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Méré's Paradox

  1. Sep 18, 2004 #1
    TASK: Show that throwing at least one "6" within 4 throws of a die is more likely than throwing a "double 6" within 24 throws of two dice.

    SOLUTION: I can do the first result two ways. Firstly, I can use the complement of "four dice all showing something other than six" which would be 1 - ((5/6)^4) = 51,7%. Or I can use the long-winded version of

    Binomial[4, 1]*((5/6)^3)*((1/6))^1) +
    Binomial[4, 2]*((5/6)^2)*((1/6))^2) +
    Binomial[4, 3]*((5/6)^1)*((1/6))^3) +
    Binomial[4, 4]*((5/6)^0)*((1/6))^4)

    = 51,7%

    which translated into words is roughly: "the chances of selecting 1 from 4 multiplied by the chances of three dice showing not six multiplied by the chances of one die showing six PLUS the chances of selecting two from 4 etc etc"

    I can show the second part by using the complement of "no six or exactly one six" which is 1 - ((5/6)^12) - (Binomial[12,1])*(5/6)^11)*(1/6)

    MY QUESTION: if I were insane enough (which I am, believe me :eek: ) to want to solve the second part using the "long-winded" binomial version, how would I do it?

    I'm gratefu for any help
  2. jcsd
  3. Sep 18, 2004 #2
    P(getting a double 6) = 1/36
    P(in 24 throws , no double throws) = (35/36)^24
    P(getting atleast 1 double throw in 24 throws)
    = 1-(35/36)^24
    approx .491 or 49.1%

    Hence the result

    -- AI
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