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 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)

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 re: Full manual solution (most probable) | Comment 14 of 23 |
(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

 Posted by Charlie on 2005-08-24 13:43:13
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