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
known as the Poisson distribution.
Probability and Statistics with Reliability, Queuing, and Computer Science
Kishor S. Trivedi