Probabilities in the board-game 'Risk'

merlin2007

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Dec 25, 2006
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This isn't really a problem from school or a math contest I know of, but I've been trying to figure it out for a while.

Consider a particular encounter in Risk. The attacker rolls three dice, and the defender rolls two. The highest die of the attacker is matched with the highest die of the defender. The next-highest die of the attacker is matched with the only other die of the defender. The defender wins if his dice are greater than or equal to the attacker's dice.

What is the probability that the attacker wins both?
What is the probability that the defender wins both?

Every time I attempt this, the calculation gets ridiculously complicated. If anyone could think of an intricate solution to this, it's be much appreciated.

Thanks
 
Ran (program) the 6^5 = 7776 possibilities:

ties: 2611 (2611/7776 = .33577...)
defender: 2275 (2275/7776 = .29256...)
attacker: 2890 (2890/7776 = .37165...)

If you get those, you're ok :shock:

For fun, ran a simulation; results from 3 runs of a million games each:
(ties - defender - attacker)
335734 293228 371038
335391 292538 372071
334938 292617 372445
 
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