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Old 08-03-2011, 03:15 AM   #15
Third Eye Third Eye is offline
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Join Date: Apr 2007
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My best guess, given the wording, is that we are working with a geometric distribution with p=.25 and trying to determine the minimum number of trials it will take for a first success that occurs more than 2.57 standard deviation away from the mean. The mean for a geometric distribution is 1/p, so 1/.25=4. The variance is given by (1-p)/p^2 or .75/.0625=12. Thus the sd is 12^.5 or approximately 3.464. 2.57 sds is approximately equal to 8.9. So, to be at least 2.57 sds away from the mean you need at least 4 + 9 rolls, or 13.
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