Also, beware when you try to calculate the odds against, say, getting three numbers correct (the �10 prize).
The number of threes in 49 is:
49 x 48 x 47
---------------
1 x 2 x 3
= 110,544/6 = 18,424
So you would think the odds against winning �10 are 18,423 to 1.
However:
In pulling out six numbers, the machine is actually selecting 20 sets of 3 numbers. [Remember (6 x 5 x 4)/(1 x 2 x 3) = 20]. This means you have 20 lots of 3 each of which can win.
In addition, when you buy a lottery ticket you choose six numbers. Although you are only paying for one go at getting all six correct, you are actually buying 20 goes at getting three correct [again, (6 x 5 x 4)/(1 x 2 x 3) = 20].
So the probability of matching three is actually reduced by a factor of 20 for each of the reasons outlined above, so they become (18,424/400) = 45 to 1 (approx).
Many people see this as a bit rough, since Camelot only pays 9 to 1 for matching three numbers. However (as usual) all is not as it seems because for your �1 you are actually getting one go at matching all six, six goes at matching five numbers, fifteen goes at matching four and twenty goes at matching three. And this, of course, does not include the �five plus bonus ball� winner.