Bayes Theorem Coin Problem There are 3 coins. One is regular (both head and tail) and the other two only have head sides. Now, flip one coin and get head. The question is what is the probability of this coin is regular one? 
Or Twisted
There are 3 coins. One is regular (both head and tail) and the other two only have head sides. Now, flip one coin and get head. The question is what is the probability you pick up the fair coin and it is head?
P(H)=P(A)⋅P(H∣A)+P(B)⋅P(H∣B)+P(C)⋅P(H∣C)
   = (1/3)(2/2) + (1/3)(2/2) + (1/3)*(1/2) = 5/6
P(A∣H)=P(A)⋅P(H∣A)/P(H)
      = (1/3)/(5/6) =2/5
I am confused as to whether this is the correct answer or not.


I am confused as to if P(A)⋅P(H∣A) should be 2/3 or 1/3. If P(A)⋅P(H∣A) is 2/3, then the final answer would be 4/5

If $A$ is the event "a particular one of the two biased coins" then $P(A\cap H)=\tfrac 13\tfrac 22$ as you have.  Just as $P(B\cap H)=\tfrac 13\tfrac 22$.  In this case $A\cup B$ is the event "either one of the two biased coins" and so indeed $\mathsf P((A\cup B)\cap H)~=~ \tfrac 23\tfrac 22~=~\mathsf P(A)~\mathsf P(H\mid A)+\mathsf P(B)~\mathsf P(H\mid B)$.
Thus, it is okay that: $\mathsf P(H)~{=\mathsf P(A\cup B)~\mathsf P(H\mid A\cup B)+\mathsf P(C)~\mathsf P(H\mid C) \\= \tfrac 23\tfrac 22+\tfrac 13\tfrac 12\\=\tfrac 56}$
However, this brings us to your second part. $C$ is the event of "an unbiased coin was picked", and as you have already correctly determined that $\mathsf P(C)~\mathsf P(H\mid C)=\tfrac 13\tfrac 12$ then you should have:
$$\dfrac{\mathsf P(C)~\mathsf P(H\mid C)}{\mathsf P(H)}~=~\dfrac{\tfrac 13\tfrac 12}{\tfrac 56}=\dfrac 15$$

Intuitively, each of the $5$ heads is equally likely to show up, and only $1$ of these is from the unbiased coin, so if one does show up, then there is a probability of $1/5$ for it to be from that unbiased coin.


The question is what is the probability you pick up the fair coin and it is head?

PS: That is $\mathsf P(C\cap H)=\mathsf P(C)\mathsf P(H\mid C)=\tfrac 16$
