# Probability of third head in a row after two flipping a maybe fair coin

You have three coins, one with two heads, and two fair. Pick one at random, flip it twice, and get two heads. What is the probability that tossing again I get heads?

I see it this way: call $$2H$$ the event of tossing two heads and $$3H$$ the event of getting the third one, $$F$$ and $$R$$ the events of picking a fair coin or the rigged one, respectively. Determine $$\mathbb P(3H|2H)$$. Is this what they are asking?

I go from the basics: $$\mathbb P(3H|2H) = \frac{\mathbb P(3H\cap 2H)}{\mathbb P(2H)}=\frac{\mathbb P(3H\cap 2H \cap F)+\mathbb P(3H\cap 2H \cap R)}{\mathbb P(2H|F)\mathbb P(F)+\mathbb P(2H|R)\mathbb P(R)}.$$ Take $$\mathbb P(F)=\frac23$$, $$\mathbb P(R)=\frac13$$, $$\mathbb P(2H|F)=\frac14$$, $$\mathbb P(2H|R)=1$$, $$\mathbb P(3H\cap 2H \cap F)=\mathbb P(3H \cap F) = \mathbb P(3H | F)\mathbb P(F)= \frac18\frac 23$$,$$\mathbb P(3H\cap 2H \cap R)=\mathbb P(3H | R) P(R)= \frac13$$, and I get $$\mathbb P(3H|2H) = \frac{\frac1{12}+\frac13}{\frac16+\frac13}=\frac56.$$ Is this reasonable?

$$\frac{5}{6}$$ looks correct. Since $$3H \subset 2H$$, using
$$\mathbb P(3H\mid 2H)= \dfrac{\mathbb P(3H\mid F)\mathbb P(F)+\mathbb P(3H\mid R)\mathbb P(R)}{\mathbb P(2H\mid F)\mathbb P(F)+\mathbb P(2H\mid R)\mathbb P(R)} =\dfrac{\frac18\times\frac23+1 \times \frac13}{\frac14\times\frac23+1 \times \frac13}=\frac{5}{6}$$