Lottery outcomes

[Jay R. Ashworth]

On Sun, Feb 13, 2005 at 02:04:52PM -0500, Kenneth Brody wrote:
> Quoting "Jay R. Ashworth" <jra at baylink.com>:
> [...]
> > > You want _combinations_, not _permutations_.  Unless this powerball
> > > requires that you pick the numbers in the _same_order_, you need to
> > > divide by 5! for a 5-ball game.
> > >
> > >     14,463,212,400 / 5! = 120,526,770
> >
> > It's not a permutations thing.  It's a "the ball you just picked isn't
> > in the hopper anymore" thing.  The first pick is 1 ball in 53.  The
> > second pick is therefore 1 ball in 52, and so forth.  IE: the Factorial
> > already *handled* that issue, and you don't need to do it again.
> >
> > I may still be wrong, but I'm afraid it's going to take a bigger 2x4 to
> > explain it to me than that, Ken.  :-)
> 
> But it's important that the order isn't important.  ;-)

It is, but your comments seem to be reversed WRT that issue.

> The number of _permutations_ of X items out of Y is:
> 
>     Y! / (Y-X)!
> 
> The number of _combinations_ of X items out of Y is:
> 
>     ( Y! / (Y-X)! ) / X!

And indeed... themathpage.com agrees with you.

> Here's my new 2x4:
> 
> 
> 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.

> 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.

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.

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?

Cheers,
-- jra
-- 
Jay R. Ashworth                                                jra at baylink.com
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