Dice Equations.

[The Neon Peon]

Could someone post the actual equations for finding the probability of winning an attack of x vs. y ?

[BaldAdonis]

Hahaha, yeah. It's a massive Markov chain. Plot your point (x, y), and look at all the paths that move to (x, y-2), (x-2, y) and (x-1, y-1) with the proper probability weightings. Add up all the possible paths that bring you to (z, 0), remembering to adjust the probabilities near the axes (to the 3v1, 2v1, 2v2, 1v2, 1v1 odds). (Taflin does the first part as a free-state solution, then finds the odds of reaching any boundary state, then combines them).

If you follow, it's a double summation (with some extra terms near the end), and is entirely useless when the numbers get above 7, unless you are using a computer. And if you are, just use the code on Gambit.

http://www.recreationalmath.com/Risk/RiskPaper.doc

[The Neon Peon]

Thanks a ton!

Sorry, never taken statistics or calculus (I am a freshman only). Did not know any simple way to do this.

[BaldAdonis]

I wouldn't quite call that simple. And that's just to do one attack... attacking in sequence (like Gambit does) extends it to a crazy level: you get a chain that counts all the defenders as one army, and then disperses boundary conditions wherever the defense territories end (so a 20v1,2,3,4,5 attack would be run as a 20v15, but with the condition that the 1st, 3rd, 6th, and 10th defensive armies are treated as boundaries) AND then to cross that boundary, the attacker has to drop an army. There are way more amusing equations to treat those (and that's probably why his paper stopped).

[e_i_pi]

An approximate general equation is

d = 6c^1.9 x a^(0.75c^0.3)

where

d is the number of defenders (maximum)
c is the chance of success for the attacker to defeat the defender
a is the number of attacking armies

Before you ask, no, I refuse to solve for c. So, for 50/50 chances of success, you'd use

d = 0.9657a^1.0488

The number you get for d is the maximum amount of armies the defender has before you have less than a 50% chance of losing. Sounds like an awful mess doesn't it? And it's not even accurate.

I'd go with Gambit. Search in google for Gambit Battle Odds, it'll be the first hit.

[Kexor]

You might also find this useful: Running the numbers for Risk

[maroco]

You might also find this useful: Running the numbers for Risk

according to that guys spreadsheet, you have a greater chance of winning with 24 vs 6 than you have with 25 vs 6.

[phantomzero]

Here's one that I use sometimes for those long shot escallating games...

http://recreationalmath.com/Risk/index.htm

It allows you to put in your initial stack, and then up to 10 defending territs.

[lancehoch]

Someone posted this the other day: http://gamesbyemail.com/Games/Gambit/BattleOdds

[snoop.daniels]

This reply may come too late, but I just finished a javascript program to determine, once and for all, which side has a statistical advantage.

Provided that my logic is sound, the defender has a 55% to 45% advantage over the attacker in any given 3 to 2 dice roll.

Here's how I arrived at it:
If the dice are arranged in series, there are 7776 possible combinations. If not arranged serially, then not all of these combinations are unique, but for the purposes of probability this fact is irrelevant. I essentially incremented the dice like a hex number, checked for the highest dice on each side, then tallied the hits. The result was 6998 for the attacker and 8554 for the defender. The total number of possible hits would by 15557 (2 for each combination out of 7776) meaning that the defender hits 10% more often than the attacker. I've never taken a statistics class, and I have been running headlong from higher mathematics my entire life, so I have no idea if my reasoning is sound or not.

The moral of the story is, just stay put and do nothing and you'll win every time

[e_i_pi]

This reply may come too late, but I just finished a javascript program to determine, once and for all, which side has a statistical advantage.

Provided that my logic is sound, the defender has a 55% to 45% advantage over the attacker in any given 3 to 2 dice roll.

Here's how I arrived at it:
If the dice are arranged in series, there are 7776 possible combinations. If not arranged serially, then not all of these combinations are unique, but for the purposes of probability this fact is irrelevant. I essentially incremented the dice like a hex number, checked for the highest dice on each side, then tallied the hits. The result was 6998 for the attacker and 8554 for the defender. The total number of possible hits would by 15557 (2 for each combination out of 7776) meaning that the defender hits 10% more often than the attacker. I've never taken a statistics class, and I have been running headlong from higher mathematics my entire life, so I have no idea if my reasoning is sound or not.

The moral of the story is, just stay put and do nothing and you'll win every time

Alas, there are problems with your stats. Not sure where, but there are problems. The advantage is more like 53-47 att-def.

[Geger]

The moral of the story is, just stay put and do nothing and you'll win every time

I'm 100% sure you can't win with this tactic. Only in assassin game you can win... if another player kill your target

[SirSebstar]

..........................
If the dice are arranged in series, there are 7776 possible combinations.

that was correct, the rest not.
Taking http://recreationalmath.com/Risk/RiskPaper.doc as starting point, you will win 2890/7776, and loose 2275/7776.
You will tie 2611/7776.
2890 vs 2275 gives you a much better chance for the attacker then the defender. However yuo need to caclulate the odds for a tie as well. In total there is an aprox 15% advantage to the attacker who attacks with 3(land has 4) vs 2.

in short, get a big stack and be first to attack

[JoseFlang]

hello alI was looking this up the other day and this is what wiki has to say

seems right to me plus I think enough math geeks probly argued it out already

Probabilities of winning a dice roll in Risk
(various die combinations)[7]
Attacker dice
one defender

one two three
Attacker wins 1 41.67% 57.87% 65.97%
Defender wins 1 58.33% 42.13% 34.03%


two defenders

Attacker wins all 25.46% 22.76% 37.17%
Defender wins all 74.54% 44.83% 29.26%
Both win one n/a 32.41% 33.58%
I hope you can make sense of it

Thus when rolling three dice against two dice (the most each player can roll) the attacker has a slight advantage. (The defender has a 52.84% chance of winning with the highest die, but the attacker has a 55.08% chance of winning with the second highest.) In other words, for every 6 armies the defender eliminates, on average the attacker eliminates about 7. When large armies face off, a player will tend to gain a greater advantage over his opponent by attacking rather than defending



see ya out there[list=][/list]

[mrblitz]

Knowing that the attacker has a slight edge in combat; a player should probably attempt to arrange all of their major battles as attacks.

You don't want to be caught defending in a big battle.

Over time, paying attention to this slight attacker's advantage; it could produce more wins for a player.

[darkmatter8888]

you know, i wonder if youve all been taking into account that the game has loaded die. the probabilities are quite hilarioiusly far from random, just so you know.

[Woodruff]

you know, i wonder if youve all been taking into account that the game has loaded die. the probabilities are quite hilarioiusly far from random, just so you know.

Quite the qualitative and fact-based statement loaded chock-full of definitive statements. Based on your massive total of 17 games completed, no doubt?