I have an elegant analytic solution:
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Suppose you know that your coin has probability heads equal to either p or 1-p. You flip the coin n times and get h heads and t tails, with h+t=n and a difference d=h-t.
Using the standard Bayesian formula and simplifying, you can show (setting tail probability as q=1-p):
P(coin bias is p given observing h heads in n coin tosses)
= (p^d)/((p^d) + (q^d))
This is a function of the difference in the observed numbers of heads and tails, not the overall number of coin tosses n. So in the example the probabilities must be equal, since 10-0=55-45. Done.