Conditional probability of equality of uniform distributed random variable

Question

Suppose I have 3 random variables $x_1,x_2$ and $x_3$ which are uniform distributed over $[0,1]$, I want to know the probability: $P(x_1>x_2 | x_1>x_3)$ (x1 greater then x2 given that x1 is greater then x3). I got a feeling that those are not independent events but I can't understand why.

Answer 1

In general we have six cases:

$P(x_1>x_2>x_3), P(x_1>x_3>x_2),P(x_2>x_1>x_3),$

$ P(x_2>x_3>x_1),P(x_3>x_1>x_2), P(x_3>x_2>x_1)$

All with probability $\frac16$ each, due symmetry.

Now we look at these cases and conclude that the probability $P(x_1>x_3 \cap x_1 > x_2)$ is $P(x_1>x_2>x_3)+(x_1>x_3>x_2)=\frac26=\frac13$

Similar for $P(x_1>x_3)=P(x_1>x_2>x_3)+P(x_1>x_3>x_2)+P(x_2>x_1>x_3)=\frac12$

Now we can apply the Bayes theorem:

$$P(x_1>x_2|x_1>x_3)=\frac{P(x_1>x_3 \cap x_1 > x_2)}{P(x_1>x_3)}=\frac{\frac13}{\frac12}=\boxed{\frac23}$$