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5's are wild.

Posted By: Bryce
Date: Monday, 25 June 2007, at 12:13 a.m.

In Response To: 5's are wild. (Chuck Bower)

Yay math! The sum of r^i from i=1 is r/(1-r) = 1/(1-r) - 1. Taking a derivative gets you 1/(1-r)^2 = sum of i*r^(i-1); taking two derivatives gives you 2/(1-r)^3 = sum of i(i-1)r^(i-2). So your sum should be (1/18) 1/(11/36)^2 + (99/1296)*2/(11/36)^3 = 5.9504, assuming I put this into the computer's awful calculator correctly. (This is no guarantee.)

Messages In This Thread

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    neilkaz -- Saturday, 23 June 2007, at 1:33 p.m.
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    Daniel Murphy -- Saturday, 23 June 2007, at 3:40 p.m.
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      Chuck Bower -- Saturday, 23 June 2007, at 10:49 p.m.
      • 5's are wild.
        Bryce -- Monday, 25 June 2007, at 12:13 a.m.
    • Jumping a 5-prime
      Ian Dunstan -- Sunday, 24 June 2007, at 11:42 a.m.
      • Jumping a 5-prime
        Daniel Murphy -- Sunday, 24 June 2007, at 4:22 p.m.
        • Jumping a 5-prime
          Chuck Bower -- Sunday, 24 June 2007, at 11:04 p.m.
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    Stick -- Saturday, 23 June 2007, at 5:11 p.m.
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    Ian Dunstan -- Sunday, 24 June 2007, at 11:13 a.m.