A probability question regarding matching socks There is one probability question, saying there are three kinds of socks. And after putting those 6 socks into a drawer, socks are pulled two at a time (without seeing socks you pull). What are the odds of pulling a matching pair? 
The answer is 1/3. However, I was a little bit confused as I thought the odds could be 1/5. My logic was like that as the left hand pulls one sock, there are 5 socks remaining in the drawer. And the right hand must pulls the other one that matches. Thus the probability, based on what I thought, should be 1/5. Could anyone help me? Thanks a lot.
 A: You are completely correct. 
Simultaneously or not is completely irrelevant. 
There are $\binom62=15$ possible choices and $3$ of them are matching choices. 
That gives probability $\frac3{15}=\frac15$.
Also your logic is correct and in my view even more elegant.

addendum.
The correct interpretation of the probability question might have been: draw $3$ times a pair of socks without replacement. What is the probability that at least one of the pairs is matching?
It is handsome to calculate the probability of the complement. As shown above the probability that the first pair is not matching is $\frac45$. Assume that this happens and let's say that from socks $A_1,A_2,B_1,B_2,C_1,C_2$ we have drawn $A_1B_1$. Then $A_2,B_2,C_1,C_2$ are left and no matching pair will appear if the second draw will be one of $4$ equiprobable possibilities: $A_2C_1$, $A_2C_2$, $B_2C_1$ and $B_2C_2$. There are $6$ possibilities in total so there is chance of $\frac46=\frac23$ that this happens.
Then we come out on probability $\frac45\frac23=\frac8{15}$ that no matching pair is drawn. The probability that at least one matching pair has been drawn is $1-\frac8{15}=\frac7{15}$.
A: My interpretation of the problem statement is that we are asked to find the probability that at least one of the three pairs of socks drawn is a match.  The following solution uses the Principle of Inclusion / Exclusion, abbreviated PIE.
Let's say the socks are three different colors, numbered $1,2,3$.  If we draw the socks one at a time without replacement, there are $6!$ possible sequences, all of which we assume are equally likely.  Let's say the pairs are numbers$(1,2)$, $(3,4)$, and $(5,6)$ in the sequence.  We would like to count the number of sequences in which at least one pair has two socks of the same color.  To that end, say a sequence of six socks has "Property $i$" if the two socks of color $i$ are paired.  Let $S_j$ be the number of sequences which have $j$ of the properties, for $j=1,2,3$.  Then
$$\begin{align}
S_1 &= \binom{3}{1} \binom{3}{1} \cdot 2 \cdot 4! \\
S_2 &= \binom{3}{2} \binom{3}{2} \cdot 2! \cdot 2^2 \cdot 2! \\
S_3 &= \binom{3}{3} \binom{3}{3} \cdot 3! \cdot 2^3  \\
\end{align}$$
Expanation of the above: For $S_1$, we have $\binom{3}{1}$ ways to pick the color, $\binom{3}{1}$ ways to pick the pair from $(1,2)$, $(3,4)$, and $(5,6)$ , $2$ ways to order the socks of the chosen color, and $4!$ ways to arrange the remaining socks.  For $S_2$, we have $\binom{3}{2}$ ways to select the two colors, $\binom{3}{2} \cdot 2!$ ways to select the two pairs from $(1,2)$, $(3,4)$, and $(5,6)$, $2^2$ ways to order the socks of the chosen colors, and $2!$ ways to order the remaining two socks.  $S_3$ is similar.
By PIE, the number of sequences with at least one of the properties, i.e. the number of sequences with at least one matching pair, is $S_1-S_2+S_3$, and the probability of getting at least one matching pair is
$$\frac{S_1-S_2+S_3}{6!} = \frac{7}{15}$$
