In Part 1, we discussed the relative chances for attack and defense in Risk, the game of world conquest. At the end of Part 1, we concluded that the attack has a 47.15% chance of winning the battle for the first soldier and we wondered how the famous conquerors were able to achieve their feats under these conditions. We saved the discussion of the second soldier for Part 2.
To refresh our memories, in Risk, the attack rolls up to 3 dice, while the defense rolls up to 2 dice. The highest rolls of each are compared and the loser loses a soldier, with the defense winning in the case of a tie. Next, the second highest rolls of each are compared, and once again, the loser loses a soldier, with the defense winning in the case of a tie once again.
Well, here we are. Let’s dive into it.
(Here you can find code in which I confirm the below probabilities.)
Of course, regarding the defender’s probabilities, we are merely calculating the lowest roll, since he has only two dice. Therefore, the probabilities are a mirror image of the probabilities we saw regarding the highest roll. This time, there are 11 possibilities yielding a 2nd highest roll of 1, 9 for 2, 7 for 3 etc. The probability can be calculated by dividing by 36, the total number of permutations possible for the two dice of the defense.
Calculating the second highest roll among the three dice of the attacker differs significantly from the calculations of Part 1. I’ll be honest. I struggled with this a little. In the calculations that follow, two things must be borne in mind.
- We must consider both how many outcomes are possible and how many ways in which each outcome can occur. For example, an outcome of (6, 2, 3) is of course a single outcome, but it can occur in 6 ways, corresponding to which die each value occurs on. It can be any of {(2, 3, 6), (2, 6, 3), (3, 2, 6), (3, 6, 2), (6, 2, 3), (6, 3, 2)}. This outcome therefore corresponds to 1*6 = 6 permutations. For another example, an outcome with exactly two ones is actually a collection of 5 outcomes, since the remaining die can take any value between 2 and 6. And it can occur in any of 3 ways, {(1, 1, x), (1, x, 1), (x, 1, 1)}, corresponding to the 3 possible locations for the remaining die, so this outcome actually corresponds to 5*3 = 15 permutations.
- We must be careful with doubles and triples. These must be considered separately since, while there are 6 ways to obtain an outcome of (1, 2, 3), there are only 3 ways to obtain a (1, 2, 2) and only 1 way to obtain a (2, 2, 2).
With the above considerations in mind, we are ready to proceed.
Consider the probability of getting a 2nd highest roll of 1. This is relatively straight-forward. Clearly the lowest roll is a 1 as well. For now, we will disregard the case where all 3 dice are 1. The highest die can then take any value between 2and 6, and it can appear upon any of the 3 dice, since we have not specified which of the 3 dice contains the highest roll. This yields a total of 3*5=15 permutations. Adding the case of a triple 1 yields a total of 16 permutations. By a symmetrical argument, we can calculate that the same number of permutations yield a 2nd highest roll of 6.
Next, what about getting a 2 as the second highest roll? For now, we will disregard the possibility of multiple twos and assume that the highest roll was higher than 2 and that the lowest roll was lower than 2. The highest roll can take 4 values (3–6) and the lowest toll must be 1, for a total of 4 outcomes, and these can occur at any of 6 permutations of dice locations(3 possibilities for the location of the highest roll (die 1, die 2 or die 3) and the two remaining possibilities for the location of the lowest roll), for a total of 4*6=24 permutations. We will now consider double twos, but not triple twos. If there are exactly 2 twos, then the remaining die can take any of 5 values (excluding 2), and this remaining die could be any of the 3 dice, for an additional 5*3=15 permutations. Adding the final case of triple 2’s, we obtain a total of 24+15+1 = 40 permutations. A parallel argument yields the same result for a second highest roll of 5.
Finally, what about getting a 3 or a 4? Let’s start with 3. Once again disregarding the possibility of multiple threes, the highest roll can take any of 3 values (4, 5 or 6) and the lower roll can take any of 2 values (1 or 2), for a total of 6 outcomes. This can once again occur at any of 6 permutations of two dice, for a total of 6*6 = 36 permutations. In the case of exactly 2 threes, the other die could take any of 5 values (any besides 3) and could occur at any of the three dice, for an additional 5*3 = 15 permutations. Adding the last possibility of 3 threes yields a total of 36+15+1=52 permutations. A parallel calculation yields 52 permutations for a second highest roll of 4 as well. These results are summarized in the below visuals.
Note that the probabilities of attack results are exactly symmetrical. To be mathematically precise, P(x) = P(6-x). We will come back to this point.
We next compare the probabilities of attack and defense directly.
We can see that the attack has a significant advantage here. It is much more likely to obtain values of 4, 5 or 6, than defense is.