My friend likes to impress people by playing 3-5-7 which has three piles of counters of sizes 3, 5 and 7. You can remove any number of coins from a single pile, the last player to move loses.

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This is a winning position for the first player, but With a solid understanding of the game tree she wins nearly every time playing second. She says, it reduces to knowing a few winning positions.

Two piles of the same size is second player win, in the jargon of Combinatorial Game Theory. Here is the pile (5,5).

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If first player moves to (5,n) for n > 1, second player can imitate on the other pile, moving to (n,n). However, if first player moves to (5,1), second player moves to 1 and wins.

The other winning positions she remembers is (3,4,1) and (4,5,1). She can win once she recognizes these positions. Eventually (after losing many times) I told her that (n, n+1,1) is a losing position for any n...

If our game were played in normal play convention (player moving last **wins**), but real life Nim is played as a misere game. Probably the analysis is similar to normal-play Nim with some modification.

Recently there was a theory of Misere quotients where each game has a commutative monoid assoicated with it. What does that monoid looks like here? Is it finitely generated?