Choosing a random number from 1, 2 or 3, and summing them, what is the probability that the sum is 4 at one point? For example we get 1+1+1+2 which can't be 4 after this point or 1+3 which is 4 after the second number chosen.
So we don't need to sum more than 4 random numbers since the minimum sum equal to for is 1+1+1+1. So for the 4 random numbers that get chosen we have $3^4$ combinations in total. There are 7 ways to sum 4 numbers or less to 4 which are $$ 1+1+1+1$$ $$1+1+2$$ $$1+2+1$$ $$2+1+1$$ $$2+2$$ $$3+1$$ $$1+3$$ so there should be a probability of $\frac{7}{81}$ which for some reason isn't the right answer. What did I get wrong?

Lord Shark is correct. The probability of $2+2$ is not $\frac{1}{81}$ but $\frac{1}{9}.$
More generally, if you select numbers from $1$ to $n$ and want the probability that the sum at some point is exactly equal to $k$, this is the coefficient of $x^k$ in:
$$\sum_{i=0}^{\infty} \left(\frac{x+x^2+\cdots +x^n}{n}\right)^i=\frac{n(1-x)}{n-(n+1)x+x^{n+1}}$$
So if $a_k$ is the coefficient of $x^k$ then $$a_{n+1+k}=\frac{1}{n}\left((n+1)a_{n+k}-a_{k}\right)$$
We also get $a_0=1,a_1=\frac{1}{n}$ and for $2\leq k\leq n$ we get $a_{k}=\frac{1}{n}\left(1+\frac{1}{n}\right)^{k-1}$. 
In particular, then, $a_{n+1}=\frac{1}{n}\left(\left(1+\frac{1}{n}\right)^n-1\right)$
For $n=3$, this gives $a_4=\frac{4^3-3^3}{3^4}=\frac{37}{81}$.
When $n=3$, we get the exact formula:
$$a_{k}=\frac{1}{2}+\frac{1}{4}\left(\left(\frac{-1+\sqrt{-2}}{3}\right)^k+\left(\frac{-1-\sqrt{-2}}{3}\right)^k\right)$$
Which gives us an estimate:
$$\frac{1}{2\cdot 3^k}\leq\left|a_k-\frac{1}{2}\right|<\frac{1}{2\cdot 3^{k/2}}$$
For general $n$ we have $\lim_{k\to\infty} a_k=\frac{2}{n+1}.$
