OFFSET
1,1
COMMENTS
In the game of Risk, an attack occurs when a player moves some number of armies into a territory defended by the armies of another player, and the outcome of the attack is determined by repeatedly rolling 6-sided dice according to the number of attacking and defending armies present, up to a maximum of 3 and 2 dice for the attacker and defender, respectively. Some number of armies are eliminated corresponding to comparing highest versus highest roll (with the loser losing one army), and second-highest versus second-highest roll (with the loser again losing one army), with ties going to the defending side. This continues either until the aggressor decides to retreat before the next roll, or until one side has lost all of its armies in the territory (in which case the victor has possession of the territory with the remaining armies).
Example: 3 of Alice's armies invade a territory held by 3 of Bob's armies; Alice rolls a (1,2,5) and Bob rolls a (4,2). Alice's 5 beats Bob's 4, and Bob's 2 beats Alice's 2 (as Bob is defending). Each side therefore loses one army, with each side having 2 armies remaining. Note that in this example, if Alice continues the attack, she will now roll only 2 dice against Bob's 2, thereby reducing her odds of eliminating Bob's armies in that territory. It can be shown that, in an arbitrarily large set of rolls, the odds favor the attacker. When the attacking player rolls 3 dice and the defending player rolls 2, of the 6^5 = 7776 possible rolls (all equally likely to occur), there are
2890 that result in the loss of 2 armies by the defender,
2275 that result in the loss of 2 armies by the attacker, and
2611 that result in the loss of 1 army by each.
However, in a 1-vs.-1 scenario, the defense is favored 7/12 to 5/12 because defenders win ties. Thus the probabilities are not so trivial as to which matchups are favorable. This is further complicated by intermediate-size armies which require repeated rolling, such as the 3-vs.-3 example above, where there is a nearly 2/3 chance that the attacker will lose troops that land them below the three-army threshold, but a 1/3 chance that the defender will be down to 1 defender against 3 attackers, which heavily favors the attacker. This requires recursively calculating the probabilities for the further downstream outcomes.
EXAMPLE
This sequence lists the numerators T(n,k) along antidiagonals of the probability of n attacking armies winning against k defending armies. The denominator of each probability is 7776^(n+k-1). The sequence starts with T(1,1) and moves to T(1,2), T(2,1); T(1,3), T(2,2), T(3,1); etc. Hence the first value is 3240/7776 = 5/12.
A k-vs.-k attack by 4 or fewer armies, played until complete elimination, will favor the defender, but 5-vs.-5 or more will favor the attacker, with the probability of victory by the attacker steadily increasing with each additional die added to both sides.
PROG
(Python)
from itertools import product
from collections import Counter
from functools import lru_cache
from fractions import Fraction
def get_elims(o, d): return tuple(-''.join(['od'[int(o>d)] for o, d in zip(*[sorted(i)[-2:][::-1] for i in [o, d]])]).count(j) for j in 'od')
@lru_cache()
def elim_p(o_n, d_n): return {j:Fraction(k, 7776) for j, k in Counter([get_elims(i[:o_n], i[o_n:o_n+d_n]) for i in product(range(1, 7), repeat=5)]).most_common()}
@lru_cache()
def res_p(o_n, d_n): return {(o_n+i[0], d_n+i[1]):j for i, j in elim_p(min(3, o_n), min(2, d_n)).items()}
def get_final_result(o_n, d_n):
d, g = {(o_n, d_n):1}, [(o_n, d_n)]
while g:
w, p = g[0], d.pop(g[0])
for i, j in res_p(*w).items():
d[i] = d.get(i, 0)+j*p
g = [i for i in d.keys() if min(i)>0]
return d
def get_win_count(o_n, d_n): return (sum(j for i, j in get_final_result(o_n, d_n).items() if i[0]>i[1])*(7776**(o_n+d_n-1))).numerator
a = [get_win_count(o_n, t-o_n) for t in range(2, 30) for o_n in range(1, t)]
CROSSREFS
KEYWORD
AUTHOR
Nicholas Stefan Georgescu, Apr 12 2024
STATUS
approved