= Solution
{c}
Starting at port 1, the first <beam splitter> encounter sends amplitude $c$ straight out through port 0 and stores amplitude $-c$ towards M2. A return from M2 enters port $1'$; it sends the stored amplitude times $-c$ out through port 1 and times $c$ towards M1. A return from M1 enters port $0'$; it sends amplitude times $c$ out through port 0 and times $c$ towards M2. Both internal continuations carry $+c$, so two additional encounters multiply the surviving amplitude by $+c^2$.
With the fixed mirror and propagation phases absorbed as in <6/a>, the nonzero coefficients of the outgoing <single-photon state> are
$$
\boxed{\begin{aligned}
c_{0,1}&=2^{-1/2},\\
c_{0,2j+1}&=-2^{-(2j+1)/2},&&j=1,2,3,\ldots,\\
c_{1,2j}&=2^{-j},&&j=1,2,3,\ldots,
\end{aligned}}
$$
and $c_{0,2j}=c_{1,2j+1}=0$. The first nonzero terms are $c_{0,1}=1/\sqrt2$, $c_{1,2}=1/2$, $c_{0,3}=-1/(2\sqrt2)$ and $c_{1,4}=1/4$. If fixed reflection amplitudes $r_1,r_2$ are retained instead, multiply $c_{0,2j+1}$ by $(r_1r_2)^j$ and $c_{1,2j}$ by $r_2^jr_1^{j-1}$, as well as the appropriate delay propagation phases. These unit-modulus factors do not alter the following probabilities.
Distinct exit events are orthogonal because they differ in output port or arrival-time bin. This requires the incident wavepacket to resolve the delay. Summing the squared moduli as <geometric series> gives
$$
\boxed{P(0)=\sum_{j=0}^\infty2^{-(2j+1)}=\frac23,\qquad
P(1)=\sum_{j=1}^\infty2^{-2j}=\frac13.}
$$
Their sum is one. After $n$ encounters the probability of still being inside is $2^{-n}$, which tends to zero, so no trapped component is omitted.
The source drawing places the delay line in the M2 arm. A visit to M2 crosses that line in both directions. If $\tau$ is its \b[one-way] delay, then an exit after $n$ encounters has crossed it $2\lfloor n/2\rfloor$ times:
$$
\boxed{t_n=2\tau\lfloor n/2\rfloor
=\begin{cases}(n-1)\tau,&n\ \text{odd},\\n\tau,&n\ \text{even}.\end{cases}}
$$
If the phrase “delay time $\tau$” instead denotes the complete M2-arm round-trip delay, replace $2\tau$ by $\tau$ in the floor formula. The wording does not explicitly distinguish these conventions. Consecutive encounters can produce the same exit time in different ports: for example $n=2,3$ both follow one M2 visit. If the incident wavepacket were much longer than the arm delay, different histories in the same port would overlap and their <probability amplitudes> would have to be added before squaring; the asserted time-bin expansion would then not be an orthonormal event basis.
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