Probability problem

I have $3$ coins, $1$ coin has $2$ heads (HH), 1 coin has $2$ tails (TT), $1$ coin has $1$ head and $1$ tail (HT). I toss the coin, it fells on my hand, and the side i see is a tail. What's the chance that the other side is also a tail?

I got this as a teaser from a friend, possible from here, as you can see he is insisting on 1/2 as not being the correct answer, I got 1/3 as my answer, am I right?

I will assume you initially choose one of the three coins at random.

There are 6 equally likely possibilities for the side you might see: the "head side" of the HT coin, the tail side of the HT coin, one "head side" of the HH coin, the "other head side" of the HH coin, one tail side of the TT coin, and the other tail side of the TT coin.

Since you observed a tail, three of the six, equally likely, possibilities listed above are ruled out, and you know you either saw one side of the TT coin or you saw the tail side of the TH coin.

The probability that the other side of the coin is also a tail is 2/3.

Here's a different way to do it:

Let $A$ be the event the observed side is a tail, $X_{HH}$ be the event that you picked the HH coin, $X_{TT}$ be the event that you picked TT coin, and $X_{TH}$ be the event that you picked the TH coin.

The desired probability is $P(X_{TT}|A)$. We have: $$\eqalign{ P(X_{TT}|A)&={P(A|X_{TT})P(X_{TT})\over P(A)}\cr &={P(A|X_{TT})P(X_{TT})\over P(A|X_{HH})P(X_{HH}) + P(A|X_{HT})P(X_{HT}) +P(A|X_{TT})P(X_{TT})}\cr &={ 1\cdot(1/3) \over 0\cdot(1/3)+(1/2)(1/3)+1\cdot(1/3) }\cr &={1 \over (1/2)+1}\cr &=2/3. } $$

Seeing it from another angle,let T1 be the event of the first side to be a Tail and T2 the event of the second side to be a Tail. From the conditional probability we have: $$P(T2|T1)=\frac{P(T2 \cap T1)}{P(T1) }=\frac{\frac{1}{3}}{\frac{3}{6}}=\frac{2}{3} $$

That's because the probability of both sides to be Tails is 1/3 and the probability of the first side to be a tail is 3/6. I believe neither of you were right.