, give me a call.

Note: I have no idea how correct the figures actually are, so if anyone wants to do a better job go ahead. We might save some more money. Also, I think this actually happened in the US a few years ago when the prize became considerably bigger than the odds and a business over there spent millions on tickets...and won

 

[EalingBlue2]

There was a scenario with the national lottery early on when I think 1,2,3,4,5,6 was picked by about 10000 buyers and due to birthdays, lucky numbers, months etc 3 or more of the numbers were in a ridiculously high amount of tickets. In this scenario getting 3 numbers with the guaranteed £10 would pay out more than the jackpot or 4 or 5which was made up of the remaining funds with a formula.

The best bet to maximise your win is to pick numbers over 31 avoiding "special" numbers like 42 and 49 as all combinations are as likely but a combination like this minimises your chance of sharing.

That said if there are enough people who over the years figured this out and acted you might have to sneak in one number less than 32 to get the new edge ?

Ps I have never seen euro millions the above is based on the old 49 number pick 6 national lottery

 

[Dave Ewing's Back 'eader]

It's gotta be rigged. I haven't won a double roll over yet!

 

[BlueHammer85]

So a multi multi millionaire only has to spend 152 million euros to get a guaranteed 38million euros ?

 

[Dave Ewing's Back 'eader]

So a multi multi millionaire only has to spend 152 million euros to get a guaranteed 38million euros ?

How long to buy 152 million tickets? It takes me a couple of minutes to buy one! So that would be 304 million minutes. By the time yer on the 147 millionth ticket the time's up and someone has nipped in with the winning ticket four days before yer!

 

[dickie davies]

So a multi multi millionaire only has to spend 152 million euros to get a guaranteed 38million euros ?

Not necessarily
It's more than possible that there will be multiple winners of the jackpot, so if there are three (including the multi millionaire) he loses 90 million on the first prize
Of course, he will win numerous smaller prizes that could be close to making up the difference

I did hear of a scenario happening in the States where someone tried to cover all the numbers and the above result happened

 

[Hamann Pineapple]

What if Jihadi John won it ?

 

[metalblue]

So a multi multi millionaire only has to spend 152 million euros to get a guaranteed 38million euros ?

Not necessarily
It's more than possible that there will be multiple winners of the jackpot, so if there are three (including the multi millionaire) he loses 90 million on the first prize
Of course, he will win numerous smaller prizes that could be close to making up the difference

I did hear of a scenario happening in the States where someone tried to cover all the numbers and the above result happened

Wouldn't the prize fund also be much larger given someone had spent 152m plus all the other punters?

 

[aguero93:20]

Your odds of winning are 116,531,800/1 so you'd need to spend €233,063,600 to be guaranteed a share of €190,000,000.

Good luck.

 

[Prestwich_Blue]

The formula for determing how many groups of consecutive numbers are in a set is, I think, (n - x)+1, where n is the total set of numbers and x represents how many numbers you want in a group of consecutive numbers.

So for the normal lottery of 49 numbers (n) to determine how many groups of 3 consecutive numbers there are you'd calculate (49-3)+1 = 47. For groups of 4, it's 46 and for groups of 5, it's 45. so by eliminating consecutive groups of 3, 4 and 5 numbers, you'd eliminate the grand total of 138 possible combinations. But eliminating all groups of 3 automatically excludes all groups of 4 and 5 so really it only accounts for 47 combinations effectively.