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Alpha and Beta play a game with marbles Question
There are $100$ marbles in a bucket.
Alpha and Beta play a game in which they take turns removing marbles from the bucket with Alpha going first.
At each turn the player can choose to remove either $1, \ 2, \ 3 \ or \ 4$ marbles from the bucket.
Whoever empties the bucket wins the game.
Could either player have a winning strategy that would ensure that they win no matter how their opponent plays?
See also:
I posted a related puzzle here:
https://proposals.codidact.com/posts/296186
1 answer
The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| will.octagon.gibson | (no comment) | May 22, 2026 at 16:14 |
Solution
Beta can always win.
Solution logic
For a player X to win, there have to be 1-4 marbles left. For that to be forced, there need to be 5 marbles left before that for player Y.
You can keep working this backwards to how many marbles need to be left each turn for X to win. To guarantee 5 marbles left, there need to be 6-9 for player X, which means 10 marbles for player Y right before that. As we can see, for X to win, Y needs to be presented with a multiple of 5 marbles.
Since the game starts with 100 marbles, whoever starts loses if the other player always makes the optimum move.
Strategy spelled out
Each turn, Beta must remove 5 minus the number of marbles that Alpha just removed.
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