Adam StrongMorse
United States Michigan

Many of us have an intuitive sense of the odds in an ordinary Risk battle. When an attacker is rolling three dice versus the defender's two, the odds of each attack are about even. Technically, the attacker has a slight edgebut it's very small, swamped by luck except in the case of very very large battles. Therefore, attackers will typically win if they outnumber defenders significantly, will typically lose if they are outnumbered significantly, and face just about random chance if the numbers are even (or close to even).
But most of us don't have intuitions about the modified battles that come up in Risk Legacy (even in the first game). How much does a Bunker help? What about Fortifications (or the Die Mekaniker starting power)? We weren't sure, so my brother whipped up a Perl script to calculate odds for single exchanges. Here are the results, always assuming 3on2 die rolling (some of these may not be possible in Risk Legacy; other possibilities may come up in later games that we haven't seen yet. ) In every case, this is from the attacker's perspective2 wins means the attacker beats the defender twice. Expected value is average number of the 2 fights won by the attacker.
# results # no modifiers # 2 wins 2890/7776 = 37.2% # 1 win 2611/7776 = 33.6% # 0 wins 2275/7776 = 29.2% # expected value = 1.08
# high defense +1 (e.g. Attacking a bunker) # 2 wins 1860/7776 = 23.9% # 1 win 3176/7776 = 40.8% # 0 wins 2740/7776 = 35.2% # expected value = 0.89
# both defense +1 (e.g. Attacking a fortification) # 2 wins 1470/7776 = 18.9% # 1 win 2436/7776 = 31.3% # 0 wins 3870/7776 = 49.8% # expected value = 0.69
# low defense 1 (e.g. one of Enclave of the Bear's attack powers) # 2 wins 3280/7776 = 42.2% # 1 win 2951/7776 = 38.0% # 0 wins 1545/7776 = 19.9% # expected value = 1.22
# high defense +1 low defense 1 # 2 wins 2022/7776 = 26.0% # 1 win 3972/7776 = 51.1% # 0 wins 1782/7776 = 22.9% # expected value = 1.03
# both defense 1 # 2 wins 4710/7776 = 60.6% # 1 win 2016/7776 = 25.9% # 0 wins 1050/7776 = 13.5% # expected value = 1.47
# high defense 1 (e.g. Ammo Shortage) # 2 wins 4120/7776 = 52.9% # 1 win 2061/7776 = 26.5% # 0 wins 1595/7776 = 20.5% # expected value = 1.32
# low defense +1 # 2 wins 2300/7776 = 29.6% # 1 win 2256/7776 = 290% # 0 wins 3220/7776 = 41.4% # expected value = 0.88
# high defense 1 low defense +1 # 2 wins 3358/7776 = 43.2% # 1 win 2020/7776 = 26.0% # 0 wins 2398/7776 = 30.8% # expected value = 1.12
I hope people find this interesting and useful.

Adam StrongMorse
United States Michigan

Here's the base code, if you want to hack it up for other possibilities:
#!/usr/bin/perl use warnings;
# a13 are the attack dice, unsorted my $a1; my $a2; my $a3; # d12 are the defense dice, unsorted my $d1; my $d2;
# ahl are the attack dice, sorted (high, mid, low) my $ah; my $am; my $al; # dhl are the defense dice, sorted (high, low) my $dh; my $dl;
my $temp;
# w02 are number of attacker wins (0, 1, or 2) my $w0=0; my $w1=0; my $w2=0;
for $a1 (1..6) { for $a2 (1..6) { for $a3 (1..6) { for $d1 (1..6) { for $d2 (1..6) { # sort dice if ($a1>$a2) { if ($a2>$a3) { $ah=$a1; $am=$a2; $al=$a3; } else { if ($a1>$a3) { $ah=$a1; $am=$a3; $al=$a2; } else { $ah=$a3; $am=$a1; $al=$a2; } } } else { if ($a1>$a3) { $ah=$a2; $am=$a1; $al=$a3; } else { if ($a2>$a3) { $ah=$a2; $am=$a3; $al=$a1; } else { $ah=$a3; $am=$a2; $al=$a1; } } } if ($d1>$d2) { $dh=$d1; $dl=$d2; } else { $dh=$d2; $dl=$d1; } # modify dice values if ($dh>1) {$dh;} if ($dl$dl) { $w2++; } elsif ($ah

Alexander West
United States Washington

Quite interesting. Thanks for sharing!

