= Solution
An upper bound follows directly from the <Schmidt-tail entanglement monotone>
$$
E_2(\chi)=1-\lambda_{\max}(\rho_A^\chi),
$$
where $\rho_A^\chi$ is the <reduced density matrix> of a bipartite pure state. On these two-qubit states, this is the smaller squared <Schmidt coefficient>, so $E_2(\psi)=s$ and $E_2(\phi)=t$.
Here is the average-monotonicity argument needed for all <LOCC> strategies. The largest <eigenvalue> is convex: for any unit vector $v$, $\langle v|\sum_r p_r\rho_r|v\rangle\leq\sum_r p_r\lambda_{\max}(\rho_r)$, and maximizing the left side gives the claim. Hence $1-\lambda_{\max}$ is concave. If Alice makes a local measurement, Bob's conditional reductions average to his original reduction. Concavity therefore gives $\sum_r p_rE_2(\chi_r)\leq E_2(\chi)$. If Bob measures, use Alice's reductions instead. The two reductions of a pure state have the same nonzero eigenvalues by the <Schmidt decomposition>. Refining all measurement outcomes into individual <Kraus operators> keeps conditional states pure; iterating the inequality covers adaptive rounds of classical communication. The same bound persists for limits of such protocols.
If success produces the target with probability $p$, its contribution to the final average is $pt$. Failure branches have nonnegative $E_2$. Thus
$$
pt\leq s,\qquad
\boxed{p_{\max}\leq\frac{s}{t}=\frac{\sin^2\alpha}{\sin^2\beta}.}
$$
This is a bound on arbitrary local protocols, not merely on a particular filtering measurement.
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