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The odds in the game risk

  1. Jan 16, 2013 #1
    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. jcsd
  3. Jan 16, 2013 #2
    Start with die in the singular and dice in the plural.
     
  4. Jan 17, 2013 #3

    haruspex

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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: 63
    6+5: 53+3.1.52 (attacker rolls no 6s or one 6)
    6+4: 43+3.2.42 (attacker rolls no 5s nor 6s, or just one such)
    :
    6+1: 13+3.5.12 (attacker rolls nothing above 1 or just one such)
    (remember to count all above except 6+6 twice)
    5+5: 53
    etc.
    Summing, we get sum for r = 1 to 6 for each of:
    r3, 2r3(6-r), 6r2(6-r) = -2r4+7r3+36r2
    Sum the series to r and plug in r=6.

    Similarly, for case +-:
    6+5: 13+3.12.5
    6+4: 23+3.22.4
    etc.
     
  5. Jan 25, 2013 #4
    Thanks man. I'm a little confused by what for example 3.1.5^2 means but thanks for the help
     
  6. Jan 26, 2013 #5
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