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Here's a simple game at which a human can out-fox even the cleverest algorithm.

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Please help :-)

I have tried to find out the possible configurations with only two marbles left and I am finding that the number of those is quite larger than the five possibilities with only one marble left.

The article author writes that "finishing with two marbles might be a little bit, but not much, easier than finishing with one", which is not very precise.

My argument is based on the fact that the last two marbles left should occupy holes labeled "f" and "g" and then I start discarding possibilities based on the symmetry of the board. For example all the "f" and "g" holes that are converted into "h" by a symmetry will be discarded.

But I still get too many to agree with the author's words (sorry for not giving the total number in my count. I stopped the counting when I realized something must have been wrong).

Thank you,

Marcos