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I need some help please with this question:

A player decides whether to stop playing or not just after the first loss -or- after 10 games.

the probability for winning a single game is p.

all the games are independent.

Let X be the amount of total games which the player Participated.

I'm trying to look for the distribution of X [P(X=k)].

I let k be the number of games and if k is not ten the distribution is (p^k-1)*(1-p)

k-1 victories and one loss, I just can't see how to Involve the 10 detail into this Equation.

Thank you.

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For $k \lt 10$ it should be $p^{k-1}(1-p)$. Please use parentheses (or, even better, $\LaTeX$) as exponentiation binds more tightly than subtraction –  Ross Millikan Apr 3 '11 at 18:15
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1 Answer

up vote 1 down vote accepted

You already know the formula for $\mathbb{P}(X=k)$ for $1\leq k\leq 9$.

The player plays a total of ten games if and only if the first nine games are all victories. So $\mathbb{P}(X=10)=p^9$.

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@Bryon: yeah but shouldn't I have this two things in one general equation? –  user6163 Apr 3 '11 at 17:48
    
@byron2: Do you think that he must win in those ten games? I think it may be interapted like he can play 10 games with whatever result, Don't you think? –  user6163 Apr 3 '11 at 18:00
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@Nir the player's gonna stop playing after 10 games no matter whether he wins or looses the 10th game. If he wins then he stops because he said he would after 10 games & if he looses then the the k-case worked with k=10. So the 10th game's result doesn't really affect the probability distribution of the interest. –  Amit L Apr 3 '11 at 18:10
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