View Full Version : Non-Poker Probability Help

The Dude
05-30-2003, 03:12 AM
To all you math geniuses, I need a little help with a non-poker question of probability. It has to do with one's odds of winning in different situations in the board game Risk.

I am curious how to figure out the exact odds of winning an attack 3on2, 3on1, 2on2, 2on1, etc. I can figure out the simple ones of 1on2 and 1on1, but need a little help for the larger ones.

For those who aren't familiar with the game Risk, I will describe the rules below to figure the odds. If you already know the game, feel free to skip the rest of the post.

Risk is a strategy game, in which an attacker and a defender each roll dice to determine the winner. The attacker may roll from 1-3 dice, and the defender may roll 1 or 2 dice, depending on thier respective number of armies.

The way it works is, the highest die cast by the attacker vs. the highest die cast by the defender (and, if both are rolling two or more, the second highest vs. the second highest). All ties go to the defender.

For example, if the attack is 3on1, and the roll is 5-2-1 vs a 4, the defender loses since the attacker's highest (the 5) beats the defender's highest (the 4). If the roll is 4-4-3 vs. 4, however, the defender wins since the attacker's highest (the 4) loses to the defender's highest.

Another example: The attack is 3on2 and the roll is 6-4-1 vs. 5-3. The result is the attacker wins twice, since his highest (6) beats the defender's highest (5) and his second highest (4) beats the defender's highest (3). However, if the roll is 5-3-2 vs. 4-3, the result is a split (each lose one army), since the attacker's highest (5) beats the defender's highest (4), but the attacker's second highest (3) loses to defender's second highest (3).

A third example: If the attack is 1-on-2 and the roll comes 5 vs. 4-2, the attacker wins since his highest beats the defender's highest.

It may help to think of it this way. If either the attacker OR the defender is rolling only one die, then there are only two possible results. Either the attacker will win or the defender will win. If they are both rolling two or more, then there are three possible results. The attacker could win twice, it could be a split, or the defender could win twice. I am curious to know how to calculate the odds in every possible scenario.

I hope I explained this clearly, but if not ask questions - I'll be happy to elaborate. Thanks.


The Dude
05-30-2003, 03:41 AM
It might help clarify the rules if I show my calculations for the odds of a 1-on-2 attack and a 1-on-1 attack.

First, 1-on-1.
There are 6 possible rolls for the attacker, and for each 6 possible rolls for the defender.
If the attacker rolls a 6, he will win 5 times and lose 1 time. If he rolls a 5, he will win 4 times. If a 4, he will win 3 times. If a 3, he will win twice. If a 2, once, and if a 1, he will win zero times.
Overall, he has 15 ways to win out of 36 possibilities. The attacker will win 41.67% of the time.

Now for 1-on-2.
For each of 6 possible rolls for the attacker, there are 36 possible rolls for the defender (a total of 216 possibilities). If the attacker rolls a 1, there are zero possibilities to win. If the attacker rolls a 2, there is only 1 possibility to win. If he rolls a 3, there are only 4 possibilities to win. If a 4, there are 9 chances to win. If a 5, there are 16. And if a 6, there are 25. In total, there are 55 chances to win out of 216 possibilites - a 25.46% chance of the attacker winning.

If the attack is 2-on-1:
there are 6 possible rolls for the defender, and 36 possible rolls for the attacker. If the defender rolls a 6, there are zero chances for the attacker to win. If a 5, there are 11 chances. If a 4, there are 20 chances. If a 3, there are 27 chances. If a 2, there are 32 chances. If a 1, there are 35 chances. In total, there are 125 chances to win out of 216 possibilities - the attacker will win 57.87% of the time.

For the more complex calculations of 3-on-2 and 2-on-2, I need a better formula to plug in. Calculating it individually would be way too tedious.

I am curious if anyone can help.

05-30-2003, 11:43 AM
It shouldn't be so hard to find the probability of finding the highest number number for 2 or 3 dice. And then it should be easy...

Viz for 2 Dice P(1 high)= 1/36

P( exactly 2 high)=3/36

P(exactly 3 high)= 5/36

P(exactly 4 high)= 7/36

P( exactly 5 high) = 9/36

P(exactly 6 high) =11/36.

Now it should be easy to compute the 1-1 1-2 and 2-2 match ups.

05-30-2003, 03:46 PM
It's simple:
Never attack with 2:2 or 1:2 unless you REALLY REALLY need a card /forums/images/icons/wink.gif

05-30-2003, 10:12 PM
Risk Board game probabilities are more complicated than you think: