mustangsally
mustangsally
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June 21st, 2011 at 8:54:19 PM permalink
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silly
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guido111
guido111
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June 21st, 2011 at 9:26:52 PM permalink
This was addressed in another thread.

DorothyGale started that thread this way:
"While I was milking the pigs today, I thought of a little problem..." LOL

The thread you are looking for is
HERE
mustangsally
mustangsally
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June 21st, 2011 at 9:56:19 PM permalink
Thanks.
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silly
I Heart Vi Hart
pacomartin
pacomartin
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June 22nd, 2011 at 1:06:05 AM permalink
Quote: DorothyGale

For example, with N = 2, the answer is 2*(1/2) + 3*(1/4) + 4*(1/8) + 5*(1/16) + ...
--Ms. D.



MustangSally

To be really honest you haven't derived the formula, you just plugged in.
You should express the answer as an infinite sum first, since that is the natural outcome. Then you should derive the answer as a closed form expression.

For example with a coin the solution is:
2*( Probability you get both head and tales in 2 ) +
3*( Probability you get both head and tales in 3 ) +
4*( Probability you get both head and tales in 4 ) +
...
i*( Probability you get both head and tales in i ) +
...
infinity

But
i*( Probability you get both head and tales in i ) = i*p^(i-1) where p = 1/2

S== infinite sum from i=2 to infinity of [ i*p^(i-1) ]

S = 2*p + 3* p^2 + 4* p^3 + ... [equation 1]
S/p = 2 +(3* p^1 + 4* p^2 + ...) [equation 2]

S/p -S = 2 +p + p^2 + p^3+ ... [equation 3]=[equation 2]- [equation 1]

Define GS as a standard Geometric Sum
GS = p + p^2 + p^3+...
GS/p -1= p +p^2+...=GS
GS - p = p*GS
GS= p /(1-p) [equation 4]

S/p -S = 2 +p/(1-p) [equation 4] into [equation 3]
S(1/p-1) = (2-p)/ (1-p)
S(1-p)/p = (2-p)/ (1-p)
S = p*(2-p)/(1-p)^2 [algebraic simplification]
when p= 1/2 then S=3
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