Numbers of double-headed, double-tailed, and normal coins based on one toss per coin Suppose I have a large jar of evenly weighted coins. I know that some of the coins are double-headed, some of them are double-tailed, and some of them are normal, but I've no idea how many. I draw 100 coins at random and toss each coin once, throwing 90 heads and 10 tails. What can be said about the probable numbers of double-headed, double-tailed, and normal coins in my sample?
My untutored intuition is that there are probably more double-headed than double-tailed coins, but that's as far as I can take it. I'm not even sure how to calculate that probability.
(To my mind, this is related to the question I asked yesterday about survey responses, but I'm unsure about this, and I mention it only as evidence that I'm not fishing for an answer to a homework question.)

This can be done using Bayes' Theorem. It requires you to have a (somewhat subjective) prior distribution for how many of each type of coin the bucket contains. For example, let $A$ be the event that the bucket contains 90 double-headed coins and 10 normal coins, and let $B$ be the event that you observe exactly 90 heads and 10 tails. You would plug in your prior probabilities for $A$ and $B$ into this formula to get an updated posterior probability for $A$:
$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{2^{-10}P(A)}{P(B)}$
