Probability in number guessing game? An oracle picks a number $W$ randomly on the uniform distribution $U(0,10)$. This is the "winning" index. $N$ other players each pick a single number over the same distribution. 
I pick a specific number $S$ from the same distribution (but I pick some number consciously, mine is not random - thanks to joriki pointing out ambiguity ).
The winner is the player with the least number picked that is $>=W$ (if any).
What is the probability that my specific number $S$ wins?
My first thought was this would just be the probability $a<=S<=b$ where $a,b$ are the first and second order statistic of the distribution, representing the oracle and the least "other player" greater than $W$, but I'm convinced that's completely wrong.
Ideas?
 A: The change in the question has broken the nice symmetry of the question in which $S$ was picked from the same distribution.
Now you win if there is at least one number less than yours and the greatest number less than yours is $W$. The probability that there is at least one number less than yours is $1-(1-S/10)^{N+1}$, and if there is, then the probability that the number right below yours is $W$ is $1/(N+1)$. Thus now your winning probability is
$$
\frac1{N+1}\left(1-\left(1-\frac S{10}\right)^{N+1}\right)\;.
$$
We can check that this integrates to $1/(N+2)$ if we go back to picking $S$ from $U(0,10)$:
$$
\int_0^{10}\frac{\mathrm dS}{10}\frac1{N+1}\left(1-\left(1-\frac S{10}\right)^{N+1}\right)=\frac1{N+1}\left(1-\frac1{N+2}\right)=\frac1{N+2}\;.
$$

This is the answer to the original question, in which $S$ was picked from the same distribution as the other numbers.
The game is symmetric among the players, so your winning probability is $1/(N+1)$ times the probability that someone wins. Someone wins unless $W$ is the greatest of $N+2$ numbers, with probability $1/(N+2)$. Thus your winning probability is
$$
\frac1{N+1}\left(1-\frac1{N+2}\right)=\frac1{N+2}\;.
$$
The same result can be obtained with even less arithmetic by closing the interval cyclically and adding an $(N+3)$-rd number that determines where the circle is cut to form the interval. All rankings of the $N+3$ numbers are equiprobable. Whatever the rank of $W$, there is now exactly one of the remaining $N+2$ ranks such that you win if you get it.
A: Edit This is an answer to the original version of the question.

Let's rephrase the question so that the oracle is considered to be one of the $N+2$ players.  Then the probability you are after is
$$P(\hbox{you win})=P(\hbox{you win}\mid\hbox{oracle doesn't win})\,P(\hbox{oracle doesn't win})$$
which by symmetry is
$$\frac{1}{N+1}\Bigl(1-\frac{1}{N+2}\Bigr)=\frac1{N+2}\ .$$
