Given a 5.67% chance for success and using its Expected Value, shouldn't this binomial formula equal 50% 1-(1-0.0567)^(100/5.67)? I play a game.  It is RNG-based and as such, has a chance of success for each sample.
Data has shown that in a particular part of the game where one may obtain a character, there is a success rate of 5.67% per sample for a certain type of character.
Based upon this data, the Expected Value of obtaining this type of character is believed to be one every ~17.64 samples (100/5.67).
We can now assume that after 17.64 samples that we have a 50% chance to obtain said character because we know 17.64 is the average number of samples it takes to obtain said character.
Since the success rate is 5.67% and one has a 50% chance to obtain said character at 17.64, shouldn't we be able to also calculate the number of samples (17.64) from a binomial formula that equals 0.5, the average?
To reiterate, the formula 1-(1-P)^x = 0.5.  Shouldn't x result in 17.64? A 50% chance to obtain the character.
It doesn't seem to work for me.
1-(1-0.0567)^(17.64) results in ~0.64 or 64% chance to obtain the character.
1-(1-0.0567)^x = 0.5 results in x = ~11.8749 or about 11.8749 samples to obtain the character.
What am I missing?  Is the average value not 50%, but instead 64%?
 A: Most of your question is premised on a misunderstanding of how these probability models work. I'm not going to directly engage with that -- instead, I'll give examples of how to work out these problems correctly.
Suppose that we model this process as a binomial distribution. This model assumes that each trial is independent and has a finite, fixed integer number of trials and a fixed probability of success for each trial. All trials are attempted, and the number of successes are totaled up at the end.
The probability of obtaining the monster in a sequence of $n$ trials is $\text{Binom}(n, 0.0567),$ which has density
$$
f(x;n,p)=\binom{n}{x}p^x(1-p)^{n-x}
$$. The density formula gives the probability of obtaining exactly $x$ monsters in $n$ trials.
For example, the probability of obtaining it in one trial is 0.0567. But in the more general case, you need to be more specific -- when you say "obtain" do you mean obtain exactly once? Or at least once? If you mean at least once, the probability of obtaining the monster is $1-f(0;n, p)$, i.e. the complement of the event that you obtain the monster 0 times in all $n$ trials.
In the scenario where you take 17 trials, the probability of obtaining the monster at least once is $1-f(0;17,0.0567)=0.629279$.
On the other hand, the outcome I think you're really interested in might be better modeled as a negative binomial distribution. This model has a fixed probability of success for each trial, the trials are independent, and the trials continue until the desired number of successes are obtained (specifically, the last trial is a success).
So we want to know how many trials to carry out to have some probability of having one success at summoning the monster. We can do this computation in R using the qnbinom function. The quantile function takes a probability, i.e. a value of the CDF, and returns the corresponding value of the random variable (in the case of discrete r.v.s, the smallest value of the r.v. that includes the probability).
qnbinom(0.5, size=1, prob=0.0567)=11
This means that there's a 50% shot of summoning the monster on the 12th attempt: 11 failures followed by the successful summoning.
A: I think you have mean and median mixed up.
The mean (or expected value) is, indeed about 17.  But this value is skewed high by the small odds of taking a very large number of turns.
The formula you use looks like a continuous approximation of the discrete CMF (cumulative mass function) $$1-(1-p)^n$$
But solving the CMF = 0.5, does NOT give you the expected value.  It gives you the median (equal chance of greater or lesser outcome) value, which is indeed about 12.
A: I believe you have confused the meaning of expected value.

We can now assume that after 17.64 samples that we have a 50% chance to obtain said monster because we know 17.64 is the average number of samples it takes to obtain said monster.

This statement is incorrect, and doesn't make fundamental sense (you cannot do 17.64 trials)
An expected value of 17.64 in this case does not mean this, (I used 18 as approximation):
X~B(18, 0.0567)
P(X=1) = 0.5 or P(X>=1) = 0.5.
The expected value of 17.64 means that if you draw many many trials from this process, we expect 1 success with the average interval of 17.64 trials. Note that this says nothing about the probability of seeing success or not.
In fact to find the probability that in 18 trials you will observe 1 or more successes:
X~B(18, 0.0567)
P(X=1) = 0.378357850443551
P(X>=1) = 0.65029888269312
i.e. your chance of getting exactly one success for 18 trials is 38%, and getting 1 or more is 65%.

However if you are insistent on defining an "average" as number of trials needed for the cumulative chance of 50% to obtain at least 1 success or more, then General Abrial's answer is the perfect way to calculate it. i.e.
"Average" = What is the needed number of trials n, to have a 50% shot at getting at least 1 success? 
Although, from a gamer perspective, why is it so important to use 50% cumulative chance? Personally I would rather go for something more significant, like 80~95% cumulative chance to get at least 1 success.
Say if the summon (presumably powerful) is needed to beat a boss (or win a PvP), would you rather calculate for 50% chance of making it or 80% chance of making it.
A: There is actually a binomial formula to use. This is due to a correspondence between the geometric distribution and the binomial distribution.
Waiting time
The geometric distribution relates to the waiting time
(more general is the negative binomial distribution that Sycorax mentions).
$$\begin{array}{rccc} 
P(\text{wait until $k$-th trial}) &= & P\left(\substack{\text{you had not yet}\\\text{ a monster}\\\text{ in the $k-1$-th trial}}\right)& P\left(\substack{\text{get monster}\\\text{ in the $k$-th trial}}| \substack{\text{you had not yet}\\\text{ a monster}\\\text{ in the $k-1$-th trial}}\right) \\
&=&(1-p)^{k-1}&p
\end{array}$$
The distribution looks like below:

Number of success
The binomial distribution relates to the number of monsters
Say you have $n$ trials then with the the value of $p$ you can estimate the probability of $1, 2, 3, \dots , n$ successes.
$P(\text{$k$ successes}) = {n\choose k} p^k (1-p)^{n-k}$
This distribution looks like below:

The number of trials for the median of the geometric distribution ($n=12$) corresponds to the binomial probability being approximately equal to $P(K=0) = 0.5$.
So the binomial formula $1-p^n = 0.5$ is correct, but it gives you the median and not the average (expected value).

Relations
The following are approximately equivalent

*

*'Median of waiting time is $k$'
versus
'probability of 0 events in $k$ trials is 0.5'
The following are approximately equivalent

*

*'average waiting time is $\mu$'
versus
'average number of events in $k$ trials is $k/\mu$'
To find the number $x$ in
"with $x$ samples we have a 50% chance to obtain said character"
You need the first case and that is either the median of the waiting time or the probability $P(K=0) = p^n$ for a binomial distribution.
