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Research Topic: Probability of Risk Attack Dice Rolls
By Ian Mander BSc, 3 November 2003.
Question: What are my chances when making an attack in Risk?
Answer: There are not very many different situations. One, two or three attackers, and one or two defenders.
| Probability of Successful Attack |
Defenders | ||
| 1 | 2 | ||
| Attackers | 1 | 41.7% | 25.5% |
| 2 | 57.9% | ? | |
| 3 | 66.0% | ? | |
Note that after each attack, two armies will be destroyed. Whether or not they are replaced by other armies in the countries involved, if the battle continues it will be one of the six situations above.
One Defender:
Reasonably straight forward for calculating, since the Defender only gets one die to roll. If the roll is equal or greater than the highest roll by the Attacker the attack fails.
One on One: 15/36 = 41.7%
Really easy. Attacker and Defender both have one die, so 6 x 6 = 36 total possibilities.
| Attacker's Roll | Number of results | Formula for deriving # of results | # of ways Defender wins | # of way Attacker wins |
| 1 | 1 | 1 | 6/6 | 0/6 |
| 2 | 1 | 1 | 5/6 | 1/6 |
| 3 |
1 | 1 | 4/6 | 2/6 |
| 4 | 1 | 1 | 3/6 | 3/6 |
| 5 |
1 | 1 | 2/6 | 4/6 |
| 6 |
1 | 1 | 1/6 | 5/6 |
| Totals: | 6 | 1 | 21/36 | 15/36 |
Two on One: 125/216 = 57.9%
Attacker rolls two dice, Defender rolls one, so 6^2 x 6 = 216 total possibilites. Easy to work out - just separate attack rolls into groups according to the highest number rolled.
| Attacker's Roll | Number of results | Formula for deriving # of results | # of ways Defender wins | # of ways Attacker wins |
| 1,1 | 1 | 1 | 6/6 | 0/6 |
| 1,2; 2,1; 2,2 | 3 | 2 x 2 - 1 | 15/18 | 3/18 |
| 1,3; 3,1; |
5 | 2 x 3 - 1 | 20/30 | 10/30 |
| 1,4; 4,1; 2,4; 4,2; 3,4; 4,3; 4,4 |
7 | 2 x 4 - 1 | 21/42 | 21/42 |
| 1,5; 5,1; |
9 | 2 x 5 - 1 | 18/54 | 36/54 |
| 1,6; 6,1; |
11 | 2 x 6 - 1 | 11/66 | 55/66 |
| Totals: | 36 | 2n - 1 | 91/216 | 125/216 |
Three on One: 855/1296 = 66.0%
Attacker rolls three dice, Defender rolls one, so 6^3 x 6 = 1296 total possibilites. Again, reasonably easy to work out - just separate attack rolls into groups according to the highest number rolled.
| Attacker's highest roll - Defender must also roll: n |
Attacker's roll: max value n |
Formula for # of results: | # of results for Attacker | # of results for Defender: | # of ways Defender wins | # of way Attacker wins |
| 1 | 1,1,1 | 1 | 1 | 6 | 6/6 | 0/6 |
| 2 | 1,1,2; 1,2,1; 1,2,2 2,1,1; 2,1,2; 2,2,1; 2,2,2 |
3 x 2 x 2 - 5 | 7 | 5 | 35/42 | 7/42 |
| 3 | 1,1,3; 1,3,1; 3,1,1; |
3 x 3 x 3 - 8 | 19 | 4 | 76/114 | 38/114 |
| 4 | 1,1,4; 1,4,1; 4,1,1; ... 4,4,4 |
3 x 4 x 4 - 11 | 37 | 3 | 111/222 | 111/222 |
| 5 | 1,1,5; 1,5,1; 5,1,1; |
3 x 5 x 5 - 14 | 61 | 2 | 122/366 | 244/366 |
| 6 | 1,1,6; 1,6,1; 6,1,1; |
3 x 6 x 6 - 17 | 91 | 1 | 91/546 | 455/546 |
| n | Totals: | 3n^2 - (3n - 1) | 216 | 7 - n | 441/1296 | 855/1296 |
Two Defenders:
The situation with at least two attacking armies and two defending armies is a tad more complicated than with only one defender - not only because more dice are involved, but because if the Attacker uses two or three dice, the Defender's rolls are matched against the Attacker's - highest rolls with each other, second highest rolls with each other. This often isn't considered when calculating these figures.
One on Two: 55/216 = 25.5%
Attacker rolls one die, Defender rolls two dice, giving 6 x 6^2 = 216 possible results. The Defender loses if they roll any combination less than the Attacker's roll.
| Attacker's Roll | Number of results | # of ways Defender wins |
# of ways Attacker wins | Formula for deriving # of ways |
| 1 | 1 | 36/36 | 0/36 | 0 |
| 2 | 1 | 35/36 | 1/36 | 1 |
| 3 |
1 | 32/36 | 4/36 | 2^2 |
| 4 | 1 | 27/36 | 9/36 | 3^2 |
| 5 |
1 | 20/36 | 16/36 | 4^2 |
| 6 |
1 | 11/36 | 25/36 | 5^2 |
| Totals: | 6 | 161/216 | 55/216 | (n - 1)^2 |
Two on Two:
This really is quite tricky. The dice are no longer independent. They are always in order from highest roll to lowest roll, so there are actually not 6^2 x 6^2 = 1296 possible results, but there are 1296 ways of getting the reduced number of results we can get. (Technically I think it's the difference between combinations and permutations. And I thought I'd never want to know that stuff.) Consider: A roll of 1 and 2 is actual a repeat of a 2 and 1 roll. (The same result but two ways to get it.)
When creating our table we also need to remember it's possible to have a "half successful" attack.
| Attacker's Roll | # of ways of getting results | Number of different results | # of ways Defender wins 2 | # of ways both win 1 | # of ways Attacker wins 2 |
| 1,1 | 1 | 1 | 36/36 | 0/36 | 0/36 |
| 2,1; 2,1 | 2 | 1 | |||
| 2,2 | 1 | 1 | 25/36 | 10/36 | 1/36 |
| 3,1; 3,1 |
2 | 1 | |||
| 3,2; 3,2 |
2 | 1 | |||
| 3,3 |
1 | 1 | 16/36 | 16/36 | 4/36 |
| 4,1; 4,1 | 2 | 1 | |||
| 4,2; 4,2 | 2 | 1 | |||
| 4,3; 4,3 | 2 | 1 | |||
| 4,4 | 2 | 1 | 9/36 | 18/36 | 9/36 |
| 5,1; 5,1 |
2 | 1 | |||
| 5,2; 5,2 |
2 | 1 | |||
| 5,3; 5,3 |
2 | 1 | |||
| 5,4; 5,4 |
2 | 1 | |||
| 5,5 |
1 | 1 | 4/36 | 16/36 | 16/36 |
| 6,1; 6,1 |
2 | 1 | |||
| 6,2; 6,2 |
2 | 1 | |||
| 6,3; 6,3 |
2 | 1 | |||
| 6,4; 6,4 |
2 | 1 | |||
| 6,5; 6,5 |
2 | 1 | |||
| 6,6 |
1 | 1 | 1/36 | 10/36 | 25/36 |
| Totals: | 36 | 21 | ?/1296 | ?/1296 | ?/1296 |
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