Odds, not probability. Probability p means odds of p:(1-p) or, if you prefer writing it as a fraction, p/(1-p).
(Note 1. The odds of a thing are the ratio Pr(thing) : Pr(not thing). You can generalize this to any mutually exclusive and exhaustive set of things: the odds are the ratio of the probabilities. The fact that there may therefore be more than 2 such things is the reason why I prefer not to turn odds into fractions as above.)
(Note 2. Bayes' theorem is, as others have mentioned, much nicer when you work with odds rather than probabilities for your prior and posterior probabilities. If you're comfortable with logarithms, it's nicer still when you work with logarithms of odds. Now you're just adding the vector of log-likelihoods to the prior odds vector to get the posterior odds vector. Which is how I think of the question above, at least if I'm allowed to be sloppy and imprecise. You start with almost exactly 10 bits of prior prejudice for "fair" over "two-headed", then you get exactly 10 bits of evidence for "two-headed" over "fair", at which point those cancel out almost exactly so you should assign almost equal probabilities to those two possibilities.)
(Note 1. The odds of a thing are the ratio Pr(thing) : Pr(not thing). You can generalize this to any mutually exclusive and exhaustive set of things: the odds are the ratio of the probabilities. The fact that there may therefore be more than 2 such things is the reason why I prefer not to turn odds into fractions as above.)
(Note 2. Bayes' theorem is, as others have mentioned, much nicer when you work with odds rather than probabilities for your prior and posterior probabilities. If you're comfortable with logarithms, it's nicer still when you work with logarithms of odds. Now you're just adding the vector of log-likelihoods to the prior odds vector to get the posterior odds vector. Which is how I think of the question above, at least if I'm allowed to be sloppy and imprecise. You start with almost exactly 10 bits of prior prejudice for "fair" over "two-headed", then you get exactly 10 bits of evidence for "two-headed" over "fair", at which point those cancel out almost exactly so you should assign almost equal probabilities to those two possibilities.)