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.)
What do you mean by probability, if not “someone’s estimate of something-or-other”?
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.
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.)