[risk] Combat Function

[Richard Rognlie <rrognlie@gamerz.net>]

I know I like the Schmittberger larger force loses as many as
the smaller force (less their difference in size), but that
does not scale to more than 2 players...

It's no huge secret that I think the current algorithm of
"largest army wins but loses as many armies as the next
larger force" is arbitrary, and *nasty*...  but no one has 
proposed anything else...

I have a brain storm... what do you think of this? (please
remember, I'm typing out loud...)

largest force wins, but loses a number of armies proportional to
the number of *other armies lost* / *total armies*.  (round up) 

A 2 -> B 1		A loses  1/3 * 2 == 2/3 == 1 army. 
A 3 -> B 1		A loses  1/4 * 3 == 3/4 == 1 army.
A 3 -> B 2		A loses  2/5 * 3 == 6/5 == 2 army.
A 4 -> B 1		A loses  1/5 * 4 == 4/5 == 1 army.
A 4 -> B 2		A loses  2/6 * 4 == 8/6 == 2 army.
A 4 -> B 3		A loses  3/7 * 4 ==12/7 == 2 army.
 
But the interesting bit is the multiple player battles...

A2 -> B1 C1		A loses 2/4 * 2 == 4/4 == 1 
A3 -> B1 C1		A loses 2/5 * 3 == 6/5 == 2 
A3 -> B2 C1		A loses 3/6 * 3 == 9/6 == 2 
A3 -> B2 C2		A loses 4/6 * 3 ==12/6 == 2 
A4 -> B1 C1		A loses 2/6 * 4 == 8/6 == 2 
A4 -> B2 C1		A loses 3/7 * 4 ==12/7 == 2 
A4 -> B2 C2		A loses 4/8 * 4 ==16/8 == 2 
A4 -> B3 C1		A loses 4/8 * 4 ==16/8 == 2 
A4 -> B3 C2		A loses 5/9 * 4 ==20/9 == 3 
A4 -> B3 C3		A loses 6/10* 4 ==24/10== 3 

So depending on the destribution, it approximates the "kludge"
but its a bit more generic...  Or am I smoking something?!?

Richard

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