# Conditional probability of equality of uniform distributed random variable

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.

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:

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

• I would like to add another question, now consider instead of x1, some constant k between 0 and 1, what is the probability: P(k>x2|k>x3) ? Commented Feb 24, 2018 at 8:07
• @דניאלניסחיזוב Assuming that $x_1, x_2$ and $x_3$ are independent then we apply the bayes theorem again $\frac{P(k>x_2 \cap k>x_3)}{P(k>x_3)}=\frac{P(k>x_3)\cdot P(k>x_2)}{P(k>x_3)}=P(k>x_2)=k$ Commented Feb 24, 2018 at 9:26
• @דניאלניסחיזוב If they are not independent, then I don´t see a way to evaluate a specific value. There are further information needed. Commented Feb 24, 2018 at 9:31