Well, there's your problem. Its in French.
$P(T|A) = 1/2,
P(T|B) = 2/3,
P(T|C) = 1/1$
$P(A) = P(B) = P(C) = 1/3$
$P(TA) = P(T|A) P(A) = 1/6$
$P(TB) = P(T|B) P(B) = 2/9$
$P(TC) = P(T|C) P(C) = 1/3$
$P(TA \lor TB \lor TC) = P(TA) + P(TB) + P(TC) = 1/6 + 2/9 + 1/3 = 13/18$
$P(T(A \lor B \lor C)) = P(T) = 13/18$
So when you randomly pick the first coin and then flip it, you stand a 13/18 chance of getting Tails.
This does not necessarily mean that if you throw the same coin a second time the probability is the same. The reason is because you are not randomly picking a (new) coin. These probabilities must get factored out of consideration.
We need to be careful because you - or me - stand a very good chance of the Monty Hall fallacy. I'm probably going to be guilty here in a moment even though I recognize it is a risk.
However, getting a Tails on this flip does not provide us with any new information and we are left not knowing what coin we have - this is effectively the same scenario than randomly picking a new coin: Total uncertainty.
If we kept this one coin for the long haul and tallied up results, we can use frequentist reasoning to deduce which coin we have. If we get a Heads, we know its not coin C and this is new information.
So Im inclined to agree with your reasoning. 13/18 is the probability.
Additional: Im trying to understand the reasoning of the other answers. I thin the reasoning is that they are reversing the conditional probabilities using Bayes Theorem.
$P(TA) / P(T) = P(A|T) = (1/6)/(13/18) = 3/13$
$P(TB) / P(T) = P(B|T) = (2/9)/(13/18) = 4/13$
$P(TC) / P(T) = P(C|T) = (1/3)/(13/18) = 6/13$
Im not sure what the next step ought to be then. We now know that the conditional probability of each coin is given that we got a tails. I dont know what to do with this information though. Multiplying each of these by $P(T)$ is redundant information. Someone multiplied each $P(C|T)$ by the associated $P(T|C)$ and Im not sure what that does for us.