What is the probability of choosing real numbers over two intervals and them being equal? You are given two intervals $[0,n_1]$ and $[0,n_2]$. What is the probability that real number chosen independently from both the intervals will be equal?
 A: If $X\sim\mathcal U[0,n_1]$ and $Y\sim\mathcal U[0,n_2]$ are independent random variables, then the joint distribution of $(X,Y)$ is uniform over the rectangle
$$R = \{(x,y)\in\mathbb R^2: 0\leqslant x\leqslant n_1, 0\leqslant y\leqslant n_2 \}.$$
For any Borel set $A\subset R$, it follows that $\mathbb P((X,Y)\in A)$ is the area of $A$. In other words, the probability measure $\mathbb P\circ(X,Y)^{-1}$ is 2-dimensional Lebesgue measure restricted to $R$. Assuming without loss of generality that $n_1\leqslant n_2$, the set $\{X=Y\}$ is the line segment
$$L=\{(x,x): 0\leqslant x\leqslant n_1\}. $$
It is clear that the area of a line segment is zero. Therefore the probability is zero.
A: Usually the probability of two independent events is the product of the probability of each event. So the answer depends mostly on your choice of probability distributions.
If you are using any distribution that is continuous, the probability is exactly $0$.
Intuitively, this is because there are just too many real numbers on any non-zero interval. There are so many that the probability of selecting a particular one is exactly $0$. So selecting the same twice in a row is just as unlikely.
More formally, the probabilities are obtained by integrating a probability density function $p_1(x)$ over an interval. The function is non-negative and it has the property:
\begin{equation}
\int_0^{n_1}p_1(x)dx=1
\end{equation}
The probability of selecting a real in an interval $[a,b]\subset[0,n_1]$ is then given by:
\begin{equation}
\mbox{Prob}(x\in[a,b])=\int_a^{b}p_1(x)dx
\end{equation}
You can see now that, given a real $x_0 \in [0,n_1]$ we have:
\begin{equation}
\mbox{Prob}(x = x_0)=\mbox{Prob}(x\in[x_0,x_0])=\int_{x_0}^{x_0}p_1(x)dx=0
\end{equation}
This is valid for all continuous distributions.
If your distribution is discrete however, this is another story.
