 Solution

Take equal arm lengths, no plate, and coherent illumination. At a coated beam splitter, choose the real-amplitude convention where transmission has amplitude $1/\sqrt{2}$ and reflection has amplitude $+1/\sqrt{2}$ from one side and $-1/\sqrt{2}$ from the other; which reflection carries the minus sign is a basis convention. The two paths interfere constructively at detector 1 and destructively at detector 2 in the diagram's labeling. ThusFor example, writing the beam splitter as $B=2^{-1/2}\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$ and the two equal-reflection mirrors as $M=-\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, one obtains $BMB(1,0{)}^T=(0,-1{)}^T$. The first output component propagates downwards to detector 2; the nonzero second component propagates upwards to detector 1. The overall minus sign is an unobservable global phase. The redistribution is interference of amplitudes, rather than two successive independent coin tosses.