What is the probability of two integers having identical values where each integer is summation of 4000 same or different integer values? There are two sets of 32 bit unsigned integer. Each set has 4000 numbers. Two sets could either be identical or they can different values. All 4000 numbers from each set are added and then the resulting sums from both sets are compared with each other. What is the probability of these two sums being identical? I am not good in math so haven't been able to try out any answer. My apologies in advance. Could you please help?
Thank you.
 A: Not a complete answer, but a start
Warning: I know a bit of probability theory, but it's certainly not my strong suit. You're getting an answer here that's a lot like getting an answer to a calculus question from someone who just finished a first calculus course. I hope someone else can give a more authoritative answer based on what I've started here. 
OK...with that caveat, here goes: 
You've got 4000 32-bit natural numbers, uniformly distributed, but without repetition, in each of two sets $A$ and $B$. You're asking what is
$$
P(\sum_{a \in A} a= \sum_{b \in B}b)?
$$
I can't give you an exact answer, but I can give you a start on a very good approximation. 
First approximation: I'm going to assume that the 4000 numbers in each set are picked uniformly randomly and independently, so that either set $A$ or set $B$ might contain duplicates. The probability of no duplicates in set $B$, for instance, is then
$$
\frac{N-1}{N} \cdot \frac{N-2}{N} \cdots \frac{N-3999}{N}
$$
where $N$ is shorthand for $2^{32}$. That works out to about 99.8 percent, so this is a very strong approximation. 
Next, the random variable 
$$
S_A = \sum_{a \in A} a
$$
is very nearly normally distributed: if you made a histogram of the values of $S_A$, it'd look a lot like the normal distribution. (This follows from something called the Central Limit Theorem; I've added that to the list of tags for this question). 
The random variable $a$ is uniformly distributed on integers between $0$ and $N-1$, so its mean is 
$$
m = \frac{N-1}{2}
$$
and its variance is (roughly)
$$
v = \frac{1}{12} (N-1)^2.
$$
Using independence of the individual numbers, we find that the mean of $S_A$ is $4000m$, and the variance is $4000v$. The same holds for $S_B$. 
The difference $D = S_A - S_B$ is also a nearly-normal random variable, whose mean is therefore $0$, and whose variance is $8000v$. (See http://mathworld.wolfram.com/NormalDifferenceDistribution.html). 
We get a "collision" -- the values $S_A$ and $S_B$
 are the same -- exactly when $D = 0$. 
Now $D$ is really a discrete random variable --- it takes on only integer values --- but we're approximating it with a continuous random variable -- let's call that $C$. I'm going to say that the probability that $D$ is $0$ is close to the probability that $C$ is between $-\frac12$ and $\frac12$. 
Sadly, we can't just look this up. But we can define
$$
C' = \frac{1}{\sqrt{8000v}} C
$$
and observe that $C'$ is normally distributed with mean $0$ and variance $1$, i.e., a normal random variable. And if $C$ is between $-1/2$ and $1/2$, then $C'$ is between 
$$
\pm\frac{1}{2\sqrt{8000v}}
$$
And the probability of THAT happening is written down. It's known as the "error function" (erf), and all we need to do is look at 
$$
erf(\frac{1}{2\sqrt{8000v}}).
$$
Fortunately, Matlab's willing to give us that value (or at least a good approximation of it), and it comes up with 3.3342e-07, i.e., about $3 \times 10^-7$, which is about 1 in 3 million. 
So: the odds of a collision (if my very rusty probability theory is right) come out to be about 1 in 3 million. 
