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.
Seven pieces cost six cuts and the schedule pays correctly, so the six-cut answer breaks one constraint and nothing else. Because the worker can hand pieces back, the contract is on his holding rather than on the transfer, and the ledger turns out to be a three-bit counter. Brute force finds 1-2-4 is the only three-piece solution.
La respuesta refleja ronda las 180, la mitad del calendario. El umbral real es 23, porque una coincidencia necesita una pareja y 23 personas cargan con 253. El mismo razonamiento sitúa la pregunta « ¿alguien comparte MI cumpleaños? » en 253 personas, once veces más gente, y el umbral en raíz cuadrada que hay detrás de ambas explica que un identificador aleatorio de 64 bits se repita tras cinco mil millones de extracciones y no tras dieciocho trillones.