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The odds in the game risk 
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#1
Jan1613, 03:04 PM

P: 10

In risk, the attacking party rolls 3 die and the top two numbers of the 3 die rolled get put up against 2 die rolled by the defender. If the die are equal then the defender wins. For example if the offender rolls 5 5 2 and the defender rolls 4 3, then 2 defender men die. If the offender rolls 5 5 2 and the defender rolls 5 4 then they trade kills because when die are equal the defender wins.
If offender rolls 5 5 2 and the defender rolls 6 6 the defender wins. Who has the advantage? Attacking or defending? What is the comparative advantage? 


#2
Jan1613, 05:55 PM

P: 229

Start with die in the singular and dice in the plural.



#3
Jan1713, 08:00 PM

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Here's one way to approach it.
Break it into four cases from defender's perspective: ++ win on both + win on high dice, lose on low + etc.  Case ++: For each defender roll, count attacker possibilities: 6+6: 6^{3} 6+5: 5^{3}+3.1.5^{2} (attacker rolls no 6s or one 6) 6+4: 4^{3}+3.2.4^{2} (attacker rolls no 5s nor 6s, or just one such) : 6+1: 1^{3}+3.5.1^{2} (attacker rolls nothing above 1 or just one such) (remember to count all above except 6+6 twice) 5+5: 5^{3} etc. Summing, we get sum for r = 1 to 6 for each of: r^{3}, 2r^{3}(6r), 6r^{2}(6r) = 2r^{4}+7r^{3}+36r^{2} Sum the series to r and plug in r=6. Similarly, for case +: 6+5: 1^{3}+3.1^{2}.5 6+4: 2^{3}+3.2^{2}.4 etc. 


#4
Jan2513, 04:52 PM

P: 10

The odds in the game risk



#5
Jan2613, 09:09 PM

P: 28




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