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Igor has eleven coins in his pocket. The first has 0% chance of landing heads, the second 10% chance, the third 20%, and so on up to the tenth, which has 90% chance, and the eleventh, which has 100% chance. He picks one out without letting me know which, and he starts to toss it. After the first 10 tosses, it has landed tails 5 times. How confident should I be that the coin is fair? That is, how confident should I be that it is the sixth coin from Igor's pocket; the one with 50% chance of landing heads? According to the Bayesian, the answer is calculated as follows:$$P_E(H_5) = P(H_5 | E) = \frac{P(H_5)P(E | H_5)}{\sum^{10}_{i=0} P(H_i) P(E|H_i)}$$where

- $E$ is my evidence, which says that 5 out of 10 of the tosses landed heads,
- $P_E$ is my new posterior updating credence upon learning the evidence $E$,
- $P$ is my prior,
- $H_i$ is the hypothesis that the coin has $\frac{i}{10}$ chance of landing heads,
- $P(H_0) = \ldots = P(H_{10}) = \frac{1}{11}$, since I know nothing about which coin Igor pulled from his pocket, and
- $P(E | H_i) = \left ( \frac{i}{10} \right )^5 \left (\frac{10-i}{10} \right )^5$, by the Principal Principle, and since each coin toss is independent of each other one.

So, upon learning that the coin landed heads five times out of ten, my posterior should be:$$P_E(H_5) = P(H_5 | E) = \frac{P(H_5)P(E | H_5)}{\sum^{10}_{i=0} P(H_i) P(E|H_i)} = \frac{\frac{1}{11} \left ( \frac{5}{10} \right )^5\left ( \frac{5}{10} \right )^5}{\sum^{10}_{i=1}\frac{1}{11} \left ( \frac{i}{10} \right )^5 \left (\frac{10-i}{10} \right )^5 } \approx 0.2707$$But some philosophers have suggested that this is too low. The Bayesian calculation takes into account how likely the hypothesis in question makes the evidence, as well as how likely I thought the hypothesis in the first place, but it doesn't take into account that the hypothesis explains the evidence. We'll call these philosophers explanationists. Upon learning that the coin landed heads five times out of ten, the explanationist says, we should be most confident in $H_5$, the hypothesis that the coin is fair, and the Bayesian calculation does indeed give this. But we should be most confident in part because $H_5$ best explains the evidence, and the Bayesian calculation takes no account of this.