= Solution
The two identical $M$ gates are independent <Bernoulli random variables>, so their open count has a <binomial distribution>. Multiplying by the independent $H$ factor gives the normalized channel densities
$$
\boxed{S_{ij}=\binom2i m^i(1-m)^{2-i}h^j(1-h)^{1-j}.}
$$
Explicitly,
$$
\begin{aligned}
S_{00}&=(1-m)^2(1-h),&S_{01}&=(1-m)^2h,\\
S_{10}&=2m(1-m)(1-h),&S_{11}&=2m(1-m)h,\\
S_{20}&=m^2(1-h),&S_{21}&=m^2h.
\end{aligned}
$$
The <binomial coefficient> $\binom21=2$ counts which of the two distinguishable $M$ gates is open. At open count $i$, the total $M$-opening rate is $(2-i)\alpha$ and the total closing rate is $i\beta$. Dividing the <expected value> of that count by two therefore recovers the same single-gate <ordinary differential equation>:
$$
\boxed{\dot m=\alpha(1-m)-\beta m,\qquad
\dot h=\gamma(1-h)-\delta h.}
$$
For the specified <initial conditions> and common rate sum $\lambda$, the all-open fraction is
$$
S_{21}(t)=\frac{\alpha^2\gamma}{\lambda^3}(1-e^{-\lambda t})^3.
$$
Thus, assuming positive opening rates,
$$
\boxed{t_{1/2}=\frac{-\log(1-2^{-1/3})}{\lambda}.}
$$
Three gates take longer to reach half their final all-open fraction than two gates: all three activation factors must build up simultaneously.
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