The Sock Drawer

(a) Minimum number of socks:

Let:

• $y \in \mathbb{N}$ = number of yellow socks
• $b \in \mathbb{N}$ = number of blue socks
• $n = y + b$ = total number of socks

The probability that both of the drawn socks are yellow is:

$$ \frac{y(y-1)}{n(n-1)} $$

We want the smallest $n$ such that some $y < n$ satisfies this equation:

$$ \frac{y(y-1)}{n(n-1)} = \frac{1}{2} $$

Try small values of $n$:

• $n = 3$, $y = 2$ $\Longrightarrow$ $\frac{y(y-1)}{n(n-1)}=\frac{2(1)}{3(2)} = \frac{1}{3}$ ❌
• $n = 4$, $y = 3$ $\Longrightarrow$ $\frac{y(y-1)}{n(n-1)}=\frac{3(2)}{4(3)} = \frac{1}{2}$ ✔️

Answer: The smallest number $n$ of socks there can be in the drawer is $n = 4$.

(b) Minimum number of socks with an even number of blue socks:

We now look for the smallest $n$ such that:

• $\frac{y(y-1)}{n(n-1)} = \frac{1}{2}$
• and $n-y$ (=the number of blue socks) is even

You can find valid $(n, y)$ pairs using this brute-force Python script:

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for n in range(1, 100):
    for y in range(2, n):
        if y * (y - 1) == (n * (n - 1)) / 2:
            if (n - y) % 2 == 0:
                print(f"n = {n}, y = {y}")

There is only one such pair : $$(n,y)=(21,15)$$

$n-y=6$ is even so the answer is:

Answer: The smallest number $n$ of socks there can be in the drawer with an even number of blue socks is $n = 21$.