Re: Probability question

On May 9, 5:35 pm, "mbg.nimda(at)gmail.com" <"mbg.nimda(at)gmail.com">
wrote:
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)

This is a version of the so-called geometric distribution.

(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?

Yes. Rewrite it as

p*(1-p) Sum 1 to inf (n*p^(n-1))
(factor out a p, and discard the n=0 term which is zero)
= p*(1-p) Sum 1 to inf d/dp[p^n]
= p*(1-p) d/dp[Sum 1 to inf p^n]
(differentiating under the infinite sum requires justification but
turns out to be allowable in this case)
= p*(1-p) d/dp[1/(1-p)]
= p*(1-p) * (1/(1-p)^2)
= p/(1-p).

- 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.

No, it's not a Poisson distribution. There are some connections
between the geometric and Poisson distributions but I think it's more
likely that your intuition is just off :)

.

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