complies with a constant risk of metapopulation extinction per time step, which we call $r$. The chance, that a metapopulation has not gone extinct after $T$ time steps then is $(1-r)^T$. For every combination of parameters we perform $M$ simulations with $M=1,000$ in our case[^1]. It is then straightforward to calculate the probability $P(M|r)$ that all $M$ simulations are viable, $$\label{eq:pMr}
  P(M|r)=(1-r)^{M T}\quad.$$ Accordingly the chance that a simulation goes extinct during the $T$ simulation steps is $1-(1-r)^T$, resulting in the probability $P(0|r)$ of observing $0$ viable simulations of $$\label{eq:p0r}
  P(0|r)=\left(1-(1-r)^{T}\right)^M\quad.$$ More interesting, however, would be the expressions $P(r|M)$ and $P(r|0)$, the probability distributions of the metapopulation extinction risk $r$ given the fact that either all or none of the simulations are viable. These expressions straightforwardly can be calculated using Bayes’ theorem. Using uniform prior distributions we obtain $$\begin{aligned}
  \label{eq:prM}
  P(r|M)&=&\left(\int_0^1dr' (1-r')^{M T}\right)^{-1}(1-r)^{M T}\quad
  \mbox{and}\\
  \label{eq:pr0}
  P(r|0)&=&\left[\int_0^1dr' \left(1-(1-r')^{T}\right)^M\right]^{-1}\left(1-(1-r)^{T}\right)^M\quad.\end{aligned}$$ Using a maximum likelihood approach confidence intervals for $r$ can be calculated. Applying a confidence level of $95\%$ the upper bound for $r$ in cases where all simulations are viable is $5.1\times
10^{-8}$. As a lower bound for $r$ for cases where all simulations went extinct we obtain $0.057$. Since the latter 