You're clev'rest arou-ound, 65-ish percent of the time!

You are very clever when it comes to solving Riddler Express puzzles. You are so clever, in fact, that you are in the top 10 percent of solvers in Riddler Nation (which, as you know, has a very large population). You don’t know where in the top $10$ percent you are — in fact, you realize that you are equally likely to be anywhere in the topmost decile. Also, no two people in Riddler Nation are equally clever.

One Friday morning, you walk into a room with nine members randomly selected from Riddler Nation. What is the probability that you are the cleverest solver in the room?

Assume that the cleverest scale is continuous and runs from $0$, utter dunce, to $1,$ utter genius. Each member in Riddler nation can be assumed to have a cleverness level that is uniformly distributed on the unit interval from $[0,1].$ So let's say that your score is $C$ and your nine fellow Riddlerites in the room each have cleverness scores $C_i, i = 1, \dots, 9,$ with $C$ and $C_i$ i.i.d. as $U(0,1).$ The probability that you are the cleverest in the room, conditional on the the fact that $C \in [0.9,1]$ is given by \begin{align*}\mathbb{P} \{ C \geq C_i, i = 1, \dots, 9 \mid C \geq 0.9 \} &= \frac{1}{\mathbb{P} \{ C \geq 0.9 \}} \int_{0.9}^1 \left(\prod_{i=1}^9 \int_0^c \, dc_i \right) \, dc\\ &= 10 \int_{0.9}^1 c^9 \, dc = \left[c^{10} \right]_{c=0.9}^{c=1}\\ &= 1 - (0.9)^{10} = 0.6513...\end{align*}