Solution:

Answer: $\frac{-1}{2}+\frac{(1007)\binom{2013}{1006}}{2^{2012}}$

We want to calculate $\Gamma=\sum_{i=0}^{\infty} i \cdot P(\max$ profit $=i)$, where we consider the maximum profit Rosencrantz has at any point over the first 2013 coin flips. By summation by parts this is equal to $\sum_{a=1}^{2013} P(\max$ profit $\geq a)$.

Let $p_{a}$ be the probability that Rosencrantz' max profit is at least $a$ and let $E_{n, a}$ be the set of sequences of flips such that Rosencrantz first reaches a profit of $a$ on exactly the $n$th flip. Let $Q_{a}^{+}, Q_{a}^{-}, Q_{a}$ be the sets such that after all 2013 flips Rosencrantz' final profit is (respectively) greater than, less than, or equal to $a$.

Then,
$$\begin{aligned} \Gamma & =\sum_{a=1}^{2013} P(\text{ max profit } \geq a) \\ & =\sum_{a=1}^{2013} \sum_{n=a}^{2013} P\left(E_{n, a}\right) \\ & =\sum_{a=1}^{2013} \sum_{n=a}^{2013} P\left(E_{n, a} \cap Q_{a}^{+}\right)+P\left(E_{n, a} \cap Q_{a}^{-}\right)+P\left(E_{n, a} \cap Q_{a}\right) . \end{aligned}$$
By symmetry, $P\left(E_{n, a} \cap Q_{a}^{+}\right)=P\left(E_{n, a} \cap Q_{a}^{-}\right)$ because, for any sequence in $E_{n, a} \cap Q_{a}^{+}$, we can reverse all the flips after the $n$th flip to get a sequence in $E_{n, a} \cap Q_{a}^{-}$, and vice-versa.

Furthermore, $\sum_{n=a}^{2013} P\left(E_{n, a} \cap Q_{a}^{+}\right)=P\left(Q_{a}^{+}\right)$ and $\sum_{n=a}^{2013} P\left(E_{n, a} \cap Q_{a}\right)=P\left(Q_{a}\right)$.

So we have
$$\sum_{a=1}^{2013} P(\text{ max profit } \geq a)=\sum_{a=1}^{2013}\left(P\left(Q_{a}\right)+2 P\left(Q_{a}^{+}\right)\right)$$
Since by symmetry $P\left(Q_{a}\right)=P\left(Q_{-a}\right)$ and we have an odd number of flips, we have $\sum_{a=1}^{2013} P\left(Q_{a}\right)=\frac{1}{2}$.

Also $P\left(Q_{a}^{+}\right)=\frac{1}{2^{2013}} \sum_{k=\left\lceil\frac{2014+a}{2}\right\rceil}^{2013}\binom{2013}{k}$.

So the rest is just computation. We have:
$$\begin{aligned} \Gamma & =\frac{1}{2}+\frac{1}{2^{2012}} \sum_{a=1}^{2013} \sum_{k=\left\lceil\frac{2014+a}{2}\right\rceil}^{2013}\binom{2013}{k} \\ & =\frac{1}{2}+\frac{1}{2^{2012}} \sum_{k=1008}^{2013} \sum_{a=1}^{2 k-2014}\binom{2013}{k} \\ & =\frac{1}{2}+\frac{1}{2^{2012}} \sum_{k=1008}^{2013}\binom{2013}{k}(k+k-2013-1) \\ & =\frac{1}{2}+\frac{1}{2^{2012}} \sum_{k=1008}^{2013} 2013\binom{2012}{k-1}-2013\binom{2012}{k}-\binom{2013}{k} \\ & =\frac{1}{2}+\frac{2013\binom{2012}{1007}-2^{2012}+\binom{2013}{1007}}{2^{2012}} \\ & =\frac{-1}{2}+\frac{(1007)\binom{2013}{1006}}{2^{2012}} . \end{aligned}$$
So the answer is $\frac{-1}{2}+\frac{(1007)\binom{2013}{1006}}{2^{2012}}$ (for reference, approximately 35.3).