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<blockquote data-quote="Flatus Maximus" data-source="post: 4324922" data-attributes="member: 55114">As malraux indicated above, sigma(i=0 to infinity){1/n^i}=1/(1-1/n) is the special case of the geometric series where a=1 and r=1/n.&nbsp; If you want to start the series at i=1, just subtract off the i=0 term's value from both sides of the given equation.As above, you have a geometric series (starting at i=1), this time with a=1 and r=m/n, so you get (m/n)/(1-m/n), which simplifies to m/(n-m).&nbsp; It should be noted that we must have |r|&lt;1, otherwise the formula gives nonsense; assuming m and n are positive, you'd need m&lt;n.&nbsp; Finally, if m=9 and n=20, then you'd get 9/11=.818181....Regarding the average number of 'ticks': A random variable that gives the number of failed &quot;Bernoulli trials&quot; before the first success has a geometric distribution.&nbsp; Here, the Bernoulli trials are &quot;roll a d6,&quot; success (which stops the process) is not rolling a 6, and failure (which keeps the process going) is rolling a 6.&nbsp; If r is the probability of success (here 5/6), then the average number of rolls is 1/r=6/5.Hope that helps.</blockquote>

[QUOTE="Flatus Maximus, post: 4324922, member: 55114"] As malraux indicated above, sigma(i=0 to infinity){1/n^i}=1/(1-1/n) is the special case of the geometric series where a=1 and r=1/n. If you want to start the series at i=1, just subtract off the i=0 term's value from both sides of the given equation. As above, you have a geometric series (starting at i=1), this time with a=1 and r=m/n, so you get (m/n)/(1-m/n), which simplifies to m/(n-m). It should be noted that we must have |r|<1, otherwise the formula gives nonsense; assuming m and n are positive, you'd need m<n. Finally, if m=9 and n=20, then you'd get 9/11=.818181.... Regarding the average number of 'ticks': A random variable that gives the number of failed "Bernoulli trials" before the first success has a geometric distribution. Here, the Bernoulli trials are "roll a d6," success (which stops the process) is [i]not[/i] rolling a 6, and failure (which keeps the process going) is rolling a 6. If r is the probability of success (here 5/6), then the average number of rolls is 1/r=6/5. Hope that helps. [/QUOTE]

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