The North Pole really is a solution, so the trap is only the words "and nowhere else". A mile north of the parallel whose whole lap measures 1/n of a mile, the eastward mile is n exact revolutions, which puts a starting circle 1.159 miles from the South Pole for one lap, 1.080 for two, 1.053 for three. Every point of every circle works, so the honest count is uncountable rather than infinite.
Also asked aswalk south east north · one mile south one mile east
Requiring the fifty-fifty at every amount you could open forces the weights to satisfy f(x) = f(x/2)/2, whose only solutions are proportional to 1/x, and that integrates to infinity at both ends. Conditional on the pair, the swap gains the smaller amount or loses it with equal chance, which is zero and needs no assumption at all. The article carries a proper spread where the conditional answer is genuinely x/2, and the infinite-mean spread where swapping really is right at every observable amount.
Sorting the list finds the gap and is merely wasteful, which is why the article says so rather than striking it out. The subtraction works because 5050 is a closed form available before the list is read, and the two conditions carrying it are distinctness and a known range. The article adds the duplicate-hunting mirror image, the sum-of-squares route when two values are absent, and the exclusive-or accumulator for when the total would overflow.
Twenty-four possible answers against 3^3 = 27 outcome sequences leaves just enough room, and four against four splits those answers into exactly 8, 8 and 8. Six against six always tips, so it wastes the level outcome and leaves twelve answers for nine remaining sequences, which makes halving impossible rather than merely slow. The counting argument bounds outcome sequences rather than strategies, so it rules out every adaptive continuation at once, and the explicit schedule closes the positive half.