I tried it, cool! But how come 6vs6 and 8vs8 *both* have lower attacking percentage than 7vs7?

Here's a link to a crude screenshot:
http://img245.imageshack.us/img245/9562/statsbk5.png

...and (hopefully) here's a hyperlink to the above screenshot:

<a href="http://img245.imageshack.us/img245/9562/statsbk5.png ">
http://img245.imageshack.us/img245/9562/statsbk5.png </a>

to answer your question joby.d: it is because the %'s are from running the matchups 100,000 times. even though that is a large number there can still be some random variations about the mean. here is the code (python):

import random

def rolldie():
return random.randrange(6)+1

iterations = 100000.0 # must be integer
resultvector = []
for attacker in range(8):
resultvector.append([])
for defender in range(8):
numberwon = 0
for x in range(int(iterations)):
asum = 0
dsum = 0
for a in range(attacker+1):
asum += rolldie()
for b in range(defender+1):
dsum += rolldie()
if asum>dsum:
numberwon+=1
resultvector[len(resultvector)-1].append(numberwon/iterations)

print resultvector

outfile = file('result.csv', 'w')

outfile.write('A/D #,1,2,3,4,5,6,7,8,')

for x in range(len(resultvector)):
outfile.write('\n'+str(x+1)+',')
for y in range(len(resultvector[0])):
outfile.write(str(resultvector[x][y]*100.0)+',')
outfile.close()

whoa, that looks ugly. it didn't keep the tabs.

nice screenshot, i should have thought of that.

okay, i was assuming a classical approach

i don't suppose anyone wants to chart how many times you should attack for "reserves vs #of territories?" ;)

I've noticed that when you attack one specific enemy territory with 8 dice territories and he is defending with 8 dice, the first two attacks always fail and the third attack always succeeds. Has anyone else noticed this?

I calculated the odds using all possible permutations, rather than random matchups, so these are more precise, aside from rounding to the nearest tenth of a percent.

Attacking: 1 2 3 4 5 6 7 8
1 defending: 41.7 83.8 97.3 99.7 100 100 100 100
2 defending: 9.3 44.4 77.9 93.9 98.8 99.8 100 100
3 defending: 1.2 15.2 45.4 74.3 90.9 97.5 99.5 99.9
4 defending: 0.1 3.6 19.2 46 71.8 88.4 96.2 99
5 defending: 0 0.6 6.1 22 46.4 70 86.2 94.8
6 defending: 0 0.1 1.5 8.3 24.2 46.7 68.5 84.4
7 defending: 0 0 0.3 2.5 10.4 26 46.9 67.3
8 defending: 0 0 0 0.6 3.7 12.2 27.4 47.1


Code, in OpenOffice's basic:

sub main

dim possibledice(8,48)
dim ratios(8,8)

For i = 1 to 8
for j = 1 to 48
possibledice(i,j) = 0
next
next

For d1 = 1 to 6
possibledice(1,d1) = possibledice(1,d1) + 1
for d2 = 1 to 6
possibledice(2,d1+d2) = possibledice(2,d1+d2) + 1
for d3 = 1 to 6
possibledice(3,d1+d2+d3) = possibledice(3,d1+d2+d3) + 1
for d4 = 1 to 6
possibledice(4,d1+d2+d3+d4) = possibledice(4,d1+d2+d3+d4) + 1
for d5 = 1 to 6
possibledice(5,d1+d2+d3+d4+d5) = possibledice(5,d1+d2+d3+d4+d5) + 1
for d6 = 1 to 6
possibledice(6,d1+d2+d3+d4+d5+d6) = possibledice(6,d1+d2+d3+d4+d5+d6) + 1
for d7 = 1 to 6
possibledice(7,d1+d2+d3+d4+d5+d6+d7) = possibledice(7,d1+d2+d3+d4+d5+d6+d7) + 1
for d8 = 1 to 6
possibledice(8,d1+d2+d3+d4+d5+d6+d7+d8) = possibledice(8,d1+d2+d3+d4+d5+d6+d7+d8) + 1
Next
Next
Next
Next
Next
Next
next
next

Open "c:\kdice.out" For Output As #1
for player1dice = 1 to 8
for player2dice = 1 to 8
waystolose=0
waystowin=0
for player1 = 1 to 48
for player2 = 1 to 48
if player1 >= player2 then
waystolose = waystolose + possibledice(player1dice,player1) * possibledice(player2dice,player2)
else
waystowin = waystowin + possibledice(player1dice,player1) * possibledice(player2dice,player2)
endif
next
next
ratios(player2dice,player1dice) = cint(((waystowin*1000.0) / (waystolose+waystowin)))/10
next
print #1,player1dice;" defending: ";ratios(1,player1dice);" ";ratios(2,player1dice);" ";ratios(3,player1dice);" ";ratios(4,player1dice);" ";ratios(5,player1dice);" ";ratios(6,player1dice);" ";ratios(7,player1dice);" ";ratios(8,player1dice)
next
close #1
end sub

6v2 the attacker wins 99.803 percent of the time.

Woohoo! I got a role that only occurs .197 percent of the time and caused me to lose!