Determining the probability of being offered a certain number of resources There are $32$ kingdoms. They each have a $4\%$ chance that they will offer me resources every year. Of the $32$ kingdoms, what is the probability that at least one kingdom will offer me resources. 
How would I then go about figuring out subsequent amounts of kingdoms offering resources, (at least 2/32 kingdoms; at least 3/32 kingdoms; and so forth)?
I thought that it would be something like this: 
If the chance of ALL kingdoms offering resources is 1/800, 1/(25x32), then the math would be something like 32/25 - 1/800 = 1023/800 = 1.27 = 127% of at least ONE kingdom happening? If this math is completely wrong, please edit this portion out, but if I am somewhere near the right answer, please help my figure out how to solve this.
 A: This is a binomial distribution
$$P(X=k)={n \choose k}p^k(1-p)^{n-k}$$
where in our case
$$n=32$$
$$p=0.04$$
We have
$$\begin{align*}
P(X\geq1)
&=1-P(X=0)\\\\
\end{align*}$$
Similarly
$$\begin{align*}
P(X\geq2)
&=1-P(X=0)-P(X=1)\\\\
\end{align*}$$
and so forth.
Can you go from here?
A: Okay so I'll adopt a weird notation for this: pretend that the probability that you get resources from a particular kingdom is just R, if they offer you nothing, I'll use N.
One Kingdom
So if there was one kingdom, the possible outcomes are, you either get resources (a 4% chance) or you don't (96% chance); but we know that probabilities have to sum to one, so:
$$ R + N = 0.04 + 0.96 = 1$$
So this is all well and good
Two Kingdoms
Now if there are two kingdoms, you could use a tree diagram like:
-----|--K1_resources----|-----K2_resources
     |                  |-----K2_nothing
     |
     |---K1_nothing-----|-----K2_resources
                        |-----K2_nothing

So there are four possible outcomes, and you can calculate the associated probabilities just fine, eg: k1_resources * k2_resources = 0.04 * 0.04 which it seems that you are pretty comfortable with:
A simper Alternative
Its actually much simpler to just enumerate the terms like this:
$$ (R + N) ( R + N) = (R+N)^2 = R^2 + 2 RN + N^2 $$
Now the probability that you get aide twice is just $R^2 = 0.0016$, that you get aide once $2RN=2 * 0.04 * 0.96= 0.0768$ and that you get no aide $N^2 = 0.96*0.96 = 0.9216$.
This is much simpler as you don't need to worry about which kingdom is actually giving you the aide.
Lots of Kingdoms
Now you can probably figure that if I have lots of kingdoms, we can say 32 as you did in your example, thats equivalent to expanding:
$$ (R + N)^{32} $$
This can easily be done by using pascals triangle, the answer will just be:
$$ (R+N)^{32} = \sum_{k=0}^{32} \binom {32}{k} R^k N^{32-k} $$
If you haven't seen this result before, I'd point you to this explanation which I think is excellent!
But then if you wanted the probability that you got aide from three kingdoms for example, that would just be:
$$ \binom{32}{3} R^3 N^{29} = 0.09716893  $$
Hope that helps :)
A: I suspect that you're having trouble conceptualising what's going on. First and foremost, the probability of something happening can never be more than $1$. In the same vein of thought, it can never be less than $0$. So, to that end, your answer is incorrect. 
Supposing that kingdoms hand out resources independently, you can use the binomial distribution (as @Remy shows) to answer your question. By 'independently' I mean that if kingdom 1 gives you resources, it doesn't change the probability of whether or not kingdom 17 will give you resources, or whether kingdom 21 will or kingdom 31 will...and so forth. 
To break down your question lets consider the classic example. You're playing basketball, and you're not such a good shot - so you only make a basket with probability $0.2$. 
Suppose then you make three consecutive and INDEPENDENT shots.
You can either make; $0$ shots, $1$ shot, $2$ shots or $3$ shots.
Now, ask yourself how many ways you can make $0$ shots in $3$ attempts. Well, there's $1$ way. Or, $3 \choose 0$. 
Now, ask yourself how many ways you can make $1$ shot in $3$ attempts. Well, there's $3$ ways. Or, $3 \choose 1$. 
Now, ask yourself how many ways you can make $2$ shots in $3$ attempts. Well, there's $3$ ways also. Or, $3 \choose 2$.
Now, ask yourself how many ways you can make $3$ shots in $3$ attempts. Well, there's $1$ way. Or, $3 \choose 3$.
To get the probability of making $1$ shot, just multiply the probability of you making a given shot by the number of ways you can make $1$ shot AND the probability that you miss $2$ shots. 
So, ${3 \choose 1} \cdot 0.20 \cdot 0.80^2$
To get the probability of making $2$ shots, just multiply the probability of you making a given shot by the number of ways you can make $2$ shots AND the probability that you miss a shot. 
So, ${3 \choose 2} \cdot 0.20^2\cdot 0.80$
Hopefully you can now see the classic form of the binomial distribution forming, 
$$Pr(X = k) = {n \choose k}p^k(1-p)^{n-k}$$
Here, ${n \choose k}$ rears it's when you're trying to count the number of ways something can occur when you don't care about how they're ordered. Read more here. 
Here, $X$ is just what we call a random variable. It can be, in your 'kingdom scenario', any number from $0$ to $32$. $k$ is just the value that $X$ takes on. $p$ is the probability of success, which in your case, is being given resources. $1 - p$ is the probability of not being given any resources (read about the complement rule (i.e. $1-p$) here). Finally, $n$ is the total number of trials. In your case, it's just the number of kingdoms.
