Summation Equation, Trying to solve this recurrence forumla.

by bobbybob
Tags: equation, forumla, recurrence, solve, summation
bobbybob is offline
Mar5-06, 06:12 PM
P: 2
Hello. I've searched around a bit for a math forum where I could get help with this and this seems like the one I found where I could get some help with this. I was posed the following problem. Now I must admit it is over my head (as is most of the math on this forum) I was hoping that someone here could help give me an answer for this person. Or if not an answer, a reason why his probem makes no sense.

I want to predict the number of cups to be hit in beer pong each round, based upon shot percentage.

It's not just number of shots * percentage, because if you make 2 shots in a row, you get an extra turn. Even your extra turns can get extra turns..

Therefore, Number of cups per round is the recursive formula

C(t) = C(a^2 * t) + 2at where 0 < a < 1 for accuracy

So if you want to find the number of cups in 1 round, calculate c(1)
This can actually be reduced to the summation equation

for i = 0 to infinity -> 2a^(2i+1)

How do I solve this summation and get a formula?
I'd really appreciate it,

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AKG is offline
Mar5-06, 07:01 PM
Sci Advisor
HW Helper
P: 2,589
You either miss your first shot, OR make your first and miss the next, OR make your first two and miss the next OR ... OR make the first n and miss the next OR ...

The probability that you make the first n and miss the next is an(1-a).

So what you want is the sum of (getting n cups) x (probability of getting n cups) for all n, i.e.:

[tex]\sum _{n=0}^{\infty}na^n(1-a) = (1-a)\sum _{n=0}^{\infty}na^n = (1-a)\sum _{n=1}^{\infty}na^n = a(1-a)\sum_{n=1}^{\infty}na^{n-1} = a(1-a)f'(a)[/tex]


[tex]f(x) = \sum_{n=1}^{\infty}x^n = \sum_{n=0}^{\infty}x^n - 1 = \frac{1}{1-x} - 1 = \frac{x}{1-x}[/tex]


[tex]f'(x) = \frac{(1-x) - x(-1)}{(1-x)^2} = \frac{1}{(1-x)^2}[/tex]


[tex]f'(a) = \frac{1}{(1-a)^2}[/tex]

Finally, the desired number is:

[tex]a(1-a)\frac{1}{(1-a)^2} = \frac{a}{1-a}[/tex]
bobbybob is offline
Mar5-06, 07:18 PM
P: 2
thank you very much

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