How is this the expected number of of probes in open-address hashing?
It is the definition of expectation. When the probability of success in exactly $i$ probes is $p(i)$, it means you've been unsuccessful $i-1$ times before! Hence, the expected of unsuccessful steps is $$\sum_{i = 1}^{\infty}(i-1)\times p(i) = \sum_{i = 1}^{\infty}i\times p(i) + \sum_{i = 1}^{\infty}p(i) = 1 +\sum_{i = 1}^{\infty}i\times p(i)\,.$$