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This is a very interesting Nim variant. I have figured out a strategy for winning. But you have to memorize a few numbers. Well, 23 numbers, to be exact. And here they are:
2 13 45
112 144 155 222 233 247 256 346 357
1113 1145 1223 1246 1257 1333 1347 1356
2245 2344 2355
These numbers assume a layout of 2 3 5 7 = 17 stones. Here's how to win: 1) If your opponent presents you with a balanced board (every power of 2 in each group has its mate in another group), pass your turn and name the object of the game (miserè or standard). It doesn't matter which object you call unless you're very close to the end.
2) If your opponent presents you with an unbalanced board (most of the time), you should take a normal turn and present them with an unbalanced board. The configuration should be one of the 23 numbers above.
If no one calls an object by the time one player presents the other with a single pile of two stones, we have a problem. The active player can simply take the two stones and end the game  without anyone calling an object! This leaves the outcome in limbo. So I propose that when the game comes down to a single pile of two stones, the active player may simply play according to the normal rules, i.e., take one stone or take both stones or pass their turn and call an object. But if the player chooses to take both stones, their opponent still gets another go. Since there are no more stones left on the table, the opponent's choice is simple: pass and call miserè or pass and call standard. If they call standard, they lose because the other player has already taken the last stones. If they call miserè, they win for the same reason.
I hope you all have fun with this game.