What is the time between groups of events when single events have a Poisson distribution? I'll ask this with a concrete example to be clear.
Let's say I have a Poisson process that tends to produce one event every two minutes. Then the probability of getting an event in a given minute is about 30%, and the probability of getting two events in a given minute is about 7.6%, assuming I'm applying the distribution correctly. 
I'm interested in the time distribution between groups of events: How long do I have to wait to see two events within a minute? What is the general form for this?
 A: I have been following this Question for about a day now, and am
still not exactly sure what you are asking. Here are some
comments based on sequences of one-minute time intervals, for which the number of $X$ of events in one minute is Poisson with
mean 1/2.
EXACTLY ONE event in such an interval: $P(X = 1) =  0.3032653.$
Geometric mean waiting time for the first interval with exactly one event: reciprocal = 3.297443. By independence, given you have just seen such an interval, this is the waiting time for the next one.
Negative binomial mean waiting time for two such intervals:
double =  6.594885.
AT LEAST ONE event in such an interval: $P(X \ge 1) = 1 - P(X = 0)
= 0.3934693.$ Geometric mean waiting time for the first interval with
at least one event:  2.541494. Negative binomial mean waiting time for two
such intervals:  5.082988. Perhaps coincidentally, this is
the same as in the Comment where @Did speculates  $E(D) = 1/[\lambda(1-e^{-\lambda})] = 5.082988,$
for $\lambda = 1/2.$
AT LEAST TWO events in such an interval: $P(X \ge 2) = 0.09020401.$
Geometric mean waiting time for the first interval with at least two
events: 11.08598.
