Lottery outcomes

[Kenneth Brody]

Quoting "Jay R. Ashworth" <jra at baylink.com>:
[...]
> > Let's simplify the original problem.  Rather than 5 items out of 52,
> > let's take 5 items out of 5.
> >
> > According to your logic, "the first ball is one in 5", "the second
> > ball is one in 4", and so on, giving us 1 in (5*4*3*2*1) or 1 in 120.
> > I think you can see that this is obviously incorrect, since how many
> > different sets of "winning numbers" would there be?  There can only
> > be one set of "winning numbers" picked, as you picked every number.
>
> Correct, and there *is* only one set of winning numbers; order *is not*
> important in ball-pick lotteries: the numbers are *sorted in ascending
> order* before being publicized.  So, in a 5 ball 5/5 lottery, there
> *is* only one winning series: 1-2-3-4-5.

... though there are 120 different ways to pick the 5 numbers.

> > If the _order_ made a difference (ie: you specify "this is the first
> > number to be picked, this is the second, and so on") then there would
> > be a 1-in-120 chance.  However, since order is not significant, and
> > you only say "this number will be picked at some point", you need to
> > divide by X!.
>
> But we aren't *talking* about *odds*, we're talking about sequences.

Remember, if we're talking about a lottery, then order doesn't matter.
So we are talking _combinations_, not _permutations_.

> I *did* divide: I divided 52! by 47!, because there are only 5 balls
> participating in that block.  52!/47! = 52*51*50*49*48.

(FYI - Looking back at the OP, it's 5-of-53, not 5-of-52.)

> No one has yet explained to me in detail why that latter math problem
> does *not* correctly characterise the set of winning sequences that can
> be drawn in a 5/52 lottery, in light of the fact that order does not
> matter.  I think you *think* you have, but I don't think you have.  :-)
>
> I'm not sure why I don't think so, though, which makes it harder to
> defend.  More correctly: the math clearly doesn't match the formulae,
> but it appears intuitively to match the actual situation.  Is there a
> math major in the house?

That'd be me.

Let's step back a bit.

Taking order as significant, how many different permutations of
5-of-53 are there?  As you said, 53*52*51*50*49 = 344,362,200.
(Times the separately-picked 1-of-42 gives the 14,463,212,400
you arrived at.)

Now, generate all 344,362,200 permutations, and then sort each
sequence, as would be done in a lottery drawing.  You will see
that each combination now appears 120 times, leaving 2,869,685
unique sets.  Multiply by the separately-drawn 1-in-42 number,
and you have 120,526,770 unique sets of possible winning numbers.

--
KenBrody at BestWeb dot net        spamtrap: <g8ymh8uf001 at sneakemail.com>
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