<s>[INST] Coin flips with two fair coins and one biased coin I encountered this problem:

We have three coins, two fair coins and one coin with heads on each
  face.  One coin is picked randomly among those three coins and is
  flipped two times. We see the sequence: H, H.  What is the probability
  of obtaining heads if we flip this coin one more time?

To tackle it, here's my reasoning:


*

*Find the probabilities of the coin being fair and unfair conditional on the sequence: H, H.

*Find the probability of getting a heads in the additional flip conditional on the sequence: H, H.


1)
Using Bayes formula we have:
\begin{align}
P(fair|HH) &= \frac{P(HH|fair)P(fair)}{P(HH|fair)P(fair)+P(HH|unfair)P(unfair)}
\\ &= \frac{\frac{1}{4}\frac{2}{3}}{\frac{1}{4}\frac{2}{3}+1\cdot\frac{1}{3}}
\\ &= \frac{1/6}{1/6+1/3}
\\ &= \frac{1}{3}
\end{align}
We get at the same time:
\begin{align}
P(unfair|HH) = 1 - 1/3 = 2/3
\end{align}
2) Let's denote $A$ the event of obtaining of heads at the additional flip. We have:
\begin{align}
P(A) = P(A|fair)P(fair) + P(A|unfair)P(unfair)
\end{align}
and conditioning on the sequence $H,H$ we get:
\begin{align}
P(A|HH) &= P(A|fair,HH)P(fair|HH) + P(A|unfair,HH)P(unfair|HH)
\\&= P(A|fair)P(fair|HH) + P(A|unfair) P(unfair|HH)
\end{align}
We then get:
\begin{align}
P(A|HH) &= \frac{1}{2}\frac{1}{3} + 1\cdot\frac{2}{3}
\\ P(A|HH) &= \frac{5}{6}
\end{align}
Is my reasoning correct? Is there maybe a quicker way to get the result?
 [/INST] Your answer is correct. A shorter way of writing it:
Let $H_n$ denote the event that we get $H$ at the $n$-th coin toss. Hence, we want the probability $$P(H_3 \vert (H_1 \cap H_2))=\frac{P(H_1 \cap H_2 \cap H_3)}{P(H_1 \cap H_2)}=\frac{P(H_1 \cap H_2\cap H_3 \vert fair)\cdot P(fair)+ P(H_1 \cap H_2\cap H_3 \vert unfair)\cdot P(unfair)}{P(H_1 \cap H_2 \vert fair)\cdot P(fair)+ P(H_1 \cap H_2 \vert unfair)\cdot P(unfair)}=\frac{(\frac{1}{2})^3 \cdot \frac{2}{3}+1^3\cdot \frac{1}{3}}{(\frac{1}{2})^2 \cdot \frac{2}{3}+1^2\cdot \frac{1}{3}}=\frac{5}{6}.$$ 
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