What's the probability of getting at least one pair in sequence after drawing 10 socks (5 different pairs)? This has been rattling my brain for a while now.
I've got 5 pairs of socks, so 10 in total.
I'm pulling them out of the laundry one by one.
By the time I'm done with my laundry, what is the probability of getting two socks from the same pair in a row (at least once)?
An example line-up:
A B C D D E A B C E
I know the number of total possible combinations is 10!, divided by 25, but finding the number of desired outcomes (at least one pair anywhere in the lineup) has proven too challenging for me.
Thanks!
 A: If we assume that two socks in a pairs are identical :
By P.I.E ,we want the cases where the sequence contains $AA$ or $BB$ or $CC$ or $DD$ or $EE$.

*

*The number of sequences containing $AA$  : $$\frac{9!}{2^4}$$
The foregoing expression is also valid for the rest.

*

*The number of sequences containing two consequtive pairs such as $AA$ and $BB$ : $$\frac{8!}{2^3}$$


*The number of sequences containing three consequtive pairs such as $AA$ and $BB$  and $CC$: $$\frac{7!}{2^2}$$


*The number of sequences containing four consequtive pairs such as $AA$ and $BB$  and $CC$ and $DD$: $$\frac{6!}{2}$$


*The number of sequences containing four consequtive pairs such as $AA$ and $BB$  and $CC$ and $DD$ and $EE$: $$5!$$
By P.I.E : $$\binom{5}{1}  \frac{9!}{2^4} -\binom{5}{2}  \frac{8!}{2^3}+\binom{5}{3}  \frac{7!}{2^2}-\binom{5}{4}  \frac{6!}{2}+\binom{5}{5}  5!$$
So , the probability is $$\frac{\binom{5}{1}  \frac{9!}{2^4} -\binom{5}{2}  \frac{8!}{2^3}+\binom{5}{3}  \frac{7!}{2^2}-\binom{5}{4}  \frac{6!}{2}+\binom{5}{5}  5!}{\frac{10!}{2^5}}$$
NOTE: If two socks in a pair were non-identical , then the selection order of $A_1$ and $A_2$ must have been taken into account . Moreover , the denominator part would be $10!$ , because all socks are different.
