How to calculate time complexity of a randomized search algorithm?
Best case is $O(1)$, that is when you find the element in the first check.

Worst case is $O(n)$, and it happens when the element is in the first position, and in each check you get the last position, thus going through the whole array.

Average case is a little bit more tricky to get, since we need to solve a recurrence to get the expected value.
Let $T(n)$ be the expected time for an array of size n. The pivot is chosen randomly, so each posible pivot has a probability $p = \frac{1}{n}$. So, we have this:

$T(n) = T(1)/n + T(2)/n + ... + T(n-1)/n + 1$

We add that 1 at the end, because that's the constant time we have for checking at each step. Now we multiply each side by n:

$nT(n) = T(1) + T(2) + ... + T(n-1) + n$

Which also gives:

$(n-1)*T(n-1) = T(1) + T(2) + ... + T(n-2) + (n-1)$

So we can subtract those equations and we get

$n*T(n) - (n-1)*T(n-1) = T(n-1) +1$ 

$n*T(n) = (n-1)*T(n-1) + T(n-1) +1$

$n*T(n) = n*T(n-1) + 1$

$T(n) = T(n-1) + 1/n$

And "unrolling" this equation, we have:

$T(n) = 1 + 1/2 + 1/3 + ... + 1/n$

And the right side of the equation is the nth harmonic number. We can show that the nth harmonic number is $\Theta(\log n)$
So we have that the average time for the algorithm is $\Theta(\log n)$