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 | [reply] |
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] |