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Thread: Lottery disaster

  1. #1
    Feb 2010

    Lottery disaster

    I would consider this basic statistics, i think, though i still need a bit of help with it as a confirmation.

    The national lottery in england is: 6 numbers out of 49. simple. I filled in 2 lines of different numbers. i.e. 12 DIFFERENT numbers. the day of the lottery the 6 winning numbers were NONE of what i chose. I was so annoyed I started calculating the odds of:
    out of a total of 49 numbers, pick any 12. what is the chance that would not have got ANY of the 6 chosen randomly?

    I make this:
    which to my horror turns out to be about 16%. So i should not be surprised there are no prises for not getting ANY numbers on the lottery! )

    can some1 please confirm my thoughts and calculations are indeed correct??

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  2. #2
    Senior Member Sambit's Avatar
    Oct 2010
    You will not get any of the 6 numbers out of 49; so you are choosing 12 numbers from the (49-6)=43 numbers--which can be done in $\displaystyle {{43}\choose{12}}$ ways (since the order does not matter),so the number of ways you won't get ANY of the 6 numbers is: $\displaystyle {{43}\choose{12}}$. Hence the required probability will be:$\displaystyle \frac{{{43}\choose{12}}}{{{49}\choose{12}}}$

    Note: It comes out to be 16.66%, like what you got.
    Last edited by Sambit; Mar 21st 2011 at 03:31 AM. Reason: Note added
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