The finite counting game up to a certain number, what is it?
Often played by people up to high school ,there’s this game which you may have seen. I’ve seen it described as a finite counting game :shrugs: but here goes
the one i was taught …15 dots, cross out 1, 2, or 3
whoever ends up with the last dot wins( OR, sometimes, loses)
Apparently, its a alt version of some ORIGINAL, nameless( DO YOU KNOW THE NAME???), game where peope take turns saying 1, 2,or 3 numbers, counting up till someone hits 20 for the winning number
Now, a long time ago , i figured out the strategy to the 15 version that i played. you’d want to leave your opponent with 11 , 7 , 3….
ack, my mind fails me, i forgot the strategy..
anyway,
1. is there a NAME to this game?
#2. someone i met said the strategy was that you were playing to get to a end number , N. Now, could you explain this for someone who is having a dum moment?
3. He was able to name any goal number, and figure out what to leave me at to lose. ). How’d this happen? for example, he did a game up to 90. . we could count up to 7. (of course i figrued out b4 we hit 20 that he was holding me at a set number every time i named a number, i think 6 or 2 numbers per round.
i found only one website(doesnt give me the name of the game_ , that has something on it, and mentions this, but it’s for going to a goal number of 20, and 21 , not 15. —–> http://riverbendmath.org/modules/Finite_Games
How can i get to the point so i can have someone name any number, and then KNOW how to set them up? At least i can tell when someone tries to play a rigged game against me, but i’d still like to master this.
K, new info that might help.
I found a old cheat sheet, with thison it-:::
For the game of 15 dots, crossing out one, two, or 3,
(always you go first)
if the goal is whoever gets the last dot LOSES , and you DONT want it, you leave the other guy with 1+4 , which meant 5, 9 , 13
if you WANTED the last ne, because whoever ended up with the last dot WON, it was 0+4 where you leave the other guy with 4,8,12
How does this relate, to the original finite counting game, and do the formulas i put on the cheat sheet …do they have anything to do with the remainder thing on that site i posted?
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One comment
mcbengt on May 31, 2010 at 12:57 am
There is a whole family of games, the prototype of which is called “Nim,” that your game is an example of. In a general Nim-like game, two players begin with a number of rows of dots, and in each turn you choose two things: (1) a row to remove dots from, and (2) a fixed number of dots to remove from that row [usually subject to a restriction on the maximum number you can take away; 3 is pretty common]. Play continues in this way and the person who takes the last dot either loses or wins, depending on how you decide to play.
You get an example of a Nim game by fixing a certain number of rows, and a starting number of dots on each row, and a maximum number of dots to take away. The games you’re describing fix one row (so that at each step you only decide how many to take away), and start with 20 dots (or 15 dots), and have 3 as the maximum number of dots you can take away. So your game is a special case of Nim. The Wikipedia page for Nim (see below) calls this variant “The subtraction game” for the number 20 (or 15). Googling around a bit I find that it tends to also be given different names depending on the starting number.
The complete theory of Nim games was worked out about 100 years ago. It is somewhat complicated for multi-row variations of the game, but for taking dots away from a single row of dots, it does more or less follow the lines you describe.
Suppose we start with N dots, and in each turn a player can remove between 1 and 3 dots. The strategy depends on whether the person who takes the last dot wins, or loses.
If taking the last dot loses (this is most common), your goal at each turn is to remove a number of dots so that the number left has a remainder 1 when you divide it by 4. (If you started with 15, then, you’re looking for 5, 9, and 13.) If you can do this at any point in the game, then you can keep doing it, until the pile gets down to 5 dots, when the person you’re playing against is stuck (whether they remove 1, 2, or 3, you can still leave them with 1 and they lose). The only way you can lose with this strategy is if you start with a pile that has (A multiple of 4) + 1 dots on it to begin with— for example, do not begin the game with 17 dots, as the first player— because then it is _you_ that are stuck, unless your opponent makes a mistake.
If taking the last dot wins (this is less common, but you could still play the game this way), your goal at each step would be to leave the other person with a multiple of 4, with the same idea and explanation of why it works. (The only way you can lose is if it’s your move and you have to take from a collection of dots whose length is a multiple of 4. So don’t play this variation as the first player with 16 dots to begin with; that would be a fixed game for the other player.)
To make a long story short, whether or not you can force a win in a game like this depends on only a very small number of things: (1) whether taking the last dot wins or loses, (2) the starting number of dots, (3) whether you move first or second. If you are in a position to apply the strategies above, then you can win; otherwise all you can do is hope that the person you’re playing against makes a mistake.
For more on this and other Nim-like games consult the links in Wikipedia. A whole subject of math— combinatorial game theory— begins with the study of this game (and others like it).