TITLE: How much space do I need to sort my socks?
QUESTION [11 upvotes]: In my pile of finished laundry, there are $2n$ socks of $n$ types, comprising $n$ easily distinguishable pairs. I sort the socks into pairs by picking one sock at a time randomly from the pile and either (1) laying it in a row of distinct socks or (2) pairing it with a sock already in the row, if there is such a sock, and putting the pair away. What is the expected maximum number of socks in the row?
Let $N$ be the peak number of socks in the row. Clearly $1\leqslant N\leqslant n$. And obviously, if $n=1$, then $N=\mathrm E[N]=1$. For $n=2$, I get $\mathrm E[N]=\frac53$. As $n$ increases, the calculation gets more complicated, and I don't know whether $\mathrm E[N]$ has a reasonable explicit general formulation. If not, perhaps it still has a nice asymptotic expression for large $n$. My guess is that there are positive constants $\alpha$ and $\beta$ such that $\lim_{\,n\rightarrow\infty}\mathrm E[N]/n^\beta=\alpha$.

REPLY [7 votes]: Expanding my comments into an answer per request.
Let $N_n$ be the random variable for the peak number of socks for a given $n$. It is known that
$$\overline{N}_n := \mathbf{E}[N_n] = 1, \frac{5}{3}, \frac{7}{3}, \frac{311}{105}, \frac{3377}{945}, \frac{3943}{945}, \ldots
$$
or in general $\displaystyle\overline{N}_n = \frac{a_n}{b_n}$ where 
$a_n$ = OEIS A225177 and $b_n = (2n-1)!! = \prod\limits_{k=1}^n (2k-1)$.
As of this moment, closed form of the sequence is an unknown. 
For more infos, one can follow the reference in the OEIS A225177 or consult the M/J2 section of this article.
There are two things in above article I'll like to comment.

It described an algorithm to compute the probability distribution for $N_n$.
It mentioned as $n \to \infty$, the probability distribution of $N_n$ becomes normal with mean $\frac{n}{2}$ and variance $\frac{n}{4}$.

I have implemented an algorithm similar to what described in the article and performed some 
numerical experiment of the probability distribution. 
Let $\mathbf{Var}_n = \mathbf{E}\left[ \left(N_n - \overline{N}_n \right)^2 \right]$  be the variance for $N_n$. Following table summarizes what I get:
$$\begin{array}{|r|rr:rr:r|}
\hline
N_n & \overline{N}_n & \overline{N}_n - \frac{n}{2} &
\mathbf{Var}_n & \mathbf{Var}_n/n & \Delta_n(2)\\
\hline
 200 &   105.107856 &     5.107856 &    40.733642 &    20.366821\% &     2.175188\%\\
 400 &   206.615246 &     6.615246 &    84.774930 &    21.193732\% &     1.600965\%\\
 600 &   307.674159 &     7.674159 &   129.707716 &    21.617953\% &     1.291395\%\\
 800 &   408.517714 &     8.517714 &   175.150639 &    21.893830\% &     1.124633\%\\
1000 &   509.230343 &     9.230343 &   220.941243 &    22.094124\% &     1.009333\%\\
1200 &   609.853430 &     9.853430 &   266.990895 &    22.249241\% &     0.923493\%\\
1400 &   710.410713 &    10.410713 &   313.244447 &    22.374603\% &     0.856246\%\\
1600 &   810.917224 &    10.917224 &   359.664606 &    22.479038\% &     0.802341\%\\
1800 &   911.383157 &    11.383157 &   406.224651 &    22.568036\% &     0.757826\%\\
2000 &  1011.815776 &    11.815776 &   452.904598 &    22.645230\% &     0.719742\%\\
\hline
\end{array}$$

As one can see, as $n$ becomes large, the leading behavior of $\overline{N}_n$ is indeed $\frac{n}{2}$. 
The difference between $\overline{N}_n$ and $\frac{n}{2}$ is a very slowly
growing function. The difference seems to grow like $O(n^\beta)$ with $\beta \sim 0.37$.
The variance $\mathbf{Var}_n$ does seem scale as $n$. However, the actual coefficient in front of $n$ is smaller than $\frac14$. It is not clear whether this is a finite size effect or the limiting coefficient is indeed different from $\frac14$.

About the claim the probability distribution of $N_n$ is normal, it actually works surprisingly well.
Let $\sigma_{n} = \sqrt{\mathbf{Var}_n}$ and $P_n(k) = \mathbf{Prob}[ N_n = k ]$ be the probability that $N_n$ equal to a particular value $k$. At the end is a picture comparing $P_n(k)$ versus a normal distribution for various $n$.

The $x$ axis is $\displaystyle\;x = \frac{k - \overline{N}_n}{\sigma_n}$.
The $y$ axis is the scaled probability $\sigma_n P_n(k)$. 
The $\color{orange}{\verb/orange/}$ curve is the PDF $\displaystyle\;\rho(x) = \frac{e^{-x^2/2}}{\sqrt{2\pi}}$ for a normal distribution of unit variance.

As one can see, for all the cases tested, the scaled probability $P_n(k)$ nearly fall right on top of $\rho(x)$. To measure how close the computed probability differ from the PDF of a normal distribution. Let us define following quantity
$$\Delta_n(s) = \max\left\{ \left|\frac{\sigma_nP_n(k)}{\rho(x)} - 1\right| : |x| < s \right\}$$
This measures the biggest difference between the ratio $\frac{\text{scaled probability}}{\text{reference PDF}}$ and $1$ for $x$ within $s$ standard derivation.
As shown in last column of table above. As long as we are within 2 standard derivation, 
the difference between the ratio and $1$ is at most a few percents and improves as $n$ increases.