Isn't the best strategy always to take off one coin?

btw, love your captcha for comments.

If it is, since there are 14 coins he won

It may be me wrongly parsing the rules, but how does this game actually end?

If you are not allowed to take more coins from one pile than remaining there – and it's in my opinion obvious the "remaining there" applies to the state of the pile after I took the coins – how do you take a coin from a one-coin-pile?

If you take the coin, there is no coin remaining there, which is less than one, and if you don't take a coin…well, you don't take one then.

Help?

Ok, so I'm confused by the statement "up to the number of coins remaining in that pile"... is it the coins remaining after you take some, or before?

If "remaining" refers to the number of coins AFTER you take some, then I take it to mean that for a number of coins taken, N, there must be a number of coins remaining, M where M >= N. If that is the case you could never take the last coin.

If "remaining" refers to the number of coins BEFORE you take some, then the statement itself is redundant, because it follows logically from these two statements: "they may remove any number of coins from one of the piles" and "they can only take from a single pile on a given turn".

one pile: grab the whole pile.
two piles: do not grab a whole pile, then the other person has won. you may grab all but one from one pile, so the other person cannot use the last coin. so he grabs all but one from the other pile, you grab one, he grabs one ... does not work out for you,

so you leave two on the pile, he grabs at least one from one of the piles ... hmmmmm

I've always played this 4 piles of 1,3,5,7

I've called a similar and very fun game Nim and Cross-out. 5 lines of 1, 2, 3, 4, and 5 lines each. You can cross out however many in one line. If you cross out the middle 3, let's say, in the line of 5, then the other two left over become separate lines and can't be crossed out together.

If we take the nim value of this game you get 0. This means that the 2nd player will always win if he plays smart. No matter what the 1st player does the 2nd player can win.

If you play the game of 1 3 5 7 then the 2nd player will win again. Nim value of 0.

The only hope is if the 2nd player stuffs up. If I was the 1st player I would just take all the coins and run

I don't get the joke. I believe Randall has a winning strategy, where Mike does not.

Try playing nim with the edges of a graph instead --- you can remove any positive number of edges you wish, so long as all the edges have a vertex in common. Much more challenging game....

A strange game. The only winning move is to not play.

sounds like a job for binary addition with no carrying.

this show helps explain how to win, along with why you'll always win. http://revision3.com/scamschool/nim/

But don't get forced into a duel against a professional!

The optimal strategy is when you say "You can go first" :D

Oh, and why xkcd?

the captcha however has always remained same and it is remembered by the autocomplete in whatever browser i try it with... or i could be a bot.

I guess xkcd is rational...

Haha, I used to use this game to win free drinks in an inconspicuous manner.

Where is Mike?????

3 ^ 5 ^ 6 = 0 [XOR of three gives a zero] , that means no matter what the first player removes from one of the piles , he is sure going to lose :D

Your Friday challenge invaded my head while I was in bed on Sunday night, and I couldn't get to sleep until I had solved it for three piles! That took quite a while because I was very tired. This made me more tired. :(
The next day I challenged my Mum to a game. :D

2 things:

- Unfunny comic,
- CAPTCHA is already filed out with correct answer

hahaha binary.

just with the captcha; couldn't an unscrupulous person hook it into wolfram|alpha?