Excalibur sword probability Here is a story of King Arthur who's in a battlefield with his $7$ soldiers. He left his sword Excalibur to one of his soldier, also the other $6$ soldiers got an identical looking sword (for deceiving).
It happens that they encountered a dragon who can only be killed by using the Excalibur sword, however in a rush Arthur doesn't have time to check which is the real sword, and since he can only carry two swords he picks randomly two sword from his soldiers.
This is quite a special dragon who will break the sword if it's a fake and nothing will happen, and there is only a $90\%$ probability that he'll be killed if Arthur uses the Excalibur.
What is the probability that the dragon will be killed if Arthur goes in alone?
My idea is first to find the probability that Arthur will pick the Excalibur among those swords or if he doesn't pick it. Then to see what will happen with them.
Since he must carry two swords, and only one is real, one of the swords must be a fake so this leaves two cases picking the right one, and picking a fake one.
$$P(\text{picking Excalibur}\ \& \ a fake)=\frac{\binom{2}{1}\binom{6}{1}}{\binom{7}{2}}=\frac{4}{7}$$
$$P(\text{picking two fakes})=\frac{\binom{2}{0}\binom{6}{2}}{\binom{7}{2}}=\frac{5}{7}$$
And now I am struggling to finalize. But since the probability of killing if it's a fake is $0$ we can ignore that, right?
$$P(\text{killing the dragon})=P(\text{chance of killing the dragon}\  \cap\ \text{having excalibur})=\frac{9}{10}\cdot \frac47 $$ Is that it? Something seems wrong to me since the probability is kinda high and I think I am missing something obvious.

Edit: Further problem that was suggested in comments by JMoravitz.
What is the probability of killing the dragon if the fake sword got a probability of killing of $\frac{2}{10}$?
Here is my thinking. Now using the correct version we have:
$$P(\text{picking Excalibur}\ \& \ a fake)=\frac{\binom{\color{red}{1}}{1}\binom{6}{1}}{\binom{7}{2}}=\frac{2}{7}$$
$$P(\text{picking two fakes})=\frac{\binom{\color{red} 1}{0}\binom{6}{2}}{\binom{7}{2}}=\frac{5}{7}$$
$$P(\text{killing the dragon})=P(\text{chance of killing the dragon}\  \cap\ \text{having excalibur})$$$$+P(\text{chance of killing the dragon}\  \cap\ \text{not having excalibur})=\frac{9}{10}\cdot \frac27 +\frac{2}{10}\cdot \frac57=\frac{28}{70}=\frac25$$

Given the dragon is successfully slain, what is the probability that Arthur is actually holding the Excalibur?
$$P(\text{Excalibur/slain})=\frac{P(\text{Excalibur/slain})P(\text{Excalibur})}{P(\text{slain})}$$
$$=\frac{\frac{9}{10}\cdot \frac{2}{7}}{\frac{9}{10}\frac{2}{7}+\frac{2}{10}\frac{5}{7}}=\frac{18}{28}=\frac{9}{14}$$
Where for $P(\text{slain})$ I have used the probability of killing with Excalibur plus the probability of killing with a fake one. 
Can anyone confirm if I proceeded correctly?

Your approach was fine and mostly correct however you made a mistake in building your expression.
In the hypergeometric distribution we have some number of objects, $N$, some number of objects of a first type, $K$, implying the number of objects of a second type as being $N-K$, and we wish to choose $n$ objects and ask the probability of $k$ of those objects being of the first type.
The probability of this happening is $\dfrac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$
Here, there are $N=7$ swords in total.  $K=1$ of those swords are the excalibur.  The remaining $N-K=7-1=6$ of those swords are fakes.  We take $n=2$ swords and we ask for the probability that $k=1$ of the swords is the excalibur.
The probability is $\dfrac{\binom{\color{red}{1}}{1}\binom{6}{1}}{\binom{7}{2}}=\frac{2}{7}$.
Further, given that the excalibur was among the chosen swords, actually killing the dragon would only occur $\frac{9}{10}$ of the time after that making the probability of actually slaying the dragon as $\dfrac{2}{7}\times\dfrac{9}{10}$
