Euromillions lottery question! maths boffin needed

[hellosailor]

I don't play the lottery but played the Euromillions biggest jackpot ever last night for a bit of fun...

It really surprises me that the 113 million was won by 1 person, and read this morning that the chances of winning the jackpot were something like 76 million to 1.

This got me thinking, given that the winner had to match 5 numbers out of a choice of 50, and 2 'lucky star' numbers out of a choice of 9, surely there are not 76 million different permutations of numbers you could pick? Surely there are not, relatively speaking, that many different permutations? If this is correct, why did someone minted not simply buy a ticket for every permutation, as however much that cost it would not cost 113 million surely?

or am I being a maths-peasant and in fact it would cost you more than that to buy a ticket for every possible combination?

Maths boffins please stop me puzzling by telling me - how many different number combinations would there have been?!:)

[Jah Lush]

The more people who enter the less chance you have of winning.

[david_carnell]

It's not that you have less chance of winning if more people enter, just that you'd have to share the prize and would lose any profit. Plus the logistical difficulties of buying 76million different lottery tickets makes the enterprise almost impossible. Although I do remember once seeing a tv programme about someone in the USA who did something similar.

[woofmarkthedog]

.....and those who don't enter will not win....... statistically speaking which seems mad because I get letters everyweek telling me i have won lotteris & prize drawers that I haven't even entered.....Mad.....I am a Squillionaire now you know (on paper of course), oh yes , just waiting for those cheques..... any day now.....Ahhh mmmm

W**F

* checks letter box..............again, tum tee tum*

[hellosailor]

Thank you chaps :)) but I am eager for an actual Maths whizz to tell me how many permutations there are if you are choosing 5 out of 50 numbers and 2 out of 9 lucky stars! I just think it can be nowhere near 76 million possible combinations! I need figures! Is Stephen Hawking on the forum please? He can do sums in his head and that.

[matthew123]

1 in 76,275,360 chances of winning Euromillions jackpot from one ticket - http://en.wikipedia.org/wiki/EuroMillions

The Spanish winning ticket may have been part of a syndicate - say a whole village. Looking at the 'wiki' probability the 'syndicate' could have improved their chances of winning..

[david_carnell]

OK it's something like this:

50 (balls) divided by 5 (the number of balls you might have one of) and then so on, so:

50/5 x 49/4 x 48/3 x 47/2 x 46/1 = 10x12.25x16x23.5x46 = 2073680:1 chance

But then you have the "lucky stars" balls too so the odds of getting both of those right are:

9/2 x 8/1 = 36:1chance

So the odds of getting all 7 numbers are

36x2073680=74652480:1 chance or nearly 1 in 75 million. So very slightly better than your original statement.

Edit: It appears, according to wiki that I've done something wrong, although I'm not sure what.

[Marmora Man]

The first number can be chosen in 50 ways, then 2nd in 50 ways, the third in 50 ways, the fourth in 50 ways, the first "lucky star" number in 9 ways and the second "lucky star" number in 9 ways. This gives the calculation for the number of combinations of numbers as:

50 x 50 x 50 x 50 x 50 x 9 x 9

This = 2.53125e+10 or 25,312,500,000 or in layman's terms 25 billion, 312 million, 500 thousand which is a very very large number and far in excess of the 1 in 76 million mentioned. So you'd need to be seriously rich to buy up all tickets and only win back ?112m - bad decision.

[david_carnell]

Mamora Man - surely you can't choose the same number twice (or you'd be bloody stupid to) so the second number can only be from 49 alternatives not 50 and so on and so on?

[hellosailor]

you can't choose a number twice out of the numbers 1 to 50, though because the lucky stars numbers are 1 to 9 you obviously might repeat a number in that sense.

[david_carnell]

But the lucky stars come from a separate machine/ball pit though, correct?

[Marmora Man]

D-C in mathematical terms when calculating combinations of numbers my formula is correct but, you're right, I didn't consider the lottery rules and process. The sum therefore becomes

50 x 49 x 48 x 47 x 46 x 9 x 8 = 1.8301e+10 = 18 plus billion, which remains a large number.

You buy one ticket then you buy one of those 18 billion combinations. The chance of winning depends on the rules tho'. If there's no absolute match does the "closest" win - ie all five numbers + one of the lucky star is better than all 5 numbers with no lucky star?

[hellosailor]

thanks MM, not sure it's worth shelling out ?1.50 every week in the light of that.... Tho there is a man in Spain this morning who would disagree I suppose...

