shoelaces and aglets

[Gerald McGarvey]

Here's my take on this...

Suppose you pick up 2*k aglets out of 2*n aglets, there are
C(2n,2k) ways of doing this.

That leaves 2*(n-k) aglets not-picked, we want none of the not-picked
aglets to be on the same shoelace.

How many ways are there of not-picking 2*(n-k) aglets such that no two
are on the same shoelace?
They must be on 2*(n-k) different shoelaces, which are chosen
from n shoelaces; there are C(n,2*(n-k)) ways of doing this choosing.
But each not-picked aglet can be on either end of its shoelace,
so multiply each of those ways by 2^(2*(n-k)) for a total of
2^(2*(n-k)) * C(n,2*(n-k)).

So we want 2^(2*(n-k)) * C(n,2*(n-k)) to be more than 1/2 of C(2n,2k).

Is that correct?

Gerald

At 08:49 PM 5/15/2005, Neil Fernandez wrote:
>You've got n shoelaces in a box. Each shoelace has two aglets.
>
>You pick up a pair of aglets at random, that may or may not be on the
>ends of the same shoelace.
>
>a(n) is the number of times you must do this, without replacement, to
>have a better than evens chance of removing all the shoelaces if you tug
>on all of the aglets you've got in your hand.
>
>I get the mean number of necessary times as 1, 4/3, 33/15, 102/35,...
>giving a sequence a(n):
>
>1, 2, 3, 3, ...
>
>Help would be appreciated in extending this sequence.
>
>Neil
>
>--
>Neil Fernandez