Alpha is not the actual probability that the driver is at point X; it's the driver's estimate of that probability.

What do you mean by probability, if not "someone's estimate of something-or-other"?

But to the driver, who is inside the problem, there is no correct value of alpha to use at both X and Y.

There's also no correct value of p to use both when you'll continue and when you won't. But that doesn't mean you should omit p from the calculation.

This means that the answer arrived at must be wrong at at least either X or Y. Since the correct answer is the same in both places, the answer arrived at must be wrong in both places.

The driver is computing an expectation. A value of an expectation can be wrong for X, and wrong for Y, and right for the union of X and Y.

(I agree, of course, that the formula involving alpha isn't the right computation for the original problem. But that's separate from whether it's computing something interesting.)