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Partial answer: Because Because of this, Continuing from Bass's partial answer. Before I get into the lines, I'll introduce to more critical boards. Position A: | X | ----------- X | | ----------- | O | Position B | X | ----------- O | | ----------- | X | These positions are important because This means that I ...

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So, we are playing misère Tic-tac-toe with the option of taking more than one turn in a row. Seems quite interesting, so let's take a stab at it. The flow of the narrative here is a bit messy, because the answer was not all written in one go. Since a full rewrite would be too time consuming, I've added some chapter breaks to give at least some structure to ...

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What I tried is the same game where you have 101 stones but instead of winning more than five would make you lose I tried it with 3. Working on @Rand al' Thor's strategy by picking 5 first, I figured out we can always lose if we copy our opponent's move every time. For eg. If I first play 5, then we are left with 96 stones in total. Our opponent might take 3 ...

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I choose to play My strategy is as follows: Continue until this is no longer possible or doing so would end the game. That means So if I take

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as @Bass mention I think you can lower the table to Pebbles left | you have Odd number stones | you have Even number stones 0 | W | L 1 | L | W (1) 2 | W (2) | W (1) 3 | W (2) ...

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The strategy is to do a move that In particular, for $15$ pebbles, your first move would be The reason this works is more interesting than with other single-pile nim variants.

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As with pretty much all the nim variants, this one can be solved by starting from the end and working backwards. With the original total number of stones being an odd number (15, as given in the title) the players will have the same parity whenever there's an odd number of pebbles left, so it's easy to work out the best strategies: they are the ones that put ...

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Alice can achieve a victory probability of Proof: Proof that there is no better strategy:

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