Let X and Y be two random real numbers, what is the probability that x = min(x,y)? As the questions suggets I want to find the probability that x= min(x,y) being that x and y are random numbers, my guess is it's 1/2, since x is random and could be greater than x or not, but I'm not sure, could you help me?
Let me elaborate, x and y are two independent random variables with exponential distribution, and z is the minimum of the two of them, the question is what is the probability that z= x.
 A: It's the probability that $X < Y$, so generally $\int_{-\infty}^\infty\int_{-\infty}^y\, f(x, y)\,dx\,dy$. If $X$ and $Y$ are independent and distributed exponentially, say $f(x, y) = ab e^{-ax-by}$, $x, y > 0$ you get $a/(a + b)$.
On the other hand if $X$ and $Y$ are simply i.i.d.,
\begin{align}
\int_{-\infty}^\infty\int_{-\infty}^y\, f(x)f(y)\,dx\,dy
&= \int_{-\infty}^\infty F(y)f(y)\, dy \\
&= \int_{-\infty}^\infty F(y)F'(y)\, dy \\
&= 1/2
\end{align}
as you guessed.
A: Due to independence, the joint distribution of $(X,Y)$ is the product of the individuals pdfs:
$$f(x,y)=e^{-\lambda(x+y)}\mathbb{1}_{x>0}\mathbb{1}_{y>0}.$$
What you are looking for is:
$$P(X>Y)=\int_{x>y>0} f(x,y)dxdy \tag{1}$$
Geometrically, (1) means integration over the lower half part of the first quadrant which can be expressed as a certain volume under the surface with equation $z=f(x,y)$, the total volume being $1$.
This volume, due to the symmetry of this surface with respect to line $y=x$ is equal to $1/2$.

This picture represents the joint pdf as a surface (in blue) ; I have taken the case $\lambda = 1$ ; it is to be noted that the two marginal distributions are materialized as the trace of this surface onto planes Oxz and Oyz ; the symmetry $x \leftrightarrow y$ is materialized as the symmetry of the surface with respect to vertical plane with equation $y=x$ (in red).
