The Game Poison Involves Two Teams of Any Number of People. It Begins with a Certain Amount

Ashleigh Hughes9/9/04

Poison Lab

Problem

The game Poison involves two teams of any number of people. It begins with a certain amount of counters (m). One counter is designated as the poison counter. The first step of the game is to flip a coin to see who goes first. Whoever wins the coin toss get to choose which team goes first. Then the teams must alternate choosing counters from the pile. When choosing counter each team can only choose 1 or 2 counter each turn. The objective is to leave the counter designated as poison to be the last one. Whoever has to choose that counter on their last turn looses.

Methods

In figuring out this problem, we tried many different things to see if we could find a pattern to win every time. We made many guesses at first, but many turned out to be wrong. First, we thought that even and odd numbers had something to do with winning. We thought that if the total number of counters (m) was even, the game would be played differently than if the total amount of counters (m) was odd. We also thought that whoever went first didn’t matter, because you got a choice in how many counters you took. Another guess we made had to do with getting half of the counters. We thought that that the first person to get half the counters would win. These, and many other conclusions, ended up being wrong.

We eventually found one strategy that worked when the total amount of counters we started off with was multiples of 3 plus one. The equation for this would be 3m+1 and the numbers would included 7,10,13,16, and 19… We found out that you have to go second and choose the opposite of the person who went first. So if the team that goes first chooses one counter, then you would choose two counters. You continue to do this until the last one is left, and it should be the other team’s turn to choose. Once we found this pattern we tried many other things to figure out how to win with other total numbers of counters (m). We tried going first and picking the same as the other person, but that didn’t work. After that, we thought about it for a while, and we came up with 2 more equations to include the numbers that 3m+1 did not. The two equations are 3m and 3m+2. 3m means that the total numbers of counters that you start off with is a multiples of 3; for example, 3, 6, 9, and 12… are multiples of 3. 3m+2 means that the number of counters that you start off with is a multiple of 3 plus 2, like 5, 8, 11, 14, and 17…. We get these numbers by plugging in any number to the m or the total amount of counters you start off with. We eventually found patterns that worked for these equations as well. If you start with a total number of counters that is included in the 3m equation, then you want to go first. When you go first, you should pick 2 counters, then pick the opposite amount of counters that the second team picks. If you start with a total number of counters that is included in the 3m+2 equation, you should go first and pick 1 counter. After that you should pick the opposite amount of counters that the second team picks. All of these patterns depend on whether or not you win the coin toss, because if you win the coin toss you get to choose who gets to go first.

Solution

The solution to this game would be to first figure out how many counters you are starting out with. Depending on how many counters you start out with determines how you play the game to win. Then you want to win the coin toss, because that gives you control over the game. Once you have won the coin toss, you take the total number of counters the game is starting out with (m), and you see which equation the amount fits into. If the total is a multiple of 3 (3m), then you want to choose your team to go first. On your first turn, you need to choose 2 counter. After you have chosen two counter it is the other teams turn. Remember how many counters they choose, because you need to choose the opposite number of counter when it is your teams turn. For example, if the other team chooses 1 counter, you want to choose 2. You continue to play until the last counter is left, which is the poison one, and it should be the others team’s. If the total number of counters (m) is not a multiple of 3, you do things a little bit different. When “m” is a multiple of 3 plus 2 (3m+2), you want to do things all the same except on your first turn. You want to take 1 counter instead of 2. Then you play the game the same, choosing the opposite of what the other team picks. If the total amount of counters is a multiple of 3 plus 1 (3m+1), then you want to go second after winning the coin toss. After the other team goes first, you choose the opposite of whatever they choose. If you win the coin toss and take all these steps you should be able to win every time.

Justification

The first thing that must be explained is the equations we use to determine what steps to take. We get these equations by looking at multiples of 3. For example, 3, 6, 9, and 12 are multiples of 3. We choose multiples of 3 because in the problem the two teams can choose up to 3 counters on each round. We will first look at the equation for 3m+1. The goal in this equation is to have 3 counters taken at the end of each round. This is why you what to go second and choose the opposite of the other team. If the other team goes first and chooses 1 counter, and your team goes second and chooses two counters, then the total number of counters gone is 3. Since the equation is multiples of 3 plus 1, there should be 1 counter left on the last round. Since the other team went first it should be their turn to take the last poison counter. When looking at the equation for 3m it is basically the same except you want to go first. On your first turn you choose 2 counters which now makes the problem like the 3m+1 equation. For example, if the total number of counters you start out with is 12 and you take 2 away then you have 10, which is a number that fits into the 3m+1 equation. You continue to play the game just like the 3m+1 equation, and you should win every time. The equation 3m+2 is played. almost the same way. In this equation you choose 1 counter on your first turn instead of 2. In the same way, now the problem becomes the 3m+1 equation. The reason for not using a 3m+3 equation is because it gives you the same numbers as the 3m equation. They will both give you multiples of 3. Basically the whole game depends on whether or not you win the coin toss. Once you win the coin toss, then you have complete control of over the game.