My Risk head hurts. Do my sums for me.
In the board game, Risk, the attacker may attack with up to 3 armies and the defender defend with up to 2. I am looking at the survival numbers of two neighbouring countries with huge armies.
(Rules: the attacker rolls 3 D6 dice and the defender rolls 2 D6 dice. The highest attacking dice is put against the highest defending dice and if the attacker is higher, the defender loses an army otherwise the attacker loses an army. This is repeated for the second highest dice. The third attacker's dice is not used.)
I had a program with a random number generator carrying out attacks millions of times but only concluded that the attacker loses about 85.3% of the troops the defender loses. Although this confirms the odds are in the attacker's favour (which is what I wanted to know), it left other questions.
There are 7,776 permutations of the possible dice rolls (6^5). In playing each of these once, that results in 7,161 dead attacking troops and 8,392 dead defending troops.
Is there a formula for determining these numbers?
7,161 is 3x7x11x13. What a funny number.
8,392 is 2x2x2x1049. Another funny number.
Can anyone enlighten me with relevant formulae?
For example, what if 4 attackers played 3 defenders? Or D10 dice were used? There must be a way of calculating the outcome without resorting to brute force.
(I must have been off sick when we did permutations and combinations, as I'm out of my depth. I certainly don't feel like a Brolly Bonce, that's for sure.)
(Rules: the attacker rolls 3 D6 dice and the defender rolls 2 D6 dice. The highest attacking dice is put against the highest defending dice and if the attacker is higher, the defender loses an army otherwise the attacker loses an army. This is repeated for the second highest dice. The third attacker's dice is not used.)
I had a program with a random number generator carrying out attacks millions of times but only concluded that the attacker loses about 85.3% of the troops the defender loses. Although this confirms the odds are in the attacker's favour (which is what I wanted to know), it left other questions.
There are 7,776 permutations of the possible dice rolls (6^5). In playing each of these once, that results in 7,161 dead attacking troops and 8,392 dead defending troops.
Is there a formula for determining these numbers?
7,161 is 3x7x11x13. What a funny number.
8,392 is 2x2x2x1049. Another funny number.
Can anyone enlighten me with relevant formulae?
For example, what if 4 attackers played 3 defenders? Or D10 dice were used? There must be a way of calculating the outcome without resorting to brute force.
(I must have been off sick when we did permutations and combinations, as I'm out of my depth. I certainly don't feel like a Brolly Bonce, that's for sure.)
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