Considering and reconsidering the abstract case once more, it now admittedly appears very similar to Monty Hall problem.
First, let's get rid of the red herrings and ambiguities. A random list needs no reshuffling: no need to pick an envelope. Generate the numbers, including the guess. Assume they are distinct. Now the roles of Entertainer and Contestant have become superfluous. Values of numbers are also irrelevant, only their order matters. All we are left with is six permutations:
A < B < C .......... 1
A < C < B .......... 1
B < A < C .......... 0
B < C < A .......... 1
C < A < B .......... 0
C < B < A .......... 1
The favorable outcome is one where A (revealed number) is
not between B and C (the guess). Comparing to the guess reduces permutations (to first or second half). The chances of winning are evidently 2/3. But increasing the number of guesses will asymptotically improve your outlook towards 3/4.
So there is a strategy that works. Très bizarre.