I prompted Claude for a math puzzle and was suprised to see one that I did not know about. Figured it might be worth sharing here.

The Sock Drawer Paradox

You have a drawer with some red socks and some blue socks. You know that if you pull out 2 socks randomly, the probability they're both red is $\frac{1}{2}$

Given this information, what's the probability that the first sock you pull is red?

The Surprising Answer

Most people think "well if P(both red) = 1/2, then P(first red) should be set" but this problem has multiple solutions!

Let $r$ = number of red socks and $b$ = number of blue socks.

The probability of drawing two red socks is: $$P(\text{both red}) = \frac{r}{r+b} \times \frac{r-1}{r+b-1} = \frac{1}{2}$$

After symplifying and using the quadratic formula you get: $$r = b \pm 0.5 \sqrt{8.0 b^{2} + 1.0} + 0.5$$

This can be varified with sympy.

from sympy import symbols, Eq, solve

# Let r be the number of red socks and b be the number of blue socks
r, b = symbols('r b')

# The probability of pulling 2 red socks
probability_both_red = (r / (r + b)) * ((r - 1) / (r + b - 1))

# Given that this probability is 1/2
equation = Eq(probability_both_red, 1/2)

# Solve for r in terms of b
solution = solve(equation, r)

Integers

Not all of these solutions are integer solutions though. But you can find them via:

from sympy import lambdify 
from decimal import Decimal 

r_given_b = lambdify("b", solution[1])

[{"b": _, "r": int(r_given_b(_))} 
 for _ in range(1000000) 
 if (r_given_b(_) * 1000 % 1000) == 0]

There are 8 solutions in the first one million values for $b$.

b	r	$p(r)$	$p(rr)$
1	3	0.75	0.5
6	15	0.714286	0.5
35	85	0.708333	0.5
204	493	0.707317	0.5
1189	2871	0.707143	0.5
6930	16731	0.707113	0.5
40391	97513	0.707108	0.5
235416	568345	0.707107	0.5

What's interesting here is that the probability of grabbing a red ball converges as $b$ grows but it is not constant. There is still a degree of freedom when you only know the probability of grabbing two red balls!