Re: Re: Re: Re: Picking (potentially) winning lottery numbers

RatArsed,

Calculating the chances of winning is fairly trivial. Assume the drawing has 50 balls from which 6 balls are drawn without replacement. The chance of picking:

one right is 6/50
the second one right is 5/49
third 4/48
forth 3/47
fifth 2/46
sixth 1/45

To get the chances of picking them all right, multiply all those probablities together: 6.29299e-8, or about 1 in 15.9 million.

Scott

Comment on Re: Re: Re: Re: Picking (potentially) winning lottery numbers

Replies are listed 'Best First'.
Re(5): Picking (potentially) winning lottery numbers by RatArsed (Monk) on Jul 25, 2001 at 13:37 UTC
That's the simple case for matching all six balls Now add other prizes (UK: match 3, match 4, match 5, match 5 + bonus ball, match 6) And work it out for multiple lines. The data is dependent on the other line; for example if I have "1, 2, 3, 4, 5, 6" as one choice, and "2, 3, 4, 5, 6, 7" as my other choice, I haven't greatly increased my overall chance of winning something, although my chance of the jackpot has doubled. Had the second line not included any of the numbers from the first line, I would have doubled my chance. The answer involves a lot of permutations and combinations work (nPr and nCr), which in turn is a lot of dealing with factorials. the bit I don't know how to do is to discount combinations that appear using both sets (2,3,4 is valid, and duplicated, but 1,3,7 is not valid) -- RatArsed	[reply]

Replies are listed 'Best First'.

Re(5): Picking (potentially) winning lottery numbers
by RatArsed (Monk) on Jul 25, 2001 at 13:37 UTC

Now add other prizes (UK: match 3, match 4, match 5, match 5 + bonus ball, match 6)

And work it out for multiple lines. The data is dependent on the other line; for example if I have "1, 2, 3, 4, 5, 6" as one choice, and "2, 3, 4, 5, 6, 7" as my other choice, I haven't greatly increased my overall chance of winning something, although my chance of the jackpot has doubled. Had the second line not included any of the numbers from the first line, I would have doubled my chance.

The answer involves a lot of permutations and combinations work (nPr and nCr), which in turn is a lot of dealing with factorials.

the bit I don't know how to do is to discount combinations that appear using both sets (2,3,4 is valid, and duplicated, but 1,3,7 is not valid)

--
RatArsed

[reply]