Probability - Socks in Drawer

[sanfordgeraci]

I thought of this problem today while I was doing the dreadful thing of
folding my socks!! All of my socks are identical, hence minimizing the task
of matching (and folding) them.

Suppose there exists twenty identical socks in a drawer. Socks are worn by
simply picking two of the identical socks and wearing them. If two socks
have been worn on the same day, they are considerred a pair. If the socks
were only worn once before (i.e. one pair has been worn), what is the
probability that today I wear a pair that has been worn before? What is the
probability that within this pair, they are warn on same feet as yesterday?
Clearly the answers to this question (I believe) are 1/(20C2) and
1/(2*(20C2)) respectively. Now, (for the harder problem), suppose that I
have worn a pair of socks on each of the last three days, compute the
probability that today I choose a pair that I have worn in the past three
days (not necessarily on the same feet since that issue is easy to modify)?
Can we generalize this answer for the case where pairs of socks have been
worn for the last n days?

Hint: Remember that probabilities are less than one. Also, should we
consider cases in which there has been repetition in the last n days? Would
complements be helpful?

[vishvas]

Given:

there exists twenty identical socks in a drawer.

If two socks have been worn on the same day, they are considerred a
pair.

Problem: If I have worn a pair of socks on each of the last three

days, compute the probability that today I choose a pair that I have
worn in the past three days (not necessarily on the same feet since
that issue is easy to modify)? Can we generalize this answer for the
case where pairs of socks have been worn for the last n days?

let event e(n) = a pair of socks were worn on each of the past n days,
and you select one of those pairs.

let probabilistic estimate of number of pairs worn in n days = y(n)
Note that y(n) <= 20C2 and that y(1) = 1

Then, P[e(1)] = 1/20C2
y(2) = P[e(1)] * 1 + (1- P[e(1)]) * 2

P[e(2)] = (1- P[e(1)])*2*P[e(1)] + (P[e(1)])^2
= 2*P[e(1)] - (P[e(1)])^2

Note that P[e(2)] can also be equivalently calculated thus:
P[e(2)] = y(2) * P[e(1)]

Now, y(3) = P[e(2)]*y(2) + (1-P[e(2)])*(y(2)+1)
= y(2)+1-P[e(2)]

In general, for all n greater than 1,
y(n) = P[e(n)]*y(n) + (1-P[e(n)])*(y(n)+1) = y(n)+1-P[e(n)]
And, P[e(n+1)] = y(n+1)*P[e(1)]

P[e(3)] can easily be calculated from this. The generalization may
also be derived in this way.

Please excuse the lack of complete rigour.

-Sincerely,
Vishvas Vasuki