= Solution
Within the merged third observed cell, the probability of belonging to the original fourth cell is
$$
\frac{p_4}{p_3+p_4}=\frac{1/2}{\theta^{(t)}/4+1/2}=\frac{2}{2+\theta^{(t)}}.
$$
The allocation therefore has the conditional <binomial distribution>
$$
Z\mid x,\theta^{(t)}\sim\operatorname{Bin}\left(x_3,\frac{2}{2+\theta^{(t)}}\right).
$$
Its conditional <expectation> and the filled-in missing counts are
$$
\boxed{z^{(t)}=\frac{2x_3}{2+\theta^{(t)}},\qquad
y_3^{(t)}=\frac{x_3\theta^{(t)}}{2+\theta^{(t)}},\qquad
y_4^{(t)}=z^{(t)},}
$$
with $y_1^{(t)}=x_1$ and $y_2^{(t)}=x_2$.
Taking the conditional <expectation> of the complete <log-likelihood> gives the <EM for merged multinomial cells> E-step:
$$
\boxed{Q(\theta\mid\theta^{(t)})=x_1\log(1-\theta)
+(x_2+y_3^{(t)})\log\theta+C_t.}
$$
This is the parameter-dependent part of $\log L(y^{(t)}\mid\theta)$. More precisely, $Q(\theta\mid\theta^{(t)})=\log L(y^{(t)}\mid\theta)+C'_t$, where the constant is independent of the candidate $\theta$, interpreting factorials of noninteger filled counts through the <gamma function>. The expected log-factorials are not generally the log-factorials of expected counts, so literal equality including the normalization would be false. Equality up to this constant is exactly what the M-step needs. Filling in conditional means works here because the candidate-parameter terms are linear in the missing counts; it is not a universal rule for the <expectation-maximization algorithm>.
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