Probability that $k$th value is largest than the $0$th value

Question

Let $X_0,X_1,...$ are nonnegative i.i.d. with a continuous cdf, and define $N$ to be $k$, where $X_k>X_0$ and $X_i\le X_0$ for $i=1,...,k-1$. What is the p.d.f. of $N$?
I know this can be done by a symmetry argument, by saying that out of $k$ r.v.'s, any one of them is equally likely to be larger than $X_0$ so that $P(N=k)=\frac {1}{k+1}$.
However, I want to know why the following reason does not work:
The probability that $N=k$ given that $X_0=x$ is given by $$P(N=k |X_0=x)=(1-F(x))F(x)^{k-1}$$ Where $P(X_i<x)=F(x)$. It then stands to reason, by the law of total probability that $$P(N=k)=\int_0^{\infty} (1-F(x))F(x)^{k-1}dF(x)=\int_0^1 (1-u)u^{k-1} du =\frac {1}{k(k+1)}$$ Which is off by a factor of $\frac 1k$ and I do not know why. Maybe I erroneously changed the bounds of integration, but the manipulations seem consistent with what is done in a normal calculus class. What am I doing wrong?

Answer 1

The problem is in

"any one of them is equally likely to be larger than $X_0$"

Let us play a game. The referee gets a ball from an urn with 100 balls and he picked number 70. Players get their turn one by one and pick a ball from the urn without looking. The one who picks up the ball with number greater than 70 FIRST have won. You and I are playing. My question is simple: would you like to start the game or not?

Do you know understand why the problem is in the sentence that I cited?