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.
Consider rounds of battle.
After each round of battle B will be halved
and A will be reduced by same amount.
After nine rounds of battle B will be < 1.
Last man in army B is dead.
After each round A = 200 + B/2 and B = B/2
A B
Start: 500,300
After Round1: 350,150
After Round2: 275,75
After Round3: 237.50,37.5
After Round4: 218.75,18.75
After Round5: 209.375,9.375
After Round6: 204.6875,4.6875
After Round7: 202.34375,2.34375
After Round8: 201.171875,1.171875
After Round9: 200.5859375,0.5859375
Probability suggests that 200 men would be left
in army A when the last man in army B is killed.
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Posted by Paul
on 2005-08-27 13:16:03 |