# Urgent- about probability - order statistics

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

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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