I’m revising my probability basics and got tripped up by a socks question that feels simple but my brain turned it into spaghetti.
I have a drawer with 6 black socks, 4 white socks, and 2 blue socks (so 12 total). If I pull out two socks at random without looking and without replacement, what’s the probability they are the same color?
My attempt: I figured the first sock can be anything. Then the chance the second matches depends on what the first was:
– if first was black: 5/11
– if first was white: 3/11
– if first was blue: 1/11
At first I just averaged those match-probabilities and called it a day, which I now suspect is wrong. Then I wondered if I should weight them by the chance the first sock was each color (6/12, 4/12, 2/12). I also briefly tried to treat the two draws as independent (which they clearly aren’t), so I’m probably mixing ideas.
Could someone point out the correct way to set this up and explain why my “just average the conditionals” idea fails here? Also, is there a simple combination-based way to see it, and are these approaches equivalent?
I’m mainly trying to strengthen my fundamentals, so I’d love to understand the reasoning rather than just memorize a trick.
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3 Responses
Your instincts were mostly fine: pick any first sock, then ask “what’s the chance the second matches it?” The trap is averaging those three match-probabilities (5/11, 3/11, 1/11) as if the first sock were equally likely to be black, white, or blue-nope. You’ve got more black socks, so that case happens more often; you have to weight by P(first color), i.e., (6/12)(5/11) + (4/12)(3/11) + (2/12)(1/11) = (30+12+2)/132 = 44/132 = 1/3. If you prefer combinations: total unordered pairs is C(12,2)=66; same-color pairs are C(6,2)+C(4,2)+C(2,2)=15+6+1=22; so 22/66 = 1/3. These two methods are the same thing in different clothes: the “multiply and add” is just the law of total probability, and the combinations way pretends order doesn’t matter (and, weirdly, that’s fine even though the draws aren’t independent-multiplying still works because you’re conditioning on the first pick). Why your plain average fails: the three scenarios aren’t equally likely; averaging would only be okay if the colors were balanced, like 4–4–4, where each conditional is 3/11 anyway. Quick toy example to see the weighting: suppose 2 black, 1 white. Conditional match chances are 1/2 if first is black and 0 if first is white; the wrong plain average gives (1/2 + 0)/2 = 1/4, but the right weighted result is (2/3)(1/2) + (1/3)(0) = 1/3. So for your drawer, the probability the two socks match is 1/3.
Your instinct to look at the “second sock matches the first” probabilities was spot on, but the part that trips people (including past-me) is that you can’t just average those 5/11, 3/11, and 1/11 numbers because the first sock isn’t equally likely to be black, white, or blue. You have to weight by how often each first color happens: P(same color) = (6/12)(5/11) + (4/12)(3/11) + (2/12)(1/11) = 5/22 + 1/11 + 1/66 = 22/66 = 1/3. The “just average the conditionals” idea fails because it treats the three starting colors as equally probable, which they aren’t (there are more black socks, so “first is black” happens more often and should carry more weight). There’s a tidy combinations view that says the same thing without the conditioning: count favorable pairs over all pairs. Favorable = C(6,2) + C(4,2) + C(2,2) = 15 + 6 + 1 = 22; total = C(12,2) = 66; so 22/66 = 1/3. These are equivalent because both are just different flavors of the same idea: summing over mutually exclusive cases with the right weights. I remember once averaging the match-probabilities “just to be efficient” and being baffled why my teacher’s answer was bigger-turns out efficiency without weighting is like making lemonade without sugar: technically a drink, but not what you wanted.
Your setup is on the right track: treat the first sock as “anything,” then weight the matching chance by how likely that first sock’s color is. By the law of total probability, P(same color) = (6/12)(5/11) + (4/12)(3/11) + (2/12)(1/11) = (30 + 12 + 2)/132 = 44/132 = 1/3 (quick check: 44/132 simplifies to 1/3). The reason the unweighted average of 5/11, 3/11, and 1/11 fails is that the first sock is not equally likely to be black, white, or blue; black is more common, so its conditional should count more-if the colors were equally likely (or if we were sampling with replacement), then that naive average would line up, but here it doesn’t. A clean combinations view gives the same result: total pairs C(12,2) = 66; favorable pairs C(6,2) + C(4,2) + C(2,2) = 15 + 6 + 1 = 22; so 22/66 = 1/3. Both approaches are equivalent: one conditions on the first draw, the other counts unordered pairs directly; the ordering details cancel out either way. As a quick extension, if you changed the counts to 7 black, 4 white, 1 blue, how would you update the weighted-sum or combination calculation?