Socks in a Drawer

Probability Level 2

One morning, you wake up with no power. You stumble to your dresser and try to pull out some socks. You have x x socks of one color and y y socks of another color. You pull out two socks from the drawer, without replacement, and you realize that the probability you get a matching pair of socks is exactly 50%.

For every integer n ≥ 2 n \geq 2 , does there exist a pair of numbers ( x , y ) (x, y) such that x − y = n ? x-y = n?

Yes No

Calvin Godfrey by Calvin Godfrey , 20, USA

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1 solution

Calvin Godfrey
Oct 31, 2017

I'm going to start off by saying that I don't know how to solve the full problem, however I can do it with wolfram alpha and so it's likely that some people can do it completely on their own. So based on the information given, you have the equality

x x + y x − 1 x + y − 1 + y x + y y − 1 x + y − 1 = 1 2 \frac{x}{x+y}\frac{x-1}{x+y-1}+\frac{y}{x+y}\frac{y-1}{x+y-1}=\frac{1}{2}

Each pair of fractions represents the probability that you get a pair of a specific color; the first pair of fractions represents the probability that you get two socks of color x x and the second pair represents the probability that you get a pair of color y y . From here, it's a lot of algebra.

x ( x − 1 ) ( x + y ) ( x + y − 1 ) + y ( y − 1 ) ( x + y ) ( x + y − 1 ) = 1 2 \frac{x(x-1)}{(x+y)(x+y-1)}+\frac{y(y-1)}{(x+y)(x+y-1)} = \frac{1}{2}

x 2 − x + y 2 − y x 2 + 2 x y + y 2 − x − y = 1 2 \frac{x^2-x+y^2-y}{x^2+2xy+y^2-x-y} = \frac{1}{2} 2 x 2 + 2 y 2 − 2 x − 2 y = x 2 + 2 x y + y 2 − x − y 2x^2+2y^2-2x-2y = x^2+2xy+y^2-x-y x 2 − 2 x y + y 2 − x − y = 0 x^2 -2xy + y^2-x-y = 0 ( x − y ) 2 = x + y (x-y)^2 = x+y

From here, I wrote a program to check every combination of numbers x , y x, y for x < 1000 , y < 1000 x < 1000, y < 1000 and I found that a pair of numbers satisfies the above equation if and only if they are adjacent triangle numbers, where a triangle number is the sum of all natural numbers from 1 to some number m m . The first few are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, etc. This means that since every pair of adjacent triangle numbers satisfies this equation, every integer n ≥ 2 n \geq 2 will equal x − y x-y for some pair of integers.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

You're almost there with your proof. Starting from your last equation:

( x − y ) 2 = x + y (x-y)^2=x+y

We know that x − y = n . x-y=n. Then y = x − n . y=x-n. Substituting into the equation:

n 2 = 2 x − n 2 x = n 2 + n x = n ( n + 1 ) 2 \begin{aligned} n^2 &= 2x-n \\ 2x &= n^2+n \\ x &= \frac{n(n+1)}{2} \end{aligned}

y = n ( n + 1 ) 2 − n y = n ( n − 1 ) 2 \begin{aligned} y &= \frac{n(n+1)}{2}-n \\ y &= \frac{n(n-1)}{2} \end{aligned}

We can see now that for every n ≥ 2 , n \ge 2, there exist an x x and y . y. We can also see that x x and y y are necessarily triangular numbers.

Andy Hayes - 3 years, 7 months ago

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