How to calculate probability that a random variable is less than the max of two other random variables? Suppose we have 3 random variables denoted $v_i$ that are i.i.d. on a uniform distribution such that $U[0,1]$. I want to know what is the probability that $v_1$ is higher than the maximum of the other two, in other words:$$Pr(v_1 \geq \max\{v_2,v_3\})$$
I start by noticing that we want to know what is the probability that $v_1$ is greater than $v_2$ and $v_3$, so we can rewrite this as:
$$Pr(v_1 \geq v_2 \text{ and } v_1 \geq v_3)$$
And since they are i.i.d, these probabilities are independent so,
$$Pr(v_1 \geq v_2 \text{ and } v_1 \geq v_3)=Pr(v_1 \geq v_2 )Pr(v_1 \geq v_2 )$$
But now I am stuck, since they are all random variables I cannot simply look at the CDF and see these values. Are my assumptions incorrect?
 A: Let $M= \{ v | v_1 \ge v_2 \text{ and } v_1 \ge v_3 \}$ and
$1_M$ denote the indicator function.
$P[v_1 \ge v_2 \text{ and } v_1 \ge v_3] = \int_v 1_M(v) dv = \int_{v_1=0}^1 \int_{v_2=0}^{v_1} \int_{v_3=0}^{v_1} 1 dv = {1 \over 3}$.
This is not too surprising as one of the variables $v_1,v_2,v_3$ must be
the $\max$ and by symmetry none is favoured.
A word of caution regarding independence (or lack thereof): By a similar analysis we have $P[v_1 \ge v_2 ] = P[v_1 \ge v_3 ] = {1 \over 2}$.
A: Here is a straightforward proof.
Let $f(x)$ be the pdf, $F(x)$ be the cdf.
$P(X_n \ge \max_{1\le i\le n-1}X_i) = P(X_n \ge X_i, 1\le i\le n-1) = \int_{-\inf}^{+\inf} f(x_n)dx_n \; \Pi_{i=1}^{n-1} \int_{-\inf}^{x_n}f(x_i)dx_i = \int_{-\inf}^{+\inf} f(x_n)dx_n F^{n-1}(x_n)=\frac{1}{n}F^n(x_n)|_{-inf}^{+inf} = \frac{1}{n}.$
Since the random variable is continuous, it makes no difference between equality and inequality. So $P(X_n \ge \max_{1\le i\le n-1}X_i) = P(X_n >\max_{1\le i\le n-1}X_i) = P(X_n = \max_{1\le i\le n}X_i) = \frac{1}{n}.$
Note that the answer given by @copper.hat misses the pdf part.
