Coin tossing problem My teacher gave us an homework. I solved it, but I don't think I have the right answer.
PROBLEM
We have three coins identical in appearance.


*

*Coin A falls on tails and heads with equal probability

*Coin B falls twice as much on tails as heads

*Coin C always falls on tails


We choose a coin at random and toss it. It falls on tails. 
What is the probability to get tails on the next toss, if we toss the same coin?
MY TRY
Soient les événements

A = "choose coin A" = {t, h}

B = "choose coin B" = {t, t, h}

C = "choose coin C" = {t}

E, get tails at the second throw

We are looking for $P(E)$. Knowing $P(E|A)=1/2$, $P(E|B) = 2/3$ and $P(E|C)=1$.
\begin{split}
P(E)& = P(E|A)P(A) + P(E|B)P(B) + P(E|C)P(C)\\
& = \frac{1}{2}\cdot\frac{1}{3} + \frac{2}{3}\cdot\frac{1}{3} + 1 \cdot\frac{1}{3}\\
& = 13/18 = 0,7 \overline{2}
\end{split}
ANSWER : 72%
 A: What you have to do in this case is update your priors according to your observation. Your first prior distribution for selecting a coin was (I assume) $\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$.
Now that you've seen tails, the probabilities for the coin you're holding change; using Bayes' theorem:
$$P(A|tails) = P(tails|A)\frac{P(A)}{P(tails)} = \frac{3}{13}$$
where $P(A)=\frac{1}{3}$, $P(tails)=\frac{13}{18}$ as you calculated, and $P(tails|A) = \frac{1}{2}$
Similarly, you can get 
$$P(B|tails) = P(tails|B)\frac{P(B)}{P(tails)} = \frac{4}{13}$$
$$P(C|tails) = P(tails|C)\frac{P(C)}{P(tails)} = \frac{6}{13}$$
So your new priors are $\left(\frac{3}{13},\frac{4}{13},\frac{6}{13}\right)$,
and $$P(tails|new\ priors) = \frac{1}{2}\cdot\frac{3}{13} + \frac{2}{3}\cdot\frac{4}{13} + 1\cdot\frac{6}{13} = \frac{61}{78}$$
(up to possible calculation errors)
And if you see only more $tails$ in the future, you'll have better and better reason to believe you picked coin $C$, and your probability for the next $tails$ will tend to $1$.
A: I guess it is 61/78. 
From the probabilities you get a total of 13/18 for a tail answer, but the dependent source is 


*

*A: (3/13)*(3/6))

*B: (4/13)*(4/6) 

*C: (6/13)*(6/6) 


which is (9+16+36)/(13*6) = 61/78 which is about 78%. 
A: 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.
A: The key point to consider here is that you now have new evidence that favours a certain hypothesis, and less others.
Here is a write up on dealing with exactly such scenarios.
