Probability that one soccer player will have his favourite t-shirt in the dressing room. 
In a soccer team there are $11$ players and $11$ t-shirts numbered
  from $1$ to $11$.The players go in the dressing room one at a
  time, casually. 
  Each one, once he arrives, takes a random t-shirt, except
  Danilo who prefers the numbered $8$ t-shirt and, if this last one is
  available, he will choose it. ' What's the probability that Danilo
  will be able to select his favourite t-shirt?

(Correct answer is $\cfrac{6}{11}$)
My effort
I've approached this problem considering the probability that Danilo will be the $n^{th}$ player going in the dressing room and for any $n$ this happens with probability $\cfrac{1}{11}$.
Then for each case, I've considered the probability that the players before Danilo take the numbered $8$ t-shirt but at this point is where I had some conceptual difficulty and looked the proposed solution of the problem.
For example(following the logic of the solution),if we take the case where Danilo is the $3^{rd}$ player going in the dressing room ,we have then that the probability that the two players before  him take his favourite t-shirt is $\cfrac{2}{11}$, so the probability that Danilo takes his t-shirt is $\cfrac{9}{11}.
So the overall probability would be $P=\cfrac{1}{11}\cdot \cfrac{2}{11}$ so the probability that Danilo takes his t-shirt is $\cfrac{1}{11}\cdot \cfrac{9}{11}=\cfrac{9}{121}$.
What doesn't click to my mind is why the probability of the previous two players is $\cfrac{2}{11}$.
This doesn't seem to me to be correct as they can't have the same probability since one of these two, which entered the dressing room first, will have already chosen his t-shirt so the probability that the second player chooses  Danilo's t-shirt must take care of the scenario where one  t-shirt has already been taken.
It seems to me that $\cfrac{2}{11}$ is only plausible if the players are choosing their t-shirt at the same time (though I've also my doubts on this).

Question
Can you guys help me make this clear?

 A: By the law of total probability,
\begin{equation}
\mathbb{P}(\text{Danilo gets $8$}) = \sum_{n = 1}^{11} \mathbb{P}(\text{Danilo gets $8$} \mid \text{Danilo enters $n$-th}) \mathbb{P}(\text{Danilo enters $n$-th}).
\end{equation}
Now, $\mathbb{P}(\text{Danilo arrives $n$-th}) = 1/11$ as you said. What remains is to determine the conditional probabilities.
Let me give a hint by considering a specific case. Assume that Danilo is the third person to enter the dressing room. For Danilo to get his shirt, we require that the first player and the second player to enter the dressing room, both do not pick the shirt with number $8$. So,
\begin{align}
&\mathbb{P}(\text{Danilo gets $8$} \mid \text{Danilo enters third}) \\
&= \mathbb{P}(\text{First does not take $8$}) \mathbb{P}(\text{Second does not take $8$} \mid \text{First did not take $8$}) \\
&= \frac{10}{11} \cdot \frac{9}{10} = \frac{9}{11}
\end{align}
Can you continue from here? What would be 
\begin{equation}
\mathbb{P}(\text{Danilo gets $8$} \mid \text{Danilo enters $n$-th})?
\end{equation}
If you have difficulty doing the general one immediately, try a few specific cases and see if there is a pattern to them.
A: $P(Danielo\; selects\; his\; prefered\; 8\; number\; Tshirt)=1$
$P(Danielo \; enters \; at\; any\; position )=\frac{1}{11}$
How will Danielo get his prefered T shirt??
This can happen if Danielo goes first and picks his tshirt OR goes Second and picks his tshirt but keeping in mind that the person before him selects a tshirt from the other $10$ available OR Danielo goes in third and picks his tshirt provided the other two before him pick from the other $9$bavailable....OR....so on
$P(Daniel\;goes \;first\; and \;picks\; his\;tshirt )=\frac{1}{11}.1$
$P(Daniel\;goes \;second\; and \;picks\; his\;tshirt )=\frac{10}{11}\frac{1}{11}.1$
$P(Daniel\;goes \;third\; and \;picks\; his\;tshirt )=\frac{10}{11}\frac{9}{10}\frac{1}{11}$
Similary you do it for the time he goes in 4th,5th...11th.
Since any of these cases guarentees that he picks his tshirt you just add all thse probabilities and get it as $$\frac{1}{11}+\frac{10}{11}\frac{1}{11}+\frac{10}{11}\frac{9}{10}\frac{1}{11}+.............+$$
$$= \frac{1}{11}+\frac{10}{11^2}+\frac{9}{11^2}+..........+\frac{2}{11^2}+\frac{1}{11^2}$$
$$=\frac{1}{11}+\frac{10.11}{2.11^2}$$
$$=\frac{1}{11}+\frac{5}{11}$$
$$=\frac{6}{11}$$
