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^{(Now that we've established the notation, you may want take a peek at the final spoiler block to see the surprisingly simple final conclusion.)}

^{(Now that we've established the notation, you may want to take a peek at the final spoiler block to see the surprisingly simple final conclusion.)}

Therefore, the (fastest) way to win is to

let the opponent start. Then, whenever the opponent plays on an even-numbered bowl, respond with an odd-numbered bowl (the smallest possible one), and vice versa.

Therefore, the way to win is to

let the opponent start. Then, whenever the opponent plays on an even-numbered bowl, respond with an odd-numbered bowl (the smallest possible one, if you want the fastest win), and vice versa.

you can interpret the game position as a binary number (marbles are ones, empty bowls are zeroes), and then every possible move becomes a (binary) subtraction of an arbitrary non-negative power of two. This reduces the puzzle to another game we have already solved, so we are done.

^{(Now that we've established the notation, you can take a peek at the final spoiler block if you want to see the surprisingly simple conclusion at this point.)}

There's one marble in an odd bowl, and one in an even bowl, so $\Delta$ is zero, and it's impossible to play a winning move, so the position is losing

so the (fastest) way to win is to

You can interpret the game position as a binary number (marbles are ones, empty bowls are zeroes). If you do, then every possible move gets exactly mapped, one-to-one, to every possible (binary) subtraction of a non-negative power of two. This reduces the puzzle to another game we have already solved, so we are done.

^{(Now that we've established the notation, you may want take a peek at the final spoiler block to see the surprisingly simple final conclusion.)}

There's one marble in an odd bowl, and one in an even bowl, so $\Delta$ is zero, and it's impossible to play a winning move, so the position is losing.

Therefore, the (fastest) way to win is to