All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 The folly of war part 1: Knights (Posted on 2005-08-23)
Consider two opposing armies of knights armed only with swords. The sizes of these armies are 500 and 300 knights.

When locked in combat with an enemy each knight has even odds of winning or losing. Knights, being chivalrous, prefer single combat and will not double up on their enemies. The extra knights in the larger army will wait until there is a free enemy to fight.

[Essentially the killing power of the larger army is proportional to the size of the smaller army.]

When the dust settles the smaller army is eliminated. How many knights (are expected to) remain in the larger army?

Generalize for two armies of size A and B where A>B.

 See The Solution Submitted by Jer Rating: 3.4000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: Folly | Comment 10 of 23 |
(In reply to Folly by Bob Smith)

If you use Bob Smith's method, the answer depends on the method of rounding used (or does it?)

Rounding Up:
201    to   1
200.5 to  0.5   round up
201    to   1  and stays there forever

Rounding Down:
201    to    1
200.5 to  0.5   round down
200    to    0

No rounding:
201.1719   to    1.1719
200.5859   to    0.5859
200.293     to    0.293
(whatever a battle between fractionated, but still chivalrous knights means)

 Posted by Larry on 2005-08-23 17:52:05

 Search: Search body:
Forums (1)