# Risk dice roll probabilities [closed]

Hi,

I'm wondering what probablity can tell me about dice rolling strategies in the game Risk. When a player attacks another player, they can roll up to 3 dice. The defending player can choose to roll either 1 or 2 dice in defence. They must choose how many dice they are going to roll before the attacking player rolls.

The highest valued dice by each player is compared, and if the defending player's dice has a higher or equal value then the attacking player loses a soldier, otherwise the defender does. If the defender has chosen to roll a second dice, then the second highest roll by each player is compared with the same comparison.

My question is, over time will the defending player be better off rolling 1 or 2 dice in defence?

Thanks, Matt

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It looks like you already got some answers, but MathOverflow is not normally for elementary probability questions such as this one. –  S. Carnahan May 6 '10 at 2:14
I agree with Scott. Have a look at the first two sections of the FAQ: mathoverflow.net/faq#whatquestions. In particular, in the second section ("what questions should I not ask here") there's a list of other math Q&A sites which are more hospitable to elementary questions. –  Anton Geraschenko May 8 '10 at 15:33

## closed as off topic by S. Carnahan♦, Anton GeraschenkoMay 8 '10 at 15:33

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You should use as many dice as possible while defending (so always 2, unless you have one army).

This problem has no doubt been tackled by hundreds of probability students over the years, and it is not hard to Google and find a complete calculation. A likely best source for your buck for this, and related questions, will be Ivars Peterson's MathTrek article here:

http://www.maa.org/mathland/mathtrek_7_14_03.html

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Let $D$ and $A$ be the number of defenders and attackers respectively killed during a roll. One quantity of interest is $\mathbb{E}(D-A)$.
According to the wikipedia page, if the attacker rolls 3 dice and the defender rolls 1 dice we have $\mathbb{E}(D)=0.66$ and $\mathbb{E}(A)=0.34$. By linearity of expectation, $\mathbb{E}(D-A)=0.32.$
On the other hand, if the attacker rolls 3 dice and the defender rolls 2 dice we have $\mathbb{E}(D)=1.08$ and $\mathbb{E}(A)=0.92$. It is reasonable to normalize so that the expected number of death is one (to match the first case). In this case, $\mathbb{E}((D-A)/2)=0.08.$