Fair and unfair coins There are 4 fair coins and 1 unfair coin that has only heads. We choose a coin and flip it three times. The result is HHH. What is the probability that the fourth flip is H? 

We would expect the probability to be greater than $\frac{1}{2}$. We must solve for $P(\text{Heads})$. We have
$$P(\text{Heads})=P(\text{Heads}|\text{fair})\cdot P(\text{fair})+P(\text{Heads}|\text{unfair})\cdot P(\text{unfair})$$
However, we cannot just say $P(\text{fair})=\frac{4}{5}$ because we are given that the first $3$ tosses are heads.
$$P(\text{fair|}HHH)=\frac{P(\text{fair} \cap HHH)}{P(HHH)}=\frac{\frac{4}{5}\cdot\frac{1}{2}^3}{\frac{4}{5}\cdot\frac{1}{2}^3+\frac{1}{5}\cdot\left(1\right)^3}=\frac{1}{3}$$
So we have that the probability that the coin was fair when we rolled those $3$ heads was $\frac{1}{3}$ and the probability that it was unfair is $\frac{2}{3}$.
Thus we have 
$$\begin{align*}
P(\text{Heads})
&=P(\text{Heads|fair})\cdot P(\text{fair})+P(\text{Heads|unfair})\cdot P(\text{unfair})\\\\
&= \frac{1}{2}\cdot\frac{1}{3}+1\cdot\frac{2}{3}\\\\
&=\frac{5}{6}
\end{align*}$$
Which agrees with Nicola's answer.
