# Probability of probing $t$ locations in a Cuckoo hash is $O(\frac{1}{2^{t/2}})$ locations in the worst case

Prove that the probability that an insertion into a cuckoo hash table probes $t$ array locations is $O(\frac{1}{2^{t/2}})$. Keep in mind that there are two tables, each with size $s \ge 2n$, where $n$ is the number of elements in the set.

I'm trying to use induction, but I don't know if this is the best method to go about proving this.

The worst case to probe $1$ array location would be when all $n$ element are stored in the first table, and we get probability $\frac{n}{2n} = \frac{1}{2} = O(\frac{1}{\sqrt{2}})$ for insertion.

The worst case to probe $2$ array locations would be when all $n$ elements are stored in the first table, we hit the first table, and we succeed in the second table. This has the same probability as it does to probe $1$ array location.

However, I don't know how to continue the analysis. For example, what's the worst case probability to probe $4$ array locations? The question statement implies that the worst case is $O(\frac{1}4)$, but how do we achieve this result? If the first and second tables each have $\frac{1}4$ of their array locations occupied, then the worst case to probe $4$ elements would be hitting in the first table, hitting in the second table, hitting in the first table again, then finally inserting into the second table $\Rightarrow \frac{1}4\cdot\frac{1}4\cdot\frac{1}{4}\cdot\frac{3}4 = \frac{3}{256} \not = \frac{1}4$.