Let $\bar{\beta}_1$ and $\bar{\beta}_2$ denote the two distinct grand effects. We assume that the realizations from the two effects follow the normal distributions, $$ \begin{aligned} & \beta_{1, i} \mid \bar{\beta}_1 \sim \mathrm{N}\left(\bar{\beta}_1, \phi^2\right) i=1,2, \ldots \\ & \beta_{2, j} \mid \bar{\beta}_2 \sim \mathrm{N}\left(\bar{\beta}_2, \phi^2\right) j=1,2, \ldots, \end{aligned} $$ respectively. Here, $\beta_{1, i}$ 's denote the realizations of grand effect $\beta_1$ in each $i^{\text {th }}$ experiment, which also represents the underlying true effect in each experiment. Similarly, $\beta_{2, j}$ 's are realizations for $\beta_2$. Without loss of generality, we assume that the heterogeneity for the two effects is same and quantified by the variance parameter, $\phi^2$.
Based on this generative model, the misclassification probability, $P_{\mathrm{mis}}\left(\bar{\beta}_1, \bar{\beta}_2, \phi^2\right)$, can be computed by $$ P_{\text {mis }}\left(\bar{\beta}_1, \bar{\beta}_2, \phi^2\right) $$ $$ \begin{aligned} & =\operatorname{Pr}(\text { An effect } \beta \text { is misclassified from its generative effect }) \\ & =\operatorname{Pr}\left(\beta \mapsto \bar{\beta}_1 \mid \beta \cong \bar{\beta}_2\right) \operatorname{Pr}(\beta \cong \bar{\beta} 2)+\operatorname{Pr}\left(\beta \mapsto \bar{\beta}_2 \mid \beta \cong \bar{\beta}_1\right) \operatorname{Pr}(\beta \cong \bar{\beta} 1), \end{aligned} $$ where $\beta \mapsto \bar{\beta}_1$ represents that $\beta$ is classified/mapped to $\bar{\beta}_1$; and $\beta \cong \bar{\beta}_1$ denotes that $\beta$ is generated from $\bar{\beta}_1$. Assuming the exchangeability of the $\bar{\beta}_1$ and $\bar{\beta}_2$, it implies $$ \operatorname{Pr}(\beta \cong \bar{\beta} 1)=\operatorname{Pr}(\beta \cong \bar{\beta} 2)=\frac{1}{2} $$
Furthermore, $$ \begin{aligned} & \operatorname{Pr}\left(\beta \mapsto \bar{\beta}_1 \mid \beta \cong \bar{\beta}_2\right)=\operatorname{Pr}\left(\beta \mapsto \bar{\beta}_2 \mid \beta \cong \bar{\beta}_1\right) \\ & =\int_{-\infty}^{\infty} \frac{\mathrm{N}\left(\beta ; \bar{\beta}_1, \phi^2\right) \mathrm{N}\left(\beta ; \bar{\beta}_2, \phi^2\right)}{\mathrm{N}\left(\beta ; \bar{\beta}_1, \phi^2\right)+\mathrm{N}\left(\beta ; \bar{\beta}_2, \phi^2\right)} d \beta \end{aligned} $$