The following problem was posed to me:

A gambler has in his pocket a fair coin and a two-headed coin. He selects one of the coins at random, i.e. the probability that the fair coin is selected is 0.5. When the gambler flips the chosen coin, it shows heads.

(A) What is the probability that it is the fair coin?

(B) Suppose that he flips the same coin a second time and again it shows heads. Now what is the probability that it is the fair coin?

(C) Suppose that he flips the same coin a third time and it shows tails. Now what is the probability that it is the fair coin?

I am concerned with (C).

The following solution was provided:

Let $F$ be the event that the coin is fair, $F^c$ is the complement of $F$. Let also $H$ be the event that it shows a head.

$$P(F|HHH) = \dfrac{P(HHH|F)P(F)}{P(HHH)} = \dfrac{P(HHH|F)P(F)}{P(HHH|F)P(F) + P(HHH|F^c)P(F^c)} = \dfrac{1/2 \cdot 1/2 \cdot 1/2 \cdot 1/2}{9/6} = 1/9$$

But isn't this a solution to the problem of probability that it is the fair coin when flipping the coin a third time and it showing *heads*? Shouldn't we instead be calculating $P(F|HHT)$ ?

But if we should be calculating $P(F|HHT)$, since only one of the coins (the fair coin) has a tails side, wouldn't $P(F|HHT)$ (the probability that the coin is fair instead of the two-headed coin) equal to $1$? In that case, we wouldn't even need to calculate anything.

I would greatly appreciate it if people could please take the time to clarify this.