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.