This is interesting. Naively I’d expect that each number in a shuffled array of length n has a 1/n chance of ending up in the correct position, so there ought to be on average 1 correctly placed number regardless of the array length. I might be neglecting a correlation here, where each incorrectly placed number decreases the odds for the remaining numbers. Assuming the above though, the whole problem becomes recursive, since we’d be left with an unsorted array of length n-1 on average. The expected number of sorts would then just be n. For the time complexity, we’d have O(n) for the original shuffle, plus O(n-1) for the next, and so on, yielding O(n^2) overall.
Key seems valid. I’ll check all the integers for you to see how accurate it is.