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A game display shows the name of the top player in record based on the amount of time the player spent before losing to the machine's opponent. Let's say $X_i$ be the time spent by the $i$th person to play the game, and that $X_1,X_2,\ldots$ are independent and identically distributed continuous variables with PDF $f_X$.

1) What would be the probability that the $k$th person get the top player in record?

2) Compute your result from part 1) by computing the probability: \begin{align} P(X_n\ge X_i \text{ for all }i=1,\dots,n-1) &= \\ \!\!\!\!\!\!&\int_0^\infty P(x\ge X_i \text{ for all }i=1,\dots,n-1)f_{X}(x)dx. \end{align} 3) What is the expected number of records after n games? Does this number reach a limit as n to infinity?

4) What is the probability that both the nth and n+1th person set a record? and are these independent events and why?

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What is urgent? –  Jonas Meyer Oct 23 '12 at 7:51
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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also the use of urgent is unnecessary. –  Julian Kuelshammer Oct 23 '12 at 9:11
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1 Answer

If there are n players the probability would be $1/n$. let $P(X_i)$ be the probability of player i being at the top, if the score for each player is independent and identically distributed then $P(X_1)=P(X_2)=...=P(X_n)$ and someone and only one is at the top so $P(X_1)+P(X_2)+...+P(X_n)=1$ then $P(X_1)=P(X_2)=...=P(X_n)=1/n$ for any PDF.

edit (as a response to posters edit):

1) as above.

2) i don't see a reason to go about it the hard way, the answer would be the same. $P(X_n≥X_i | i=1,...,n-1)=P(X_n-1≥X_i | i=1,...,n-2,n)= ...$ so $P(X_n≥X_i | i=1,...,n-1)=1/n$ for any n

3) records after n games $\sum_{k=1}^{n} 1/k$, this does not have a finite limit as n goes to infinity.

4) it's $\frac{1}{n*(n+1)}$ and there independent events as the probability for n+1 being a record does not change if the previous record was reached by n-1 or any other previous player.

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