Home > Riddles > Nim (3,4,5) in modo misère
Nim (3,4,5) in misère mode works when the reader accepts a simple rule and follows it to the end without losing the thread. That's where the beautiful part appears: in the clarity of the mechanism.
There are three piles with 3, 4 and 5 chips. On each turn you can withdraw any positive number of tokens, but from a single pile.
It is played in misère mode: whoever takes the last piece loses. Does the first player have a winning strategy?
If you have it, what is your first move?
Answer: Yes, the first player wins. The first correct move is to go from $(3,4,5)$ to $(1,4,5)$ (remove 2 from the pile of 3). Initial XOR calculation:
$$ 3\oplus4\oplus5=2\neq0. $$ Since we are not in the “all 1” phase, in misère it is played like normal Nim: you have to leave XOR 0. With $s=2$: - $3\oplus2=1<3$ (valid), - $4\oplus2=6>4$, - $5\oplus2=7>5$. The only play that leaves XOR 0 is reducing 3 to 1. **Misère closure:** when at the end there are only piles of size 1 left, it is adjusted to leave an odd number of piles to the rival. $$
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