TITLE: Frequency from probability
QUESTION [6 upvotes]: Assume an event has a probability $p=1/100$ of happening, per trial. Here are three statements that I believe are true:

On average, the event is witnessed in a one-hundredth of the the
trials.
The average number of trials between consecutive events is
$100$.
It is more likeley than not to witness at least one event in a
set of $69$ trials. This is found by computing the probability of the complement, i.e. the probability of no event in $69$ trials, which is $(1-p)^{69}=0.4998...$.

I can't understand how statements 2 and 3 can both be true. Is there a flaw in the reasoning somewhere ? Can they both be true? What is the "frequency" of the event ?

REPLY [2 votes]: To add onto Clement's answer: If you think of the events as being the possible decay of a radioactive atom, with one trial per second, then your statements 2 and 3 reflect the difference between the mean life of the element (which would be about $100$ seconds), and the half-life of the element (which would be about $69$ seconds).
The inter-event time follows a geometric distribution, with the time $\tau$ between events having the probability distribution
$$
P(\tau = n) = p(1-p)^n
$$
with $p = 1/100$.  (You might write $n-1$ in the exponent, depending on what you mean by "between.")  The continuous analogue of this is the exponential distribution, where the probability density function (PDF) of the inter-event time is given by
$$
f_\tau(t) = \lambda e^{-\lambda t}
$$
with $\lambda = 1/100$.  Here, $\lambda$ gives (as $p$ did, above) the event "rate," which you may notice is the derivative $f'_\tau(0)$.  It turns out that for this distribution, the mean lifetime of an atom is given by $1/\lambda = 100$.
However, the half-life of the element is given by the cumulative distribution function (CDF), which is the definite integral, from $0$ to $t$, of the PDF.  That is,
$$
F_\tau(t) = \int_{x=0}^t f_\tau(x) \, dx = 1-e^{-\lambda t}
$$
The CDF gives the probability that an event happens before time $t$; equivalently, for this scenario, it gives the proportion of the sample that has decayed by time $t$.  Thus, the half-life is governed by
$$
e^{-\lambda t} = \frac{1}{2}
$$
$$
\lambda t = \ln 2 \doteq 0.69315
$$
which is why you found the half-life and the mean life to be related in approximately a $69:100$ ratio.