Probability: Matching socks $n$ left socks are in  a drawer and n right socks are organized in a line. Each sock in the drawer has only one match in the line. On the first round one sock was randomly puled out of the drawer and matched with the first sock in the line. Same goes for all the rest. what are the chances that the first match and the last one are right pairs?
I am absolutely not sure about the way i should solve this. Thanks for your help!.
 A: Probability that the first sock will match is $1/n$. Probability that the second sock won't pick the matching sock of the last sock is $(n-2)/(n-1)$. Probability that the third sock won't pick the matching sock of the last sock is $(n-3)/(n-2)$.... Probability that $(n-2)^{th}$ sock won't match the matching sock of the last one is $2/3$. Probability that $(n-1)^{th}$ sock won't match the matching sock of the last one is $1/2$.
So the final result is:
$$\frac 1n\cdot\frac{n-2}{n-1}\cdot\frac{n-3}{n-2}\cdot\frac{n-4}{n-3}\cdot\dots\cdot
\frac23\cdot\frac12=\frac1{n(n-1)}$$
The result is the same if you consider, for example, the first and the second sock.
Simpler explanation. The chance for the first sock to get the matching one is $1/n$. the last sock can be matched with any other with equal probability. So probability that the last (second, third, seventh...) sock will be paired with the matching one is $1/(n-1)$ because you have one sock less. So the total probability is the product of these two:
$$\frac1{n(n-1)}$$
