Re: Probability question


"mbg.nimda(at)gmail.com" <"mbg.nimda(at)gmail.com"> wrote in message
news:464268a9$0$13867$ba620e4c@xxxxxxxxxxxxxxxxx
Suppose I have a probability of p of winning a certain game (I have
FreeCell in mind) and I want to calculate the probability that I win this
game n times successively (until I lose one for the first time).

Logic tells me that this probability is P(n) = p^n*(1-p)

(note that Sum 0 to inf ( P(n) ) =1 )

Suppose I want to calculate the expectation value for n:
This leads me to
Sum 0 to inf (n*P(n))
or
Sum 0 to inf ( n*p^n*(1-p) )

Now I have two questions:
- is there a formula for this sum?
- I intuitively suppose there is a relationship with a Poisson
distribution but I don't see where p and n fit in, moreover the Poisson
distribution contains a factorial n! in the numerator that I can't find
back, and this is disturbing to me.

It is the geometric distribution.
http://en.wikipedia.org/wiki/Geometric_distribution

--
rb


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