# Misere nim game - winning move for player 1

A misère version of the nim game is being played. Let there be 4 piles of coins, each having 17, 25, 55, 60 coins respectively. What is the winning move for player 1?

I figured out the normal version by calculating the nim sum and reducing the sum to 0. But I’m having trouble how to approach this particular misère version of the nim game where you’re allowed to take as many coins as you want from 1 pile.

• What I have seen asserted is that you play normally until near the end, then deviate. Certainly $2,2$ is a P position in misère as well as normal. I believe essentially all games go through this unless you get down to two piles and one has just $1$ coin. Then you take all the other pile instead of all but 1. Feb 11, 2020 at 3:26
• @RossMillikan so you don’t think there’s a winning first move? I agree about the fact that it is more clear as you get less piles and coins...
– Toby
Feb 11, 2020 at 3:32
• You make the same move as in normal nim. I think if you take $3$ from $55$ that wins in normal Nim, but I did it mentally and may have an error Feb 11, 2020 at 3:34
• @RossMillikan yeah that’s the winning move for the normal nim game. So you’re saying that’d be the winning move for misère as well?
– Toby
Feb 11, 2020 at 3:40
• Yes, that is what I said originally. You just play regular nim until close to the end. I just don't know what close to the end means, but my suspicion is that it is when you have two piles and one has only one coin. Then you change what you do by one coin. Feb 11, 2020 at 3:41

It’s straightforward to check that this is the correct strategy. So the winning move in your case is the one discussed in the comments for normal play, taking $$3$$ from $$55$$.