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euromillion odds

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helpmetoo | 12:57 Tue 05th Feb 2008 | Science
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hi boffins, could someone explain why the odds on winning the lotto euromillions seem not to work out logically? if you dont know, you choose 5 different numbers from 1 to 50 and 2 different numbers from 1 to 9. they state the odds of matching all 5 main numbers and the 2 lucky stars as 1 in 76,275,360, but the stated odds for achieving the 5 main numbers and just 1 lucky star are 1 in 5,448,240. my pea brain would compute that if the odds are 1 in 5,448,240 for 5 & 1 then as there are only 9 lucky star balls and at that stage one has been drawn the chances of having the correct second lucky star ball is 1 in 8 (or 7 to 1 against), making the chances 5,448,240 x (7:1) ??? the same sort of differences apply relative to the other win dividend ratios
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p.s
or half the odds quoted...
I make the odds on picking 5 correct balls from 50 1/211760 = 50!/(5! x 45!) = (50*49*48*47*46)/(5*4*3*2*1)

The odds on picking the correct 2 balls from 9 are 1/36 =
9!/(2! x 7!) = 9*8/2*1

Multiplying the two sets of odds together gives 1/76275360 which agrees with the overall odds quoted. So the problem if there is one is in the odds against getting 1 ball out of the 9 correct.

The odds on picking 1ball and only 1ball out of the 9 is 1/9, so with two balls to choose from the odds rise to 1 in 4.5, so I would reckon the true odds of getting 5 plus 1 is
1/211760 * 2/9 = 1/9534420
I agree with dundurn. Where do the figures you are quoting come from helpmetoo?
Question Author
the figure are off the nat lottery web site

www.national-lottery.co.uk/player/information.do?info=euroinfo
Ah wait a minute though - I am not clever enough to work out the figures myself (perhaps dundurn will come back) but the odds quoted there are probably right - the odds of getting all five main numbers and then NOT geting either of the two lucky stars will be different than the odds of simply getting the five main numbers. Likewise the odds of getting the five main numbers and one lucky star and NOT getting the other lucky star will be different to just getting the five main numbers and one lucky star.

Oh dear, badly explained, I am afraid but hope this helps a bit!
To get 1 ,and only 1, from 2 goes in 9 can be done in two ways with odds ...

(2 in 9) * (7 in 8) = 5.142856 to 1
or
(7 in 9) * (2 in 8) = 5.142856 to 1

giving (approx) 2.571428 to 1 ... of getting exactly 1 ... I think this shows their odds are correct.

Question Author
mmm, i thought it was tricky,

hey, any player has a rubbish chance of winning the jackpot,

will still have a go anyway because i in 76M and then hope you dont have to split it is still better odds that a big fat zero if you dont play at all!

thanks for your input folks,

i see now to a certain degree that there are still odds against missing both lucky stars as well as hitting either or both.
Thanks Maty. Your method is nice and logical - I had a nasty feeling I had missed something.

My personal favourite "lottery" joke was during the BSE scare. Someone pointed out that when we had 50 cases of CJD in a population of 50-60 million the odds of catching it were described by the Government as minimal and not worth bothering about but when it came to 1 in 14 million in the lottery "it could be you!".
I like it - sounds like a typical government spin.

One I heard recently (with all racial stuff removed just in case) ...

There was Man from (a land fabled with care of money) whose business was in a pretty bad way financially. So, one Friday he asked (his personal favourite deity) "Please let me win the lottery tomorrow." Sure enough nothing happened, Man did not win.

Next week, things were even worse, his wife had left him, business was dire. Man thought he would try harder "Please, I will worship regurlarly and attend (the building of his deity) every day. Please let me win the lottery." Again, nothing.

The following week the bailiffs were in: "Please, please, I am begging you, what do I have to do to win the lottery ...?"

(His personal deity) replied: "Man, I will help you, but you have to meet me half-way on this - buy a f######g ticket!!".
Different scenario - similar basis

Man falls over Beachy Head and manages to grab on to small protruding bush of the sort only scriptwriters can invent. Realising he can't hang on for ever and that it's still a long, long way down he starts to pray. "I've never been religious and never, ever been sure there's anything in it but if there really is Anyone up there, I could do with a hand. Immediately a large, soft, warm, comforting sort of voice comes into his head - "Don't worry, I've got you, just let go." A couple of moments pass and the man speaks again - "Is there Anyone else up there?"
211760 / 1 odds of picking 5 numbers

36 / 1 odds of picking 2 stars

211760 x 36 = 7 623 360??

so there is 7 623 360 possible combinations? x that by 1.50 (price of a ticket) = 11 435 040

If you played that on 100Million draw

you would have 1 ticket with all the numbers...
14 tickets with 5 + 1 star
pluss 1000;s of smaller prizes

the jackpot would need to be won by at least 12 peole before you would make a lose!!

now who wants to give me 11.5 million ?

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