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And the winner is … Click here to reveal Delta can win the game no matter how Epsilon plays. The winning strategy is … Click here to reveal Delta begins by removing $1$ marble leaving ...
#1: Initial revision
by
will.octagon.gibson
·
2026-05-28T02:12:39Z (4 months ago)
<details> <summary>And the winner is … Click here to reveal</summary> Delta can win the game no matter how Epsilon plays. </details> <br> <details> <summary>The winning strategy is … Click here to reveal</summary> Delta begins by removing $1$ marble leaving $145$ marbles. Note that $145$ is a multiple of $5.$ If the bucket has a multiple of $5$ marbles when it is Epsilon’s turn to remove marbles, it is impossible for Epsilon to win on that turn because Epsilon can only remove $1, \ 4 \ or \ 6$ marbles none of which is a multiple of $5$ marbles. If Epsilon removes $1$ marble, Delta responds by removing $4$ marbles which in total removes $5$ marbles so the bucket will again have a multiple of $5$ marbles when it is Epsilon’s turn to play. Similarly if Epsilon removes $4$ marbles, Delta responds by removing $1$ marble. Lastly if Epsilon removes $6$ marbles, Delta responds by removing $4$ marbles. In all three cases, Epsilon will be left with a decreasing multiple of $5$ marbles in the bucket when it is Epsilon’s turn to remove marbles. Eventually Delta empties the bucket ($0$ marbles which is a multiple of $5$ marbles) and wins the game. </details>