After thinking some more, this is what I came up with:

We only query the pair $$$(b,c)$$$ when $$$a < b, c$$$. So basically in every single pair, the min element has a lifetime — it lives until a bigger element is encountered (though this model disregards the fact that we also need the min element to be $$$a$$$ to spend an additional query). Now our question is: If we iterated over the array, keeping a cumulative max in the process, how many times would we expect this cumulative max to change?

Here, I imagined the process from back to front. Let's also pretend the numbers are a permutation of $$$[1, 2, ... n]$$$. Once we encounter $$$n$$$ (the largest element), the max will never change. The expected position of this element is $$$\frac{n+1}{2}$$$, so we expect our range to be reduced to about half its original size. Then we look at $$$n-1$$$. If it is positioned after $$$n$$$, it is useless, we can skip it. Otherwise, since we know it appears earlier than $$$n$$$, we will once again expect it to halve our range. Obviously, since randomness is involved, it will almost never halve the range exactly, but this is the behavior we expect to see on average. And an additional 25 queries should be more than enough for $$$1500 < 2^{11}$$$ since we also need $$$a < b$$$ to hold whenever we encounter such a "maximum" (I think that provides an additional factor of $$$\frac{1}{2}$$$?)

Though obviously, this is mostly intuition based and not rigorous at all.

It is semi-well-known that the expected number of times that the prefix maximum changes in a shuffled array is O(log(N)). See Blogewoosh #6 for proofs.