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
Full manual solution (most probable) by Dimmeh)
While 201 and 202 are the most highly probable individual numbers of knights to be left, that is not the expected value, as the expected value is the average value you'd have after repeating the experiment many, many times. That's the number that's ever so slightly above 200 and is influenced by the probabilities of other numbers being left.
The probability of being exactly 201, like that for 202, is 0.0163071442412252144, per the following table:
190 0.0146334717170926276
191 0.0148977872981895387
192 0.0151428166945413404
193 0.0153673395120880818
194 0.0155702086805644921
195 0.0157503598553809408
196 0.0159068203837456521
197 0.0160387177335279544
198 0.0161452872865746518
199 0.0162258794028137931
200 0.0162799656674898391
201 0.0163071442412252144
202 0.0163071442412252144
203 0.0162798290917424251
204 0.0162251987927768465
205 0.016143391067771249
206 0.0160346813636111732
207 0.0158994816893817367
208 0.0157383382938812462
209 0.0155519281956457661
210 0.0153410545929929421
10 for L=190 to 210
20 print L,combi(800-L-1,500-L)/2^(800-L)
30 next
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Posted by Charlie
on 2005-08-24 13:43:13 |
re: Full manual solution (most probable)
