Depending on how you count collisions, we can expect either about 2000 collisions, or about 1472 collisions. The latter is in close agreement with what you observe, so yes, 1500 collisions looks about right to me.
Counting collisions pairwise
I think the most natural way to count collisions is "pairwise". For instance, if A, B, and C all hash into the same bucket, let's count that as 3 collisions. If A, B, C, and D hash to the same bucket, let's count that as 6 collisions (not 4, but 6: the 6 collisions are AB, AC, AD, BC, BD, and CD). And so on. In general if there are $k$ items in a bucket, that causes ${k \choose 2}$ collisions, with this method of counting collisions.
Then, you can use linearity of expectation to estimate the number of collisions, assuming the hash function is good. Let $X_{i,j}$ be an indicator random variable that is $1$ if objects $i,j$ hash to the same bucket, or $0$ otherwise. Let $Y=\sum_{i<j} X_{i,j}$ be the sum of these random variables. Then $Y$ counts the number of collisions.
Can we estimate a typical value for $Y$? Well, we can compute the expected value of $Y$ in a straightforward way:
$$\mathbb{E}[Y] = \sum_{i<j} \mathbb{E}[X_{i,j}] = \sum_{i<j} {1 \over 4000} = {4000 \choose 2} {1 \over 4000} = {3999 \over 2} \approx 2000,$$
using the fact that $\mathbb{E}[X_{i,j}] = \Pr[X_{i,j}=1] = {1 \over 4000}$ since there is a 1 in 4000 chance that object $j$ hashes to the same bucket as object $i$ (assuming a good hash function). Typically, the observed value of $Y$ will be near its expected value. So, that'd predict that you see about 1500 collisions. That's a consequence of the number of buckets and the number of items hashed.
In general, if you hash $n$ items into $m$ buckets using a good hash function, you should expect to see about ${n^2 \over 2m}$ collisions, on average, using this method for counting collisions.
Counting collisions by bucket size
On reading your comment, it sounds like you are counting collisions differently. If there are $k \ge 1$ items in a bucket, it sounds like you are counting that as $k-1$ collisions. How does that affect things? Well, it makes the calculation more complicated. I'll share a much terser analysis for this case, without showing all the details.
Let $N_i$ denote the number of items in bucket $i$. If the hash function is good, the number of items in a bucket should follow approximately the Poisson distribution with parameter $\lambda = n/m$. In your case, we have $\lambda=1$. The total number of collisions (the way you are counting it) is
$$N = \sum_i \max(N_i-1,0).$$
Now,
$$\begin{align*}
\mathbb{E}[\max(N_i-1,0)]
&=
1 \cdot {\lambda^2 e^{-\lambda} \over 2!} +
2 \cdot {\lambda^3 e^{-\lambda} \over 3!} +
3 \cdot {\lambda^4 e^{-\lambda} \over 4!} + \cdots\\
&=
\lambda e^{-\lambda} + \lambda^2 - \lambda.
\end{align*}$$
(Don't ask me how I summed the series; black magic.) It follows that
$$\mathbb{E}[N] = \sum_i \mathbb{E}[\max(N_i-1,0)] = (\lambda e^{-\lambda} + \lambda^2 - \lambda) n.$$
In your case $\lambda=1$ and $n=4000$, so we predict $1/e \times 4000 \approx 1472$ collisions. That has very close agreement with what you are observing empirically, so practice and theory seem to be in close agreement.
