I don't think so. After each drawing the number of whites is congruent modulo 2 to the initial number of whites. Hence, when there are two beans left, one will be white and the other black so the last bean will be white.
I don't think so. After each drawing the number of whites is congruent modulo 2 to the initial number of whites. Hence, when there are two beans left, one will be white and the other black so the last bean will be white.