A die is rolled and a coin is flipped. What is wrong with the following reasoning? A die is rolled and a coin is flipped. What is wrong with the following reasoning?:
Let $A=\{1,3,6\}$ (event from die roll), let $B=\{H\}$ (event of heads)
then, $A\cap B=\emptyset$ (since they have no common elements)
and $\Pr(A\cap B)=\Pr(\emptyset)=0$.
I realize that the probability of the intersection of two independent events is the product of their probabilities but I want to know what is 'wrong' with this reasoning.  
 A: Events are sets of outcomes.   $\{1,3,6\}$ and $\{\mathcal H\}$ are sets of measures of outcomes.   The values of a die roll, and the enumerations of a coin toss.   They are images of events, not the events per se.
The event you want is not the union of these sets, but the intersection of their preimages as mapped by appropriate random variables.
That is, if we let random variable $X$ be the result of the die roll, and $Y$ be the result of the coin toss, then the event you want is: $X^{-1}\{1,3,6\}~\cap~Y^{-1}\{\mathcal H\}$.
We might choose to describe the outcomes of the process as ordered lists of such measures.   In which case $\Omega=\{1,2,3,4,5,6\}{\times}\{\mathcal H,\mathcal T\}$ is a sensible representation† for a sample space.   It is the set of all discernible results of rolling a die and tossing a coin.
$$\begin{align}X^{-1}\{1,3,6\}~\cap~Y^{-1}\{\mathcal H\}~&=~\{\omega\in\Omega : (X(\omega)\in\{1,3,6\})\wedge (Y(\omega)\in\{\mathcal H\})\} \\[1ex] &=~ \big\{(1,\mathcal H),(3,\mathcal H),(6,\mathcal H)\big\}\end{align}$$
$\begin{align}\\[10ex]\color{silver}{\tiny\dagger\text{ Technically this is still also only a map of the sample space, but let us not fall too far down the rabbit hole.}}\end{align}$
A: The problem is the wrong choice of sample space.
The right sample space is $\Omega=\{1,2,3,4,5,6\} \times\{H,T\}$ note that the elementary events are ordered pairs $(\omega_1, \omega_2)$.
Now consider the events
\begin{align*}
E_1&=\{(\omega_1, \omega_2) \in \Omega: \omega_1\in\{1,3,6\}\}\\
E_2&=\{(\omega_1, \omega_2) \in \Omega: \omega_2=H\}\,.
\end{align*}
You should be able to say what $P(E_1 \cap E_2)$ is, and it's not $0$.

Suppose $P$ is the uniform probability.
$$P(E_1  \cap E_2)=\frac{|E_1 \cap E_2|}{|\Omega|}=\frac{3}{12}=0.25\,.$$
A: In order to say sensible things about the probability that the die shows a face in $\{1,3,6\}$ and a flipped coin shows head we must first construct a suitable model for the situation. 
This usually is done by means of a probability space $\langle\Omega,\mathcal F,P\rangle$ together with functions $X:\Omega\to\{1,2,3,4,5,6\}$ and $Y:\Omega\to\{H,T\}$. 
In this context $\mathcal F$ is a collection of subsets of $\Omega$ that meets specific conditions. Elements of $\Omega$ are outcomes and and elements of $\mathcal F$ are events. 
In our example the sets $A=\{1,3,6\}$ and $B=\{H\}$ induce the events:


*

*$\{X\in A\}=\{\omega\in\Omega\mid X(\omega)\in A\}$ 

*$\{Y\in B\}=\{\omega\in\Omega\mid Y(\omega)\in B\}$


Note that nothing points is the direction that the events $\{X\in A\}$ and $\{Y\in B\}$ should have an empty intersection.
A: On the other hand note:
$$P(A\cap B)=P(A)\cdot P(B|A)=P(A)\cdot P(B)=\frac{3}{6}\cdot \frac{1}{2}=\frac{1}{4}.$$
The thing is this is a two step experiment with the first step of rolling a die and the second step of flipping a coin. The counting rule of the total number of experimental outcomes of the multiple step experiment is to multiply the numbers of outcomes in each step. In this case it is:
$$n(S)=n_1\cdot n_2=6\cdot 2=12.$$
Thus, all expermental outcomes make up the sample space, from which one should form the events (or subsets of the sample space). In your case, you should have:
$$X=\{1H,3H,6H\}, P(X)=\frac{n(X)}{n(S)}=\frac{3}{12}=\frac{1}{4}.$$
Note that all outcomes are equally likely.
Also one can form an event from within each step, however the intersection of the events of distinct steps in probability theory imply coexisting outcomes (unlike in set theory common elements). Moreover, the events or outcomes within a step can be mutually exclusive or not, while across steps dependent or independent. Thus, in the case of rolling a die and flipping a coin, the events $A$ and $B$ are across steps and independent.
