Take the 2-minute tour ×
Board & Card Games Stack Exchange is a question and answer site for people who like playing board games, designing board games or modifying the rules of existing board games. It's 100% free, no registration required.

I'm looking for an way to estimate the percent chance of winning a specific Risk battle. Assume the maximum number of dice will be used.

It doesn't need to be a perfect calculation, just an estimate to give a general idea of whether to attack or not. It needs to be easy to memorize.

I'm sure there is a trade off between accuracy and easy of use, so I'm hoping there is more than one method proposed. I would be using this to estimate after each die roll so I can stop attacking when my winning percentage drops below a certain threshold.

Bonus free wild card if you can also describe a method of estimating armies left.

share|improve this question
add comment

4 Answers 4

All of the detailed probability calculations and Markov analysis posted by Eric P. and ire_and_curses can be distilled into a simple set of Risk attack heuristics:

  • Large battles favor the attacker but only very slightly.
  • For small battles, attack if you have more armies, stop if you don't.

The rationale for these guidelines is outlined below.

A large battle is 3 or more attackers and 2 or more defenders. For large battles the army count has nothing to do with the outcome of any single battle, which means you only need to remember a single number: the net attrition rate, defined as the expected value of the difference between defender and attacker losses. This rate can be calculated from data in Table 2 of the Jason Osborne paper in Eric P.'s answer:

  • The expected value of the defenders lost is 1.08. The number of defenders lost by events pi_32x weighted by their probability.
  • The expected value of the attackers lost is 0.922.
  • This means the net attrition rate is 0.158 (i.e. over 10 rounds of combat, the defender will lose 1.58 more armies--on average). Over 20 rounds 3.16.

Leading us to our first heuristic:

  • Large battles favor the attacker but only very slightly. As the attacker, it takes on average 20 rounds to make up a three army deficit. Whether you should attack in any given situation is a strategic not a tactical decision, but tactically the attacking advantage is on average very slight. Remember as well that past events have nothing to do with future rolls.

For small battles, the full probability matrix is supplied above, but our second rule distills this knowledge:

  • For small battles, attack if you have more armies, stop if you don't. When outnumbered your probability of winning is no more than 0.417 and probably less. Otherwise your probability of winning is at least 0.656 but as high as 0.916.
share|improve this answer
1  
I upvoted several answers, but I think this one actually does the best at giving a usable, memorable heuristic. –  JSBձոգչ May 25 '11 at 12:53
    
NO, I this heuristic neither brief, nor memorable, nor correct. In 1v1 games (at least) the concept is to always attack if you have 3 attackers (4 or more troops on the attacking territory). There is no intrinsic difference between one large and many small battles. –  TheChymera yesterday
add comment

A good paper by Jason Osborne can be found here. (It's a correction to an earlier paper by Tan.) He uses Markov chain calculations to get the exact probabilities. You'll especially want to look at Table 3 on page 6, which has these probabilities rounded to three decimals for up to 10 armies per side. I've reproduced it below:

enter image description here

As to ease of use: just print it out and look up your odds after every roll!

share|improve this answer
    
I've often found players don't like to use such aids during play (thinking it cheating), this is why I prefer mental calculations. –  Neal Tibrewala May 26 '11 at 3:26
    
Looks like you get a reasonably good approximation by taking 0.42 + 0.11 * A - 0.09 * D. So, start with 0.42, add 0.11 for every attacker, subtract 0.09 for every defender. (I tried to get a good fit for the results between 0.1 and 0.9; if you get outside of that range it's less important anyway.) –  Erik P. May 26 '11 at 14:25
    
Your statement "look up your odds after every roll" is actually very deep. The chart assumes you'll fight until you've won or until you're down to one troop. If you stop once the odds are against you, the chart isn't accurate. If you make a chart to compensate, you can then stop when that NEW chart tells you the odds are against you, and so on, ad infinitum. I've done some inconclusive work along these lines: github.com/barrycarter/bcapps/blob/master/bc-calc-risk-odds.m –  barrycarter May 27 '11 at 4:14
    
@barrycarter - interesting work. Essentially there is a third parameter at play here: the minimal probability of winning at which you're willing to continue. You set that at 0.5, I think (from a quick glance over your code). I think in different situations, you'd want to use different probabilities here: sometimes it's worth to risk a few armies for a long shot that would help you a lot. So we would need different tables for different values of this parameters... interesting, interesting. –  Erik P. May 27 '11 at 18:33
add comment

Although the full calculation to discover whether you will win a sequence of battles is difficult to make, it is easy to calculate the chances of winning any particular combination of attacker and defender dice. I reproduce here the table of expected losses described in this paper.

                       Defender Dice
                       1           2
Attacker Dice
1                  0.58/0.42   0.75/0.25
2                  0.42/0.58   1.22/0.78
3                  0.34/0.66   0.92/1.08

The first number in each cell is attacker losses, the second defender losses. So, if you attack with 1 die against a 2 dice defence, you have a 3/4 chance of losing the battle, which in this case will cost you 1 army. Remember also that if each of you have 2 or more dice, then exactly 2 armies will be lost in the battle. This is why in 2 dice vs 2, the attacker will lose on average more than one army.

So, you could easily memorise this table, and use it to decide at each step of a battle whether to proceed or not. Or you could just remember the broad rule of thumb described in the paper:

  1. When both attacker and defender have a large number of armies, the attacker will, on average, lose armies at a 15% slower rate than the defender.

  2. Towards the end, when either the attacker or the defender must shake fewer dice, the advantage swings more strongly toward the player with the most armies.

The details of the full calculation are explained in the paper. The author also provides an online Javascript implementation of the calculations that you might find interesting to play around with.

share|improve this answer
add comment

As stated in the Risk FAQ the expected losses per attack for standard Risk rules is about 6 to 7. This means the attacker is expected to lose 6 armies for every 7 defender armies destroyed. Since we're talking about expected values, this represents the mean (average), which is most akin to a 50th percentile or 50% chance that that is what will happen. (this is not techically correct in terms of statistics, but I'm trying to explain it in more layman's terms).

As for 'chances to win a battle' this is very difficult to produce a rule of thumb for, since the numbers vary wildly depending on # of armies in play. For example, an 'even match' of 100 to 100 is won with over 85% chance, but 10 to 10 is only about 50%-50%.

It would be easier if you specified a particular odds at which you want to attack or not, then a simple matrix of attacker/defender army counts is possible to create (perhaps with an easy formula), but when you start with N attackers and M defenders the best 'answer', statistically, is a probablilty distribution function which isn't easy to calculate.

Assuming you want to attack at, at least 50% expected value (as above), use the 6 to 7 rule. This will also tell you how many you are expected to have left, so, using this rule, if you have 20 armies (to attack with, so 21 in the country), and he has 21, the 6 to 7 rule would say that you're expected to lose 18, killing his 21 (6*3, 7*3), thus leaving 2 left over.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.