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.
Andy
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.
Andy