Coin probability 
There are three coins, one with both sides heads, one with both sides tails, and the last a regular coin. A coin is randomly chosen and tossed and it is found that the result is heads. What is the probability that the other side is heads as well?

One method I used for this is Bayes' Theorem which gives
$P =\frac{\frac{1}{3}*1}{*\frac{1}{3}*1+\frac{1}{3}*\frac{1}{2}+\frac{1}{3}*0}=\frac{2}{3}$
Another way I looked at this is there are two cases in which a heads can occur on tossing and out of those two there is only one case which gives a heads on the other side as well, which gives 
$P=\frac{1}{2}$
What am I doing wrong in the second method?
 A: Possible results of the first phase (coin choice):
$$\begin{array}{|c|c|c|}
\hline
\text{two-heads}&\text{regular}&\text{two-tails}\\
\hline\end{array}$$
with equal probabilities of $1/3$ each.
Possible results of the second phase (toss):
$$\begin{array}{|c|c|c|}
\hline
\color{green} H\atop \color{red} H & \color{green} H\atop \color{red} H & \color{green} H\atop T & T\atop H &T\atop T & T\atop T \\
\hline\end{array}$$
with equal probabilities $1/2$ within respective three cases above, so $1/6$ each.
You observe the top side is heads, i.e. one of three leftmost cases happenned (green). Then you have two cases among them, which satisfy 'bottom side is heads' (red). Hence the probability $2/3$.
Edit
Your second answer $P=\frac 12$ is a result of ignoring the internal structure of the experiment, which led you to assigning wrong probabilities to elementary events (and possibly to misidentification the elementary events themselves).
The oversimplification you made, when pushed to the limits of absurd, will be similar to the conclusion 'when you fall from the sixth floor onto a pavement you can either die of survive, so a probability of surviving is $\frac 12$.'
Suppose you said: 'we have three random machines with such and such outputs possible...' – would we be able to calculate the answer? Certainly not, because the description lacks information about the elementary probabilities. But you said 'we have three coins'. The word coin, when not specified otherwise, denotes a random machine with two different outputs of equal probabilities. Of course you can say the coin is unfair and the probabilities are e.g. $\frac 25$ and $\frac 35$ instead of $\frac 12$ each; or you can say the outputs are indistinguishable, and we have effectively just one output possible with probability $1$ – but that always needs to be explicitly said, or you risk some omission leading to a wrong answer.
Let's modify a description of the problem a bit to reveal details:

There are three regular coins, with sides 'heads' ($H$) and 'tails' ($T$). We put letters $h$ and $t$ on respective sides of one coin, $h$ on both sides of another coin, and $t$ on both sides of the last one. A coin is randomly chosen and tossed and it is found that it displays the letter $h$. What is the probability that the other side is marked with $h$ as well?

The first phase is choosing a coin. No method of choosing is specified, so we assume each coin can be chosen with the same probability of $\frac 13$. Then we toss a coin. Let's consider each possibility in order.

*

*If the chosen coin is $ht$, it may land in $H$ or $T$ position, both with same probability of $\frac 12$. So we have two elementary events $Hh \atop Tt$ and $Tt\atop Hh$, both with equal probabilities $\frac 12$ under the condition the chosen coin is $ht$, so their final probabilities are $\frac 12\cdot\frac 13=\frac 16$ each.

*Similary, if the chosen coin is $hh$, it may land in $H$ or $T$ position, both with same probability of $\frac 12$. So we have two elementary events $Hh \atop Th$ and $Th\atop Hh$, both with equal probabilities $\frac 12$ under the condition the chosen coin is $hh$, so their final probabilities are $\frac 12\cdot\frac 13=\frac 16$ each.

*Finally, if the chosen coin is $tt$, it may land in $H$ or $T$ position, both with same probability of $\frac 12$. So we have two elementary events $Ht \atop Tt$ and $Tt\atop Ht$, both with equal probabilities $\frac 12$ under the condition the chosen coin is $tt$, so their final probabilities are $\frac 12\cdot\frac 13=\frac 16$ each.

So we have $6$ elementary events with equal probabilities of $\frac 16$, each representing one of two possible positions of one of three possible coins.
Given the visible letter is $h$, we know we may have one of three equally probable elementary events: $Hh \atop Tt$, $Hh \atop Th$ or $Th\atop Hh$. And in two of them the bottom side has an $h$ mark, while in one the bottom side has a $t$ mark. As a final result, the probability of bottom side being $h$ is $\frac 23$.
Now let's see the same experiment in generalized description:

There are three random machines, each having two output displays, called 'top' and 'bottom'. One of those machines, every time it is triggered, shows $h$ on both the 'top' and 'bottom' display. Another machine always shows $t$ on both displays. The last one shows either $h$ above $t$ or $t$ above $h$, both with equal probability of $\frac 12$.
A machine is randomly chosen and triggered and it is found that it displays the letter $h$ on 'top' display. What is the probability that the 'bottom' display shows $h$ as well?

As there is no specific method defined to choose a machine, we assume each one is chosen with equal probability of $\frac 13$. Then:

*

*if the first machine was chosen, it always shows $h$ on both displays, so the probability of getting 'top' $h$ shown by machine 1. is $1\cdot\frac 13=\frac 13$;

*if the second machine was chosen, it never shows $h$ on any display, so the probability of getting 'top' $h$ shown by machine 2. is $0\cdot\frac 13=0$;

*in the case of the third one, $h$ appears on the 'top' display half the time (relative probability $\frac 12$, so eventual probability equals $\frac 12\cdot\frac 13=\frac 16$), but then the 'bottom' is $t$ and never $h$.

This time we have $4$ possible elementary events: $h\atop h$ displayed by the first machine, $t\atop t$ displayed by the second, and $h\atop t$ or $t\atop h$ displayed by the third one. Their respective probabilities are $\frac 13$,  $\frac 13$,  $\frac 16$ and  $\frac 16$.
Finally we have two cases with top $h$: either it is machine 1. with probability $\frac 13$ or machine 3. with probability $\frac 16$ – but the bottom is $h$ in the first case only. Hence the probability
$$P = \frac{P(\text{it's machine 1.})}{P(\text{it's machine 1.}) + P(\text{it's top h on machine 3.})} = \frac{\frac 13}{\frac 13 + \frac 16}=\frac 23$$

*

*the same which we get from the first model.

The crucial point is that the two cases are not equally probable, so one of two satisfying the requirements does not imply the probability $1/2$.
A: Let the sides of the 2-headed coin be $H_1$ and $H_2$, and the sides of the regular coin be $H_3$ (heads) and $T_1$ (tails). Each of those $4$ sides has equal probability of being the result of the toss. Because it is given that the result is heads, there are $3$ possibilities of the result, namely $H_1,H_2,H_3$, each of which has probability $\frac13$ of being the result. The number of possibilities such that the other side is also heads is $2$ ($H_1,H_2$). So, the probability is $\frac23$.
A: When you get a head, you know that it could be either side of the double headed coin(giving 2 possibilities) or it could have been the head of the regular headed coin(giving 1 possibility)
Probability to get the double headed coin is what is required.
Therefore probability is 2/3.
