
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:
(43/49)*(42/48)*(41/47)*(40/46)*(39/45)*(38/44)*(37/43)*(36/42)*(35/41)*(34/40)*(33/39)*(32/38)
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! :o)
can some1 please confirm my thoughts and calculations are indeed correct??
thanks!

You will not get any of the 6 numbers out of 49; so you are choosing 12 numbers from the (496)=43 numberswhich 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.