We split the interval into *n* equal subintervals of length *t*/*n*.
If we make *n* large enough, we can ignore the possibility that two
requests arrive in the same subinterval, so we can model the *n*
subintervals as independent (Bernoulli) random variables with
probability of success equal to *rt*/*n*. The probability
of *k* successes (request arrivals) in time *t* is then

C(n, k) * (rt/n)^{k } * (1 - rt/n)^{n-k}

where C(n, k) is the binomial coefficient *n* choose *k*.
If we let *n* go to infinity, we can weaken the assumption that
no two requests arrive in the same interval to the assumption that
no two requests arrive at the exact same time, so we rewrite the above as

(n/n) * ((n-1)/n) * ((n-2)/n) * ... *((n-k+1)/n) * (1 - rt/n)^{-k} *
(rt)^{k}/k! * (1 - rt/n)^{n}

As *n* goes to infinity, the first *k*+1 terms approach 1 and
the next term doesn't depend on *n*. The last term approaches
exp(-*rt*), so we get the following distribution

exp(-rt) (rt)^{k}/k!

known as the **Poisson** distribution.

**Reference:**

Probability and Statistics with Reliability, Queuing, and Computer Science
Applications

Kishor S. Trivedi

Prentice-Hall, 1982.