Algebraic strategy to Poison game

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geekily

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Jan 24, 2007
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I need to find the algebraic strategy to winning the game Poison. The rules are:

Form two teams in your group. One team will play against the other team. Your instructor will give you 10 color tiles. Place the 10 tiles between the two teams, and follow these rules:
i. Decide which team will go first
ii. When it is your team's turn, you must take one or two tiles from the table
iii. Alternate turns until there are no tiles remaining on the table
iv. The team who takes the last tile, the "poison" one, is the loser.

I've played several games with my sister, and messed around with possible combinations of moves myself, and I can't figure out a winning strategy, much less an algebraic one. Any tips?

Thanks so much! This forum is such a lifesaver!
 
Whoever goes 1st is sure to win. Let A and B be the players: A goes 1st.

Strategy for A is to leave 5 after his 2nd pick; to ensure that, A picks 1;
B will pick 1 or 2 or 3; so on 2nd pick, A picks a number such that 5 are left.

B is next:
if B picks 1, 4 left: A picks 3, so 1 left
if B picks 2, 3 left: A picks 2, so 1 left
if B picks 3, 2 left: A picks 1, so 1 left

You can play with 11 instead of 10; same strategy, except A starts with 2;
or with 12: A starts with 3.
 
Thanks so much for the reply! I'm sitting here trying to play it by myself with 10 coins, and I'm just wondering about a couple of things: The version I'm playing only allows you to take 1 or 2 at a time, so will this strategy always work? It seems to, but I just want to make sure. Also, I'm supposed to vary the game by playing with 9, 11, and 12 coins. You already said that with 11 A would start with 2, but since 12 can't be 3, would A then start with 1 because 12 is even? And thus, 9 would start with 2 because 9 is odd?

Thank you so much for all your help!
 
Also, I'm a little confused: When player A takes the necessary amount to insure that there are 5 left in the middle, it seems B can still win: If B takes 1 and then A takes 2, B would take 1 so that A would take the last one. Only if B takes 2 can A definitely win. Is this because it doesn't go up to 3 in my version? Is there another strategy for only 2?

Thanks for your help!
 
Didn't realise 3 wasn't allowed; quite different if only 1 or 2:
with 10 coins, whoever goes 2nd is sure to win;
the strategy is to leave 4 (instead of 5).

To leave 4, B simply needs to pick the 3rd coin, leaving 7,
which he can do wether A picks 1 or 2.

I don't feel like typing....TRY IT!
 
Thank you so much for clearing this up, you don't know how much I appreciate it! I played around with a bunch of coins for a while, and this is the explanation I came up with for class:

With ten coins, the second player can easily win as long as they take the third coin on their second turn. If the first player takes one, the second player should take two, and if the first player takes two, the second player should take one. The second player’s second turn should take either one or two coins, depending on which number would leave four in the middle. From here, any combination of moves can come out in the second player’s favor. With a different number of coins, the strategy varies slightly. For example, with nine coins, the first player seems to have the advantage as long as he or she takes two coins during their first turn. The first player must make sure that after his or her second turn, there are four coins remaining. Again, once there are only four left, any combination of moves will work out in the first player’s favor. With eleven coins, the second player is most likely to win, but only if he takes two coins on his first turn. With twelve coins, the first player also has the advantage, as he or she is the one who will most likely be in the position to leave four coins in the middle.

Does that sound right? Thanks so much!
 
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