Author |
Topic: Coin Removal (Read 1546 times) |
|
Jeremy Randolph
Guest
|
You’re playing a game with a friend. A large number of coins are placed in front of you; each person takes turn removing 1 to n coins (the current player decides how many coins to remove). Whoever removes the last coin loses the game. Your opponent gives you the first move. Devise a strategy to maximize your chances of winning.
There are enough coins so that each person will have at least one turn before someone loses.
|
|
 IP Logged |
|
|
|
BNC
Uberpuzzler
Gender: 
Posts: 1732
|
 |
Re: Coin Removal
« Reply #1 on: Jan 26th, 2003, 1:06am » |
 Quote  Modify
|
Since "n" is not bounded in this riddle, taking all the coins - 1, leaving you poor friend with a single coin to remove should work.
|
|
IP Logged |
How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
|
|
|
jeremy randolph
Guest
|
n is a constant number for each individual game.
maybe my last statment wasn't clear, but basically there are at least 2n coins. both players will have the option of taking a maximum number of coins before someone will lose.
|
|
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Coin Removal
« Reply #3 on: Jan 26th, 2003, 11:35am » |
Quote Modify
|
if there are 2 up to n+1 coins left then you can win by taking all but one coins
for n+2 coins there is no way to win since you have to take at least 1, reducing it to the first case, but for the other player..
for n+3 up to n+n+2 coins you can win by taking all but n+2 coins.
for 2*n+3 coins there is again no way to win, since 1 to n coins will reduce to the former case in which the other player will win.
for 2*n+4 to 3*n+3 you can reduce it to the former case and win..
etc..
you can win if the number of coins is between (inclusive) (i-1)*(n+1)+2 and i*(n+1) for some i, by reducing the number of coins to (i-1)*(n+1) +1
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Misha
Newbie
Gender: 
Posts: 3
|
|
Re: Coin Removal
« Reply #4 on: Jan 26th, 2003, 5:19pm » |
Quote Modify
|
Generally speaking, most games like this allow you to remove 1 to 3 coins only. So you can take one, two, or three coins in your turn. Therefore, you won't be able to take all but one leaving one for your opponant. The best strategy in a two player game is to take the maximum each time til you can make the remainder 5 after taking your 1,2, or 3.
If your opponent takes 1, you can take 3 busting them
If your opponent takes 2, you can take 2 busting them
If your opponent takes 3, you can take 1 busting them.
The strategy here is to be sure you opponent has 5 to choose from on their turn.
I'm sure this strategy can be altered for other numbers (ie: 1 to 4 each turn, 1 to 5 each turn etc). But this is the basic principle of the game.
|
|
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Coin Removal
« Reply #5 on: Jan 26th, 2003, 11:54pm » |
Quote Modify
|
In the n=3 case taking the maximum number of coins untill there are 5 left will often make you lose.
Suppose there are 10 coins, you take the maximum, 3, so there are 7 left, your opponent take 2, leaving 5, you loose..
Whereas taking 1 makes you win.. (because 10=2*(3+1)+2 )
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
TimMann
Senior Riddler
Gender:
Posts: 330
|
|
Re: Coin Removal
« Reply #6 on: Feb 2nd, 2003, 3:06pm » |
Quote Modify
|
The winning strategy is to always make the number of coins remaining after your turn congruent to 1 modulo n+1. If the number is already of that form before your turn, your opponent is winning, so take an arbitrary number and hope he makes a mistake.
|
|
IP Logged |
http://tim-mann.org/
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print
Remove