My grandfather's socks

Question

My grandfather has a big drawer where he keeps his socks. The drawer contains more than 900 but less than 1000 individual socks. Each of his socks is black or blue, and there are more blue socks than black socks. The socks aren't paired by color within the drawer, but if he reaches into it and grabs two socks at random, then exactly half the time he will pull a matching pair.

My grandfather has always been extremely careful when doing the laundry, and has only ever lost a single sock in his entire life.

What color was that lost sock?

Answer 1

Your grandfather lost a black sock.

This is because

Let $a$ be the number of blue socks and $b$ be the number of black socks.
To get exactly 50% matching, you will need
$~~~~~ a(a-1) + b(b-1) = 2ab$
which can be rewritten as
$~~~~~ (a-b)^2 = a+b$.
As $a+b$ is a square with $900 < a+b < 1000$, we get $a+b=961$. Then $a > b$ yields $a-b=31$. The only remaining possibility is $a=496$ blue socks and $b=465$ black socks, and the lost sock was black.

Answer 2

Assuming he bought pairs of socks, then either the number of blue socks $(u)$ or the number of black socks $(k)$ is odd with the other is even.

$$u>k$$ $$p = .5 = \frac u {(k+u)}\frac k {(k+u-1)}+\frac k {(k+u)}\frac u {(k+u-1)}$$ $$p = .5 = \frac {2uk} {(k+u)(k+u-1)}$$ $$p = .5 = \frac u {(k+u)}\frac {u-1} {(k+u-1)}+\frac k {(k+u)}\frac {k-1} {(k+u-1)}$$ $$p = .5 = \frac {u^2-u+k^2-k} {(k+u)(k+u-1)}$$ $$0=u^2+k^2-u-k-2uk$$

$$u=n^2/2+n/2$$ $$k=n^2/2-n/2$$ $$u+k=n^2$$

so

$$n=31$$ $$u=496$$ $$k=465$$

This means grandpa lost a black sock.

Answer 3

The lost sock was

black

Let $B$ be the event that grandpa pulls a blue sock and $K$ be the event that he pulls a black sock. Also let $b$ be the number of blue socks and $k$ be the number of black socks. We are told that $b>k$ and $900<b+k<1000$.

There are four possibilities for the events of pulling two socks: $BB$, $BK$, $KB$, and $KK$. The probability of choosing a matching pair, i.e. pulling two socks without replacement, is $P(BB) + P(KK)$.

Calculations:

$$\begin{align} P(BB) & = \left( \frac b{b+k} \right) \left( \frac {b-1}{b+k-1} \right) \\ \\& = \frac {b^2-b}{b^2+2bk+k^2-b-k} \\ \\P(KK) & = \left( \frac k{b+k} \right) \left( \frac {k-1}{b+k-1} \right) \\ \\& = \frac {k^2-k}{b^2+2bk+k^2-b-k} \\ \\P(BB) + P(KK) & = \frac {b^2-b}{b^2+2bk+k^2-b-k} + \frac {k^2-k}{b^2+2bk+k^2-b-k} \\ \\& = \frac {b^2-b+k^2-k}{b^2+2bk+k^2-b-k}\end{align}$$

But we know that:

$$\begin{align} P(BB) + P(KK) & = \frac 12 \\ \\ \frac {b^2-b+k^2-k}{b^2+2bk+k^2-b-k} & = \frac 12\end{align}$$

Multiplying both sides by both denominators:

$$\begin{align}2b^2-2b+2k^2-2k & = b^2+2bk+k^2-b-k \\b^2-2bk+k^2 & = b+k \\(b-k)^2 & = b+k\end{align}$$

So

$b+k$ is a perfect square and $900<b+k<1000$. The only perfect square in that range is $961=31^2$, so $b+k=961$, i.e. there are 961 socks total. But this is $(b-k)^2$, so $b-k=31$.

Finally:

$$ \begin{align} b+k & = 961 \\ b-k & = 31 \\ 2b &= 992 \\ b &= 496 \\ k &= 465 \\ \end{align} $$

Since the odd number is

$k$

the missing sock was

black.

Answer 4

I'm going to feel very stupid if this is wrong, but...shouldn't the lost sock be

black since he has more blue socks than black ones, and his odds of pulling a matching pair are 50%?

Even looking at all of the computations, I can't see a reason that my logic would be insufficient.

Answer 5

"Each of his socks is black or blue, and there are more blue socks than black socks"

followed by

"My grandfather has always been extremely careful when doing the laundry, and has only ever lost a single sock in his entire life."

I stopped going further immediately, given that socks are only ever sold in pairs, it had to be black. Sorry, was this question supposed to only be looking for mathematical proofs, and not common sense?