To come up with a strategy that works every time the amount of counters must be consistent. The amount of chips that can be taken in one full turn (both teams taking their counters) is two, three, and four. For the game to be consistent every time the best strategy would be to take the opposite amount of counters as the other team (if they take one then you take two). This gives a total amount of three counters taken in a full turn. If you were to take the same amount of counters as the other team you could end up with a total of two counters (them taking one and you taking one) and four (them taking two and you taking two).
The key number to a winning strategy is three. This is the only consistent number that can be taken every time a full turn is completed. So the key step is to take the opposite amount of counters as your opponent. In the above game where you start off the game with ten counters, ten is not a multiple of three. So there has to be a way to get the counters to a multiple of three. I learned through trial and error that if you go second and take the opposite amount of counters that you always win. So now I have a winning strategy. I tried this with other numbers and it did not work. Then I realized that ten is equal to one more than a multiple of three (3×3+1). I tried other numbers that are one more than a multiple of three and it worked. I tried thirteen (3×4+1), sixteen (3×5+1) and twenty-two (3×7+1) and my hypothesis was correct. For any number that fits the formula 3x +1 will win if you go second and take the opposite. But this strategy did not work for any other numbers.
I started to test some ideas about taking the same amount of counters as the opponent but I couldn?t come with anything consistent. Therefore the path to a winning strategy could not be accurately tracked. I also tried going first and taking one on a couple of different numbers but I would sometimes win and lose. So I took one of the numbers that I won with while going first and taking the opposite and tried to test another hypothesis on it.
The number that I tried was twenty. With twenty, every time I went first, took one counter and then took the opposite amount of counters as my opponent, I would win. I looked at twenty and divided it by three and was left over with two (3×6+2). I tried this strategy out with other numbers with this same calculation and I won every time. So now I came up with a strategy that will win every time with the formula 3x +2. It is to go first, take one, and take the opposite amount of counters as your opponent.
Now that I came up with a strategy for numbers that fit into the 3x +1 and 3x +2 formulas the only numbers left would be multiples of three. Since 3x +1 works where you get the opposite amount of counters as your opponents, I figured that if I could get a multiple of three and somehow turn it into 3x +1 then it would be the simple strategy from above. So I took two away first and was left with the formula I was looking for. For numbers that are multiples of three you go first, take two and take the opposite amount of counters as your opponent. This will give you a winning strategy every time.
With these three strategies in mind, any number could be used to play this game, poison. If the number is a multiple of three you go first, take two and then take the opposite amount of counters. If the number is one more than a multiple of three, then you go second and take the opposite amount. If the number is two more than a multiple of three you go first, take one and take the opposite amount of counters as our opponent. You are guaranteed a win if you use these strategies.