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.
(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?)
201 to 1
200.5 to 0.5 round up
201 to 1 and stays there forever
201 to 1
200.5 to 0.5 round down
200 to 0
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