Given a sequence of numbers $a_i \in R$ choosen uniformly, what is the probability that $a_i > a_{i-1}$ and $a_i > a_{i+1}$
I know that by symmetry, the answer is 1/3. I'm interested in understanding why it isn't 1/4. These are two independent events, where $P(a_i > a_{i-1}) = 1/2$ and $P(a_i > a_{i+1}) = 1/2$ which makes me think the answer is $1/2 * 1/2 = 1/4$.
Thoughts?
For example, if I have the sequence $a_n := 1/n$ then the event when $1 > 1/3$ is dependent of the event when $1 > 1/2$ because the probability that $1 > 1/3$ when we know that $1 > 1/2$ is $1$.