I saw this game played first in the classic 1960's French film 'Last Year in Marienbad'. I learnt how to play it to always win, by trial and error. I believe there is also a simple maths algorithm for winning play but I have neither the skill nor patience to figure it out. Can anyone help?

 

The game:

 

For two players. Lay four rows of mathsticks: 1, 3, 5, 7.

 

I

III

IIIII

IIIIIII

 

The play: A player can take any number of sticks from any one row at a time. The one who has to take the last match loses.

 

Its always fun to play this a few times and let the patsy win, then offer to bet a couple of beers that you can win the next three games straight out. When he loses, offer double or quits. Eventually he twigs and you both enjoy the beers!

I think the game's generally called nim and the trick for always winning is based on parity.

I think the game's generally called nim and the trick for always winning is based on parity.

 

Nope, it is not based on parity, there is no trick, as any player can choose the move. It's mathematics.

 

If anyone wants, we could play a few times right here on the forum. If you want to make it more interesting, we could do a 1 - 3 - 5 - 7 - 9 instead of 1 - 3 - 5 - 7

 

The simple way ? OK:

 

I

III

IIIII

IIIIIII

 

(Just write the next 'matches' pattern on the response)

Miguel

Edited May 21, 200916 yr by Externet

The game isn't the same as nim, although the trick is the more or less the same.

Edited May 21, 200916 yr by the tree

As far as I can see the game is the same as the one here.

http://www.dgp.toronto.edu/~ajr/270/probsess/03/strategy.html

except for a reversal of the object (ie the last stick wins vs the last stick looses.

The strategy given is based on parity.

There are lots of variations on the basic game and lots of them get called nim.