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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.)

Remster
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  • A "uniformly most powerful" test would be to look at both sides of each coin before tossing it! – Dilip Sarwate Jun 28 '21 at 15:22
  • Hey, I don't make the rules. – Remster Jun 28 '21 at 15:30
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    Interestingly, the highest likelihood is always associated with having no normal coins. In your case, the maximum likelihood solution is 90 HH and 10 TT (for which the likelihood is $1:$ you can't beat that!). – whuber Jun 28 '21 at 22:21
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    @whuber Someone has given the same answer on another forum. Can you elaborate, please? I understand that the probability of throwing 90 heads and 10 tails is 1 if I have 90 double-headed coins and 10 double-tailed coins, and I understand the principle of inference to the best explanation, but how do I calculate the actual probability of 90 double-headed coins and 10 double-tailed coins given the data? – Remster Jun 29 '21 at 12:03
  • You can't. As pointed out in an answer here, you cannot even conceive of such a probability until you view the distribution of coins as governed by a probability law. Then you can apply Bayes' Theorem. The *likelihood* is the probability of the data, *given* any hypothesized set of coins. – whuber Jun 29 '21 at 12:22
  • Ah, I didn't realise that "probability" and "likelihood" had different meanings in a technical context. I'll do some reading. – Remster Jun 29 '21 at 13:58
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    See [What is the difference between likelihood and probability?](https://stats.stackexchange.com/questions/2641). – whuber Jun 30 '21 at 13:44

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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 $A$:

$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{2^{-10}P(A)}{P(B)}$

fblundun
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