Two heads from a mystery coin, which coin?

Question

A drawer has three coins that look identical but behave differently:

a fair coin, $P(\text{H}) = \tfrac12$ ;
a two-headed coin, $P(\text{H}) = 1$ ;
a weighted coin, $P(\text{H}) = \tfrac14$ .

You pick one uniformly at random, flip it twice, and get heads both times.

What is the probability you are holding the two-headed coin?

Accepted Answer

Each coin has prior probability 	frac13. Compute the likelihood of "heads, heads" (HH) under each: P(	ext{HH}\mid 	ext{fair}) = \left(	frac12ight)^2 = 	frac14,\quad P(	ext{HH}\mid 	ext{2-headed}) = 1,\quad P(	ext{HH}\mid 	ext{weighted}) = \left(	frac14ight)^2 = 	frac1{16}. Since the priors are equal, the posterior is proportional to these likelihoods. Their sum is 	frac14 + 1 + 	frac1{16} = 	frac{4}{16} + 	frac{16}{16} + 	frac{1}{16} = 	frac{21}{16}. So P(	ext{2-headed}\mid 	ext{HH}) = \frac{1}{21/16} = \frac{16}{21} \approx 0.762. Takeaway. With equal priors, posterior…

Two heads from a mystery coin, which coin?

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