Probability that after n trials the sum of values would be more than x. Suppose we have a bag with cards, each card has a number $2$,$3$ or $4$ on it, so sample space is $\{2, 3, 4\}$. Each sample point is equally likely (so $1/3$).
We than pick $n$ cards from the bag. What is the probability that sum of numbers on the cards will be more than $x$?  
Also the overall number of cards in the bag is infinite or so high it is negligible (I suppose it doesn't matter at all in this question).
I want to find general formula, but at least solve some example would be good also, I'm kinda stuck.  
So suppose we pick $12$ cards and need to find what is the probability that we will get more than $40$ if we sum all the numbers on them.  
$$n=12, p=1/3, P[X\geq40]=?$$
It looks kinda like Binomial distribution, but here $X$ does not mean "number of successes obtained in $n$ trials". In fact, I don't think we have any kind of "success" here. What should I use in this case?
 A: $$\frac{N(2x_1+3x_2+4x_3\geq 40,x_1+x_2+x_3=n)}{N(x_1+x_2+x_3=n)}=\frac{N(2x_1+3x_2+4n-4x_1-4x_2\geq 40, x_1+x_2\leq n)}{N(x_1+x_2+x_3=n)}$$
$$=\frac{N(2x_1+x_2\leq 4n-40, x_1+x_2\leq n)}{N(x_1+x_2+x_3=n)}=\frac{\sum_{i=0}^{4n-40}N(2x_1+x_2=i, x_1+x_2\leq n)}{\binom{n+3-1}{3-1}}$$
$$=\frac{\sum_{i=0}^{4n-40}\sum_{x_1=0}^{[\frac{i}{2}]}N(x_2=i-2x_1\leq n-x_1)}{\binom{n+2}{2}}=\frac{\sum_{i=0}^{4n-40}\sum_{x_1=0}^{[\frac{i}{2}]}f(i,x_1)}{\binom{n+2}{2}}$$
Where $N()$ means number of possible non-negative integer answers. $x_i$, $i=1,2,3$ respectively show number of cards with 2, 3 and 4 on them in the chosen n cards. $[z]$ means biggest integer before $z$. And finally $f(i,x_1)$ is $0$ if $i-2x_1>n-x_1$ and is $1$ if $i-2x_1\leq n-x_1$.
For $n\geq 20$ this fraction is $1$ and for $n< 10$ is $0$, which is compatible with the observation since;
$$n\geq 20\Longrightarrow 2x_1+3x_2+4x_3\geq 2(x_1+x_2+x_3)=2\times n\geq 40$$ 
$$n< 10\Longrightarrow 2x_1+3x_2+4x_3\leq 4(x_1+x_2+x_3)=4\times n< 40$$ 
Here I wrote a Maple code to compute it. I chose it return us for values $n$ from $1$ to $20$, you can run it for different values. As you can see for $n=20$ it is $1$ and for $n$ strictly less than $10$, it is $0$.
f := proc (n, x, y) 
     if x-n <= y then 1 else 0 end if; 
     end proc;
for n from 1 by 1 to 20 do
    add(add(f(n, i, j), j = 0 .. floor((1/2)*i)), i = 0 .. 4*n-40))/binomial(n+2, 2);
    end do;


