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irchans
10-20-2002, 09:09 AM
A drawer contains red socks and black socks. When two socks are drawn at random, the probabilitiy that both are red is 1/2. (a) How small can the number of socks in the drawer be? (b) How small if the black socks is even? (c) How small if there are at least 30 socks?

(Mosteller)

10-20-2002, 02:00 PM
I don't quite understand your question as worded but that has neve stopped me from answering yet! /forums/images/icons/smile.gif

A) 3
B) 4
C) 2

Jimbo

Now if you are asking what is the minimum number of socks you must draw before you are assured of retrieving 2 red socks my answer to C) would be 16. If you are asking how many before it is more likely than not that you will have 2 red socks, I'll leave that to the real mathmaticians.

PseudoPserious
10-20-2002, 07:51 PM
Nifty.

a) 4 (r=3, b=1)
b) 21 (r=15, b=6)
c) 120 (r=85, b=35)

Let (r/(r+b))((r-1)/(r+b-1)) = .5. Simplify. Find the roots that satisfy the given constraints.

PP

BruceZ
10-20-2002, 08:30 PM
This reduces to finding integer solutions to t(t-1) = 2r(r-1). I also found the above solutions using excel. I'm guessing we can use number theory to find all the solutions in general, right irchans?

irchans
10-21-2002, 05:32 AM
BruceZ, Pseudo,

You got the right answers! Number Theory is certainly involved in the "general answer" which is beyond me. Mosteller writes, "... we would be wise to appreciate that this is a problem in the theory of numbers. It happens to lead to a famous result in Diophantine Analysis obtained from Pell's equation."

Pell's equation is

n x^2 + 1 = y^2 .

I don't see how this is related to the sock problem. Mollester gives a reference (Elementary theory of numbers by LeVeque.) If I get some time tomorrow, I will look it up.

Cheers, Irchans

Bozeman
10-22-2002, 01:09 AM
Mollester gives a reference

Is this a Freudian slip?

thylacine
09-16-2003, 03:47 PM
I stumbled across this thread via the more recent thread "Probability Problem (non-poker)."

BruceZ says we are solving t(t-1) = 2r(r-1).

irchans says he reads of a connection to Pell's equation: n x^2 + 1 = y^2 .

The connection is this. Let T=2t-1, R=2r-1.

Then t(t-1) = 2r(r-1) becomes 2R^2=T^2+1.

(I don't know how to solve these or what is known.)

thylacine
09-16-2003, 04:25 PM
It just struck me that solving 2R^2=T^2+1 is related to the continued fraction for sqrt(2).

The bottom line is that if (R,T) is a solution, then so is (3R+2T,4R+3T).

Starting with (R,T)=(1,1) you generate infinitely many solutions, and I think these might be all.

So Solutions are (R,T)=(1,1), (5,7), (29,41), (169,239), ...

Then translate back to get solutions (r,t) to t(t-1) = 2r(r-1).