Sock It To Me

Probability Level 2

You have twice as many white socks as black socks in your sock drawer. If the probability of randomly picking 2 2 matched socks (without replacement) is the same as the probability of randomly picking 2 2 mismatched socks (without replacement), how many socks total are in your sock drawer?

(Assume that you have at least one sock in your sock drawer, that any 2 2 white socks match each other and any 2 2 black socks match each other, and that you have no other colored socks.)


The answer is 9.

David Vreken by David Vreken , 42, USA

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1 solution

Mark Hennings
Aug 15, 2019

If there are 2 n 2n white socks and n n black socks, then the probability of drawing two matched socks is 2 n ( 2 n − 1 ) + n ( n − 1 ) 3 n ( 3 n − 1 ) = 5 n 2 − 3 n 9 n 2 − 3 n \frac{2n(2n-1) + n(n-1)}{3n(3n-1)} \; = \; \frac{5n^2 - 3n}{9n^2 - 3n} and this probability should be equal to 1 2 \tfrac12 , so 2 ( 5 n 2 − 3 n ) = 9 n 2 − 3 n n 2 − 3 n = 0 \begin{aligned} 2(5n^2 - 3n) & = \; 9n^2 - 3n \\ n^2 - 3n & = \; 0 \end{aligned} and hence (since n ≠ 0 n \neq 0 ), n = 3 n=3 , so there are 9 9 socks in the drawer.

9 Helpful 2 Interesting 2 Brilliant 0 Confused

Let's define some variables:

  • Number of white socks: n w = 2 a n_w = 2a
  • Number of black socks: n b = a n_b = a
  • Total number of socks: n = n w + n b = 3 a n = n_w + n_b = 3a

Probability on similar socks: P ( w , w ) + P ( b , b ) = n w ( n w − 1 ) n ( n − 1 ) + n b ( n b − 1 ) n ( n − 1 ) = 2 a ( 2 a − 1 ) n ( n − 1 ) + a ( a − 1 ) n ( n − 1 ) = 2 a ( 2 a − 1 ) + a ( a − 1 ) n ( n − 1 ) = 5 a 2 − 3 a n ( n − 1 ) \begin{aligned} P(w,w) + P(b, b) &= \frac{n_w (n_w - 1)}{n (n-1)} + \frac{n_b (n_b - 1)}{n (n-1)} \\ &= \frac{2a(2a-1)}{n(n-1)} + \frac{a(a-1)}{n(n-1)} \\ &= \frac{2a(2a-1) + a (a-1)}{n(n-1)} = \frac{5a^2 - 3a}{n(n-1)} \end{aligned}

Probability on different socks: P ( w , b ) + P ( b , w ) = 2 P ( w , b ) = 2 ⋅ n w n b n ( n − 1 ) = 2 ⋅ 2 a ⋅ a n ( n − 1 ) = 4 a 2 n ( n − 1 ) \begin{aligned} P(w,b) + P(b, w) &= 2 P(w,b) \\ &= 2 \cdot \frac{n_w n_b}{n (n-1)} \\ &=2 \cdot \frac{2a \cdot a}{n(n-1)} = \frac{4a^2}{n(n-1)} \end{aligned}

These probabilities are equal. Assuming n ( n + 1 ) ≠ 0 n (n+1) \neq 0 we can write:

5 a 2 − 3 a = 4 a 2 a 2 = 3 a a = 3 \begin{aligned} 5a^2 - 3a &= 4a^2 \\ a^2 &= 3a \\ a &= 3 \end{aligned}

Therefore: n = 3 a = 9 n = 3a = \boxed{9}

Laurent Verweijen - 1 year, 6 months ago

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