= Finite-population model of raven confirmation
{title2=$L_{\mathrm{nonblack}}=(1-1/M)^{-1},\quad L_{\mathrm{raven}}=(1-1/r)^{-1}$}
Fix $r>1$ <ravens> and $M>1$ nonblack objects in a finite population. Under $H_0$, all <ravens> are black; under $H_1$, exactly one is nonblack. The total numbers $r,M$ are held fixed under both <hypotheses>, so the classification of one nonraven also changes between hypotheses. This is a deliberately specified model for <Bayesian confirmation>, not a universal rule about how bird observations supply evidence.
A uniform draw among nonblack objects returns a nonraven with <probabilities> $1$ under $H_0$ and $1-1/M$ under $H_1$. A uniform draw among <ravens> returns a black object with <probabilities> $1$ and $1-1/r$. The respective <likelihood ratios> for $H_0$ over $H_1$ are therefore $(1-1/M)^{-1}$ and $(1-1/r)^{-1}$. When $M>r$, the former is closer to one, so supplies less support under these assumptions.
With equal <prior probabilities>, <Bayes' theorem> gives <posterior probabilities> $1/(2-1/M)$ and $1/(2-1/r)$. More generally, with <prior probability> $p$ for $H_0$, replace the two expressions by $p/[p+(1-p)(1-1/M)]$ and $p/[p+(1-p)(1-1/r)]$. Previously known information or a different sampling protocol requires different <likelihoods>; selecting an object already known to be a nonraven need not supply new evidence.
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