[Huguenot]

Right not quite guys.

In reality the odds are better because you don't need to get the numbers in any particular order.

For example, you could say that the 'first' number out of the bag doesn't need to be 1 in 50, because you could have chosen it as one of your other numbers.

So a simpler interpretation would be that the first number only has to be one of your chosen six (sorry actually 5, but I don't do euro, so assume this is still relevant and continue) - e.g odds of getting the first number are 6 in 50, and the second number 5 in 49.

In reality it's not this simple: you'd need to exclude the possiblity of having got the first number wrong etc. In mathematical terms it's easier to calculate the odds in a gamble of this sort by multiplying the odds of NOT getting it and subtracting from 1.

The calculation of 1 in 76m is near as dammit right, David's was the closest, but strictly speaking a co-incidence rather than a reflection of appropriate mathematical strategy.

I have a tedious qualification in double mathematics.

The reason you wouldn't buy 76m tickets in the hope of gaining the 113m, is because it wouldn't prevent several other competitors also getting the right result.

In this situation, because the other guy would still have won as well, getting the winning ticket with a 76m gambit would have split the 113m, with a net loss of 19m euros.

Not a great idea.

[Huguenot]

To get it exactly right for you for 5 numbers in 50 and two in 9

5/50 = 0.1

4/49 = 0.0816326

3/48 = 0.0625

2/47 = 0.0425531

1/46 = 0.0217391

2/9 = 0.2222222

1/8 = 0.125

Multiply them altogether and you get 1 in 76,275,360

*snaps the fingers*

That's da bizniss boyz, read and weep

[hellosailor]

Huguenot you are my hero :))

[Huguenot]

Embarrassed kissing, but sense of righteousness

erm.... boooyaka?

[Peckhamgatecrasher]

Here's a geometric posing Huguenot.

Last week on morning dog walk I spied chaps placing three poles in equilateral triangle in middle of the park. By afternoon there was a 400m running track with six lanes. How did they do that?

[Huguenot]

But this is Mornington Crescent?

Brendan knows better, I think that gambit deserves nothing more sleek than Bromley-by-Bow?

[Peckhamgatecrasher]

Oh Cockfosters!

[lard]

Huguenot Wrote:

-------------------------------------------------------

>

> The reason you wouldn't buy 76m tickets in the

> hope of gaining the 113m, is because it wouldn't

> prevent several other competitors also getting the

> right result.

>

> In this situation, because the other guy would

> still have won as well, getting the winning ticket

> with a 76m gambit would have split the 113m, with

> a net loss of 19m euros.

>

> Not a great idea.

At some point though the increase in the prize size would mean that it would be a valid strategy to buy all the tickets.

This would be based on an estimate of the number of participants for that lottery. It all depends on the average number of players, how many this is increased by for a rollover etc.

At some point there will be a tipping point.

For example a prize fund of ?1 billion with only an increase of 50% participants would mean buying all the tickets would be wise.

If the prize fund is out of synch with the number of participants then this could arise in a situation where buying all the tickets was a valid strategy.

[Jeremy]

I wish I'd seen this earlier. Dammit.

The correct term, hellosailor, for this type of calculation is a COMBINATION, not a PERMUTATION. Because the order of the numbers are not important.

The formula is 50 C 5 x 9 C 2.

In long hand, this is (50! / (5! x 45!)) x (9! / (2! x 7!)) = 76,275,360

[dottordivago]

Actually a cool tool is the win-o-meter (the bar on top of the bet slip) http://www.euro-millions.co/play-euromillions you can see live how your winning percentages change according to amount of both numbers andbet-slips played). Pretty cool.

[david_carnell]

It did occur to me that the original idea of buying 76million odd tickets would result in more than just a guarantee of winning the jackpot. You would of course win every other sum of money available too, such as getting 5 numbers and 1 bonus, 4 numbers and 2 bonus balls etc etc right down to the minimum win of three numbers (or whatever it is).

So you'd stand to win much more although my maths skills don't allow me to tell you how much more...

The logisitical problem is still the barrier though. Buying 76m tickets would be an epic ask since you'd have to fill out forms for every permutation.

Although it has been attempted:

http://www.nytimes.com/1992/02/25/us/group-invests-5-million-to-hedge-bets-in-lottery.html?pagewanted=all&src=pm

[Jeremy]

But DC... as you said back in May 2009(!), you might have to share the jackpot... so there's no guarantee you'll make money, however large the prize fund. For all you know, 10 people might all pick the winning numbers!