That doesn't change the complexity.

Rubbish!

The point: A well implemented binary search is O(logN) worst case, but averages O(logN - 1).

But, in order to accommodate duplicates, you are forced to recind the possibility of early discovery, which forces the average case to the worst case. Eg. The complexity changes!

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In big O notation O(logN - 1) and O(log N) are equivalent. They denote the same complexity order.

Though, that does not mean that the two algorithms are equally efficient. Actually they are not: Re^3: Modified Binary Search.

A well implemented binary search is O(logN) worst case, but averages O(logN - 1).

That's wrong on multiple levels. First of all, your notation is sloppy. O(log N) and O(log N - 1) are equivalent classes. Any function that is in one class is also in the other.

I assume you mean that on average, a well implemented binary search only needs log N - 1 comparisons on average. But that's not true either. That's only true if you only search for elements that are present. Each unsuccessful search will take ceil(log N) or floor(log N) comparisons.