= Solution
The <expectation> of a complex random quantity is defined through its real and imaginary parts. Both $\cos X$ and $\sin X$ are bounded, so all expectations below exist. Expand the addition formulae and apply the <expectation of a product of independent random variables> identity:
$$
\begin{aligned}
E\cos(X+Y)&=E\cos X\,E\cos Y-E\sin X\,E\sin Y,\\
E\sin(X+Y)&=E\sin X\,E\cos Y+E\cos X\,E\sin Y.
\end{aligned}
$$
Combining real and imaginary parts proves
$$
\boxed{E e^{i(X+Y)}=(E e^{iX})(E e^{iY}).}
$$
For the <simple symmetric random walk>, write $W_n=\sum_{j=1}^n\xi_j$, with independent increments satisfying $P(\xi_j=1)=P(\xi_j=-1)=1/2$. The usual casino model uses these independent, or conditionally fair, increments. Their <characteristic function> is $(e^{iu}+e^{-iu})/2=\cos u$, so repeated factorization gives
$$
\boxed{\phi(u)=E e^{iuW_n}=(\cos u)^n.}
$$
This is real for real $u$. Equivalently, the <distribution> of $W_n$ is symmetric, making $E\sin(uW_n)=0$.
For any integer $k$, <Fourier orthogonality> gives
$$
\frac1{2\pi}\int_{-\pi}^{\pi}e^{iuk}\,du=\mathbf1_{\{k=0\}}.
$$
The <random walk> has finite support, so we may interchange its finite <expectation> with this integral. Therefore
$$
P(W_n=0)=\frac1{2\pi}\int_{-\pi}^{\pi}(\cos u)^n\,du.
$$
For even $n$, evenness and reflection symmetry about $\pi/2$ reduce this to
$$
\boxed{P(W_n=0)=\frac2\pi\int_0^{\pi/2}(\cos u)^n\,du,\qquad \gamma=\frac2\pi.}
$$
For odd $n$, $W_n$ has odd parity and cannot be zero, so \b[$P(W_n=0)=0$]. As a separate combinatorial check, for $n=2m$ the return <probability> is $\binom{2m}{m}/2^{2m}$, since exactly $m$ positive increments are required.
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