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I'm very unsure when it comes to probability, but this is my work / thought process so far.

Chances of everyone having matching socks:

Sample space size is 8!.

In order to get matching socks, first any of 8 are picked, the matching one must be picked, then any of 6, then the matching one, then any of 4, then the matching one, then any of 2, then the matching one.

So, size of event is 8 x 6 x 4 x 2 = 384

So, the probability is 384/8!

Chances of exactly one person having matching socks:

Sample space size is 8!.

Then one person chooses one sock at random, then the matching one, so 8 x 1. That should be multiplied by (6!-3!2!2!2!).

StrugglingMathStudent
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2 Answers2

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Your first answer is correct. You're distinguishing between the orders in which the people choose their two socks. Another approach would be not to distinguish between those orders, and only distinguish between the $\frac{8!}{2!^4}$ assignments of unordered pairs. Then the number of favourable assignments is simply $4!$, the number of permutations of the four coloured pairs. There result is of course the same; the additional factor $2!^4$ that you have in the numerator and denominator that counts the orders within the pairs cancels out.

In the second answer, you're only subtracting the cases in which the other three people have completely matching pairs; but you need to subtract all cases in which anyone else has a matching pair. You're also missing a factor of $4$ because any of the four people could be the one with the matching pair.

One approach to get this right is inclusion–exclusion. The probabilities $p_k$ for at least $k$ particular persons to have matching socks are $p_1=\frac17$, $p_2=\frac17\cdot\frac15=\frac1{35}$ and $p_3=p_4=\frac17\cdot\frac15\cdot\frac13=\frac1{105}$, so by generalized inclusion–exclusion the probability that exactly $1$ person has matching socks is

$$ \sum_{k=1}^4(-1)^{k+1}\binom k1\binom4kp_k=4\cdot\frac17-2\cdot6\cdot\frac1{35}+3\cdot4\cdot\frac1{105}-4\cdot\frac1{105}=\frac{32}{105}\;. $$

Alternatively, the probability for one particular person to be the only one with matching socks is $\frac17\cdot\frac45\cdot\frac23$ (that person must pick a matching sock, the next person must not pick a matching sock, and no matter which first sock the third person picks, $2$ out of $3$ choices for the second sock avoid another pair), so the probability for any of the $4$ people to be the only one with matching socks is $4\cdot\frac17\cdot\frac45\cdot\frac23=\frac{32}{105}$.

joriki
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    Interesting that the order in which the person picks the matching pair doesn't matter. For example, if the 2nd person has the matching pair then the probability sequence will be $\frac{6}{7}\cdot \frac{2}{15} \cdot \frac{2}{3} = \frac{8}{105}$ which is the same as if the first person had the matching pair. – Phil H Sep 08 '18 at 19:17
  • @PhilH: I don't see how it could be otherwise. All $8!$ permutations of the socks are equally likely; the procedures that we imagine that lead to these permutations might appear to distinguish among the people, but by definition they must all be equivalent, else they wouldn't be procedures for uniformly randomly choosing a sock assignment; the conclusion in that case would not be that different procedures lead to different probabilities, but that some of these procedures are wrong. (I assume here and in the answer that a uniformly random assignment was implied in the question.) – joriki Sep 08 '18 at 19:35
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Another method for part one is to use a pairs sample space which is $\binom{8}{2} = 28$ possible pairings.

Therefore the first person has a $4$ out of $28$ = $\frac{1}{7}$ chance of picking a matching pair. The second person $3$ out of $15$ = $\frac{1}{5}$ and the third person $2$ out of $6$ = $\frac{1}{3}$. The fourth pair is automatic.

So, $P(\text{4 match}) = \frac{1}{7}\cdot \frac{1}{5}\cdot \frac{1}{3} = \frac{1}{105}$

Phil H
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