I have 100 boxes. C of them have a gift. I can open up to 16 boxes. What is the number of C that will give me probability over 0.5 to find a gift? Details:
We start by opening a box. If nothing is in there, we open another one. Once we find a gift, we can stop. Each empty box that was opened is discarded (no revisit).
I can find the number of $C$ that will give probability over $0.5$ by writing a program to try for $C=1, C=2$ .. etc.. , but I can't solve the equation for $C$ to find a more "mathematical" and elegant answer.

My work until now is:

1) Found in 1st box: $P(1) = \frac{C}{N}$

2) Found in 2nd box: $P(2) = \frac{1-C}{N}\cdot\frac{C}{N-1}$

4) Found in 3rd box: $P(3) = \frac{1-C}{N}\cdot\frac{1-C/}{N-1}\cdot\frac{C}{N-2}$

Etc...

Adding them up makes things very complicated to solve for $C$.

Any ideas? Thank you in advance!
Start with a rough estimate: If the box contents were independent, the probability of losing would be $(1-C/100)^{16}$. Equating this to $0.5$ gives us $C\approx 4.2$. 

Hence, we boldly check the cases $C=4$:
$C=4$ leads to a losing probability of $$\frac{96\choose 16}{100\choose 16}=\frac{96!84!}{80!100!}=\frac{84\cdot 83\cdot 82\cdot 81}{100\cdot 99\cdot 98\cdot 97}\approx 0.492$$
so a winning probability slightly above $\frac12$. A close look reveals that $C=3$ leads to a winning probability below $\frac12$, so the correct answer is $C=4$. 



Note that the "true" breaking point is thus between $3$ and $4$, not between $4$ and $5$ as the rough estimate suggested - the box contents are not independent after all (namely, if you find a - rare - gift, the probability of finding a gift in another box falls dramatically).
An alternative approach is to work in base 10 logarithms.  Chance of failing on 16 tries is 

$\displaystyle f(C) = \left(\frac{100-C}{100}\right) \times
\left(\frac{99-C}{99}\right) \times
\left(\frac{98-C}{98}\right) \times \cdots \times
\left(\frac{85-C}{85}\right).
$

Assume that you've written a computer program that calculates 
$\;\log_{10}n\;$ for $n\in\{30, 31, \cdots, 100\}.$

Then it becomes a simple matter to calculate 
$\;g(C) = \log_{10}f(C).$