Rich responds:
% Decreases over what? It's not worse off than if it didn't have the
% peasants, is it?
Decreases over the "meta-round" of 4 peasant-knight and 1 dragon attack,
but possibly not over the entire battle....
Hmmm... let's try again...
Side A -- N peasants (att/def 1) & M knights (att/def 45)
Side B -- D dragons (att/def 500)
N >> M >> D. Let p=N/(N+M) & q=M/(N+M).
For simplicity, I'm going to ignore the "edge" effects and assume that
Side A is strong enough to always win with many peasants and knights
left over and that D is so much smaller that the vast majority of Side B's
attack come from the "4-in-a-row" rule -- it's just a question of how many
losses each side will take.
Each "meta-round" results (on average) in:
peasants killing 4p*(1/501) dragons,
knights killing 4q*(45/545) dragons,
dragons killing p*(500/501) peasants, and
dragons killing q*(500/545) knights
Killing D dragons will take D/(4p*(1/501)+4q*(45/545)) meta-rounds and
result in:
125 D p
------------- peasant casualties, and
p + 41.36 q
125 D q
-------------- knight casualties.
1.088 p + 45 q
Now let D=1 and M=10, then we have:
N p q peasant knight strength loss
--- ---- ---- ------- ------ -------------
40 .800 .200 11.0212 2.5329 125.0000
60 .857 .143 15.8338 2.4259 125.0000
80 .889 .111 20.2565 2.3276 125.0000
100 .909 .091 24.3347 2.2370 125.0000
200 .952 .048 40.7385 1.8725 125.0000
300 .968 .032 52.5453 1.6101 125.0000
The total losses remain 125 points, as N goes up more of these become
peasants. At this point, some real statistical trials need to be done
to determine "actual" expected results, especially since the assumption
that p & q stay relatively constant is obviously violated for "real-world"
cases.
I'd say that this example says that *I* would *not* use huge N. I may
be saving .9 knights, but it took me 41 noble-days to recruit those extra
fodder^H^H^H^H^H^Hpeasants.
Ed Bailey
--
Ed Bailey | Voice: (512) 471-4198 Fax: (512) 471-6715
Inst. for Fusion Studies | Internet: bailey@{hagar,ziggy}.ph.utexas.edu,
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