I’m playing a Playstation video game which presents the following varient of the game of NIM as a side-game: three piles of randomly chosen, but often roughly equal, sizes. You can take 1, 2, or 3 stones from any one pile. Whoever takes the last stone wins, and you always choose who goes first.
Now, even a non-mathematician like myself can see that given these circumstances, you SHOULD be able to win almost every, if not every, time. But is there a relatively easy to understand strategy I can use without knowing anything about binary numbers, or are all winning strategies so steeped in complex math that I couldn’t possibly grasp it? Any help is appreciated!
Um, if you leave one stone in pile A, one in pile B, and two in pile C, for a total of 4 stones, you’ve lost, if the computer opponent is any good (takes both stones from pile C on the next turn, in other words).
My advice, humble as it is, would be to attempt to eliminate one entire pile while always reducing the total number of stones to a multiple of four. Once you get to a two-pile version of the game, it should be a snap.
With that said, if the total number of stones in all three piles is a multiple of four at the start of the game, you don’t want to go first. Otherwise, you do. But, I’ll also assume that you don’t get to count before choosing who goes first, in which case, if the opponent is at all versed in the game, you will lose one game out of four simply due to chance.
What Playstation game is it, anyway? My wife is getting back into console gaming after a break due to the birth of our first child… and I don’t think she knows Nim (hehehe).