Probability of a two-headed coin given a few sample flips?

I think the rewording is fine, however, I believe your anwser is not correct.

Indeed, without throwing coins at all, there is a $1/5$ chance you get the double headed coin, and a $4/5$ chance you get a normal coin.

Now, the probability of getting three heads on three throws of a regular coin is $1/8$, as you can surely check. And as there is a $4/5$ chance you do get a regular coin, so you have a $1/8 \cdot 4/5 = 1/10$ chance you get three heads flipping a regular coin.

So to sum up the outcomes:

• $1/5$ chance of getting a double headed coin

• $1/10$ chance of getting a normal coin and three heads

• $7/10$ chance of getting a normal coin and anything but three heads.

However, note that you are asking: What is the probability that your coin is double-headed assuming you GET three heads? So the only options you are considering are the first two (as you know the third didn't occur).

Now there is a $1/5 + 1/10 = 3/10$ chance you get three heads. So there is a $\frac{1/5}{3/10} = 2/3$ chance of getting a double headed coin.

With the Bayesian interpretation of the question (your second phrasing), and leaving it as an exercise to figure out my notation: $$p(HHH|DH)=1$$ $$p(DH)=\frac{1}{5}$$ $$p(DH|HHH)=\frac{p(DH \wedge HHH)}{p(HHH|DH)p(DH)+p(HHH|N)p(N)}$$ $$=\frac{1/5}{1/5+(1/8)(4/5)}=\frac{2}{3}$$