Misere nim game - winning move for player 1

Question

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.

Answer 1

Wikipedia says:

When played as a misère game, Nim strategy is different only when the normal play move would leave only heaps of size one. In that case, the correct move is to leave an odd number of heaps of size one (in normal play, the correct move would be to leave an even number of such heaps).

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 $55$.