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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 generalised derivation | Comment 22 of 23 |
(In reply to Full generalised derivation by Machiveli)

Suppose we apply that formula to the case where the larger army has 2 knights and the smaller only 1, and in any individual combat the probability of either one winning is equal, for f=1/2.

The formula gives the expected number of soldiers in the larger army remaining as 2-1*1, since E(n)=(1 - 1/2) / (1/2).  So that expected value is 1.

What are the possible outcomes? Consider W as representing a win by the originally larger army and L a loss.

W p = 1/2, leaving 2 knights in the larger army
LW p = 1/4, leaving 1 knight in the larger army
LL p = 1/4, leaving 0 knights in the larger army

The expected number, overall, of knights remaining from the larger army is
1/2 * 2 + 1/4 * 1 + 1/4 * 0 = 5/4.

The expected number, given that in fact the larger army won is
(1/2 * 2 + 1/4 * 1) / (3/4) = 5/3

Neither of these is the 1 given by the formula.

 Posted by Charlie on 2005-08-28 17:42:36

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