1, 4 or 7?

Probability Level 2

You and a friend play a game where you start with a pile of 100 stones.

You alternate turns, and each turn you can take 1, 4 or 7 stones.

The player that takes the last stone (leaving nothing in the pile) wins!

Which player can guarantee a win?

The one who goes first The one who goes second Neither player

2 solutions

Geoff Pilling
Jan 3, 2018

The first player can take 4 stones, leaving 96 in the pile, which divides evenly by 8.

After that if the other player takes $n$ stones, the first player takes $8-n$ stones, leaving the pile with a number that is again divisible by 8.

This continues until the pile contains 0 stones, at which point the $\boxed{\text{player that went first}}$ has won.

The option of "neither player" would be interesting if it was possible for the game to not end, e.g. if we could only take stones of sizes 4, 7, 10, which leaves the possibility of having 1/2/3 stones left behind.

Calvin Lin Staff - 3 years, 5 months ago

Definitely more interesting!

I suppose the strategy then would be to leave the other player with a number of stones in the pile divisible by 14. I wonder if their are distributions where the strategy is more complex than "leaving the other player with a pile divisible by n".

For example, 2, 5, and 7... I think that perhaps the best either player could do then is force a draw if the original pile contains more than 11 stones.

Geoff Pilling - 3 years, 5 months ago

Steven Perkins
Jan 5, 2018

The first player can also win by first taking 7 stones. Then on each set of turns forcing that 8 stones be removed until they are left with 5 stones and it is the second player's turn.

Whether he takes 1 or 4 the first player can do the opposite and gets the last stone.

Taking 1 stone on the first move would be a mistake. I will leave it to the reader to see a winning strategy for the second player in that case.

Good point!

Geoff Pilling - 3 years, 5 months ago

1, 4 or 7?

2 solutions

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