Danny Barbuto
United States Poughkeepsie New York

I was going to post something similar but forgot. If someone is attacking my territory that has a bunker, could it more advantageous for me to roll one die instead of two?
Also, why do people on this site run scripts to find the experimental probability? Wouldn't it be easier to determine theoretical probability?

A L D A R O N
United States Cambridge Massachusetts
A L D A R O N
+++++[>+++++++++++++.>++++++++++[>++++++++++++++++++...+++++++++++++++++...

http://anydice.com/

Danny Barbuto
United States Poughkeepsie New York

CapNClassic wrote: I am not sure you understand what the scripts are doing since you are calling them "experimental" probabilities. The scripts are not randomly rolling a set of dice millions of times, the script is systematically going through every possible die combination
The last Risk odds program I used was experimental, it would "roll" dice n number of times then tell you the odds. But I see now that the script you're using is calculating the theoretical probability. I was originally mistaken.
So what exactly are the odds in the 1 die vs. 2 dice Bunker scenario?

Adam StrongMorse
United States Michigan

Okay, I think that OtubrabNad's question can be most easily answered by calculating it directly. I'm assuming 3 attacking dice.
PDefenseWin1Die= 1/6*PDefenseWinDrolls6+1/6*PDefenseWinDrolls5+...
If D rolls a 5 or a 6, D will always win with a bunker (assuming no other modifiers). Therefore
PD=1/6*1 + 1/6*1 +1/6*PDefenseWinDrolls4 + ...
PDefenseWinDrolls4 = 1P of a six on 3 dice; P of a six on 3 dice is 1/6+5/6(1/6)+25/36*1/6=91/216. So PDefenseWinDrolls4=125/216
So now we're at: PD=1/3+125/1296 + 1/6*PDefenseWinDrolls3 + ... = 557/1296 + 1/6*PDefenseWinDrolls3 + ...
PDefenseWinDrolls3= 1  P of a 5 or 6 on 3 dice P of a 5 or 6 on 3 dice= 1/3 + 2/9 + 4/27= 19/27 So PDefenseWinDrolls3= 8/27 So PD = 557/1296+ 1/6*8/27 + 1/6*PDefenseWinDrolls2 + 1/6*PDefenseWinDrolls1 1/6*8/27=4/81=64/1296, so: PD=621/1296+ 1/6*PDefenseWinDrolls2 + 1/6*PDefenseWinDrolls1 PDefenseWinDrolls2 = 1P of 4,5,6 on 3 dice= 1  7/8=1/8 So PD= 621/1296+1/48 + 1/6*PDefenseWinDrolls1 = 648/1296 + 1/6*PDefenseWinDrolls1 PDefenseWinDrolls1=1  P of 3, 4, 5, or 6 on 3 dice= 1 (2/3 + 2/9 + 2/27)=1/27 PD= 648/1296+ 1/162=656/1296
So PD=0.50617... That means that the attacker's probability of winning is 0.49382716
So, if the attacker attacks twice, then the attacker has a probability of 0.2438 of winning twice, a probability of 0.256 of losing twice, and a probability of 0.49992 of winning once and losing once; an expected value of 0.98765432. That’s roughly 10% better, so when the attacker has at least 4 troops available to attack, the defender is better off using the bunker with 2 dice. (It’s actually worse than that, because the intermediate result gives the attacker more information, which is bad from the defender’s perspective.)
If the attacker has exactly 3 troops available to attack (i.e. attacking from a 4 troop country), it gets more complicated, because if the attacker loses the first attack, there will only be 2 troops for the second attack. I’m not going to work that out right now, but my rough guess is that even then, the defender probably wants to use both dice.


