I have a bag containing N coins. What is the probability that I have a round dollar amount? In my country we have \$0.10, \$0.20, \$0.50, \$1, and \$2 coins. 
If I were to pour a bag of coins out on the table what would be the probability that I could buy a heap of \$1 snacks without needing any change? Does this change if the bag doesn't contain any whole dollar value coins?
I'm fairly sure the that  $P=0.1$ for very large values of $N$ (as there is 10 possible cent values). I'd like to be able to prove this and be able to see how the probability changes with $N$, but I cant figure out a rule for the entire series & larger values of $N$.
I've written a little Python simulation to test $N$ values $0$ through $50$ and I'll edit with the results of that when it finishes.
EDIT: Results of my script seem to confirm my thought: http://pastebin.com/cD8PeuwT
 A: You are in one of ten states; that being the number of 10c.
So treat it as a ten-vector, initially in state $v=[1,0,0,0,0,0,0,0,0,0]$.
The transition matrix is either this one, for no gold:
$$A=\left[\begin{array}{cccccccccc}
0&1/3&1/3&0&0&1/3&0&0&0&0\\
0&0&1/3&1/3&0&0&1/3&0&0&0\\
0&0&0&1/3&1/3&0&0&1/3&0&0\\
0&0&0&0&1/3&1/3&0&0&1/3&0\\
0&0&0&0&0&1/3&1/3&0&0&1/3\\
1/3&0&0&0&0&0&1/3&1/3&0&0\\
0&1/3&0&0&0&0&0&1/3&1/3&0\\
0&0&1/3&0&0&0&0&0&1/3&1/3\\
1/3&0&0&1/3&0&0&0&0&0&1/3\\
1/3&1/3&0&0&1/3&0&0&0&0&0\end{array}\right]$$
or something similar with fifths if there is gold.
The initial distribution, for no coins is $v$; for one coin is $vA$, for two coins is $vAA=vA^2$ and for $n$ coins is $vA^n$.
Look at the eigenvalues of the matrix to see how quickly the initial state approaches $[1/10,1/10,...,1/10]$
It turns out the eigenvalues are $(\omega+\omega^2+\omega^5)/3$, where $\omega^{10}=1$ is one of the tenth roots of unity.  The largest of these, when $\omega=1$, is $1$.  The next largest eigenvalues have amplitude 0.71632, so the distance from an even distribution decreases by that proportion when $N$ increases by 1.
The eigenvalues with gold coins are $(2+\omega+\omega^2+\omega^5)/5$, the largest of which (except 1) has absolute value 0.58713, so it reaches equilibrium more quickly.
A: 
I'm fairly sure the that $P=0.1$ for very large values of $N$

Here is a proof of this statement.
As $N$ tends to infinity,


*

*The amount of extra money from 10-cent coins will tend towards a uniformly random distribution on $\{0, .1, .2, .3, .4, .5, .6, .7, .8, .9\}$.

*The amount of extra money from 20-cent coins will tend towards a uniformly random distribution on $\{0, .2, .4, .6, .8\}$.

*The amount of extra money from 50-cent coins will tend towards a uniformly random distribution from $\{0, .5\}$.
To find the final probability of the total money being an integer, we just add these three distributions mod $1$.
But the uniform distribution added to any distribution mod 1 is again the uniform distribution.
So the distribution on the sum mod 1 is (as $N \to \infty$) uniform on $\{0, .1, .2, .3, .4, .5, .6, .7, .8, .9\}$, as you conjectured.
