What is the probability that an event will happen when the probability decreases exponentially? So I am just a middle student who started studying probability in my stats class and I see the questions on this site are little advanced so I don't hope this one is too basic.
Anyways lets say that the probability of an event happening was some number(like 1%), and each time a trial is run the of the event happening decreases by half(so 1% probability for the first trial, .5% probability for the second, .25% probability, and so on).
What is the probability that the event will happen after 1 trial, 10 trials, 100 trials, and an infinite amount of trials?
 A: Note that if $p,q\gt0$, then
$$
\begin{align}
(1-p)(1-q)
&=1-(p+q)+pq\\
&\gt1-(p+q)
\end{align}
$$
Inductively, we can show that if $p_k\gt0$ and $n\gt1$, then
$$
\prod_{k=1}^n\left(1-p_k\right)\gt1-\sum_{k=1}^np_k
$$
and therefore,
$$
1-\prod_{k=1}^n\left(1-p_k\right)\lt\sum_{k=1}^np_k
$$

Assuming the events are independent, the probability that the event occurs at least once in the first $n$ trials is
$$
1-\underbrace{\prod_{k=1}^n\overbrace{\left(1-\frac{0.01}{2^{k-1}}\right)}^{\substack{\text{probability that}\\\text{the event does}\\\text{not occur on}\\\text{trial $k$}}}}_{\substack{\text{probability that the event}\\\text{does not occur in the}\\\text{first $n$ trials}}}\lt0.02
$$
So the probability that the event occurs at all is less than $2\%$.
$$
\begin{array}{c|l|l}
n&1-\prod(1-p_k)&\sum p_k\\\hline
1&0.01&0.01\\
2&0.01495&0.015\\
3&0.017412625&0.0175\\
4&0.018640859219&0.01875\\
5&0.019254208682&0.019375\\
10&0.019847903640&0.01998046875\\
100&0.019867047111&0.02\\
\infty&0.019867047111&0.02
\end{array}
$$
A: Well, I don't think you'll understand most of the answers same as me but well, genereally this is what will happen first one is 1% then 0.5% then 0.25% so the probability here is
Pn= 1% * 0.5^n
can't use the maths symbols as I'm new here but there you go, n refers to the number of trial, 1% is your starting point, 0.5 is the chance decrease( n times)
A: An interesting problem, that has many practical applications.
Moreover, considering a physical model will help to fix clearly the universe of events which might be bewildering
when dealing with probability of infinite sets.
We can think in fact of having a quantity of objects (radioactive nuclei, reacting molecules, mutating viruses, etc.) of type  $A$, which
decay (transform, mutate,..) into objects of type $B$ at a constant rate, so that
$$
\eqalign{
  & A(t) = A(0)\,e^{\, - \,t/T} \quad A(0) + B(0) = A(t) + B(t)  \cr 
  & B(t) = B(0) + A(0) - A(t) = B(0) + A(0)\left( {1 - \,e^{\, - \,t/T} } \right) \cr} 
$$
Starting with a given concentration at time $t=0$, then the concentration will vary in time as
$$
c_{\,0}  = {{A(0)} \over {A(0) + B(0)}}\quad  \Rightarrow \quad c(t) = c_{\,0} \,e^{\, - \,t/T} 
$$
Now we "scan" each single object every $\Delta t$ and ask which is the probability of
revealing an $A$ object at scan $n$, letting $n$ to start from $0$.
Let's denote with $\overline P \left( n \right)$ the cumulative probability of revealing only type $B$ till  scan $n$ incluse, then
$$
\eqalign{
  & \overline P \left( n \right)
 = \prod\limits_{k = 0}^n {\left( {1 - c_{\,0} e^{\, - \,k\,\Delta t\,/\,T} } \right)}
  = \prod\limits_{k = 0}^n {\left( {1 - c_{\,0} e^{\, - \,\,k/K} } \right)}  =   \cr 
  &  = \prod\limits_{k = 0}^n {\left( {1 - e^{\, - \,\,\left( {k + a} \right)/K} } \right)}
  = \prod\limits_{k = a}^{n + a} {\left( {1 - e^{\, - \,k/K} } \right)}
  = {{\prod\limits_{k = a}^{n + a} {\left( {e^{\,k/K}  - 1} \right)} }
 \over {\prod\limits_{k = a}^{n + a} {e^{\,k/K} } }} \cr} 
$$
and taking the logarithm
$$
\ln \overline P \left( n \right)
 = \sum\limits_{k = a}^{n + a} {\ln \left( {1 - e^{\, - \,k/K} } \right)}
 \quad  \Rightarrow \quad \left\{ \matrix{
  f(x) =  - \ln \left( {1 - e^{\, - \,x/K} } \right) \hfill \cr 
   - \ln \overline P \left( n \right) = \sum\limits_{x = a}^{n + a}
 {f(x)}  \hfill \cr}  \right.
$$
where we changed the sign of the log so that $f(x) \; |\,0<x$ is positive, monotonically decreasing and convex.
Now, unfortunately, there is not a closed form for the product or sum.
The series development of
$$
P(x,n) = \prod\limits_{k = 1}^n {\left( {1 - x^{\,k} } \right)} 
$$
is quite unmanageable : see OEIS sequence A231599.
However we can squeeze the logarithm into two inequalities
$$
{1 \over 2}\left( {f(a) + f(a + n)} \right) + \int_{x = a}^{n + a} {f(x)dx}
  < \sum\limits_{x = a}^{n + a} {\,\,f(x)}
  < \int_{x\, = \,a - 1/2}^{n + a + 1/2} {f(x)dx} 
$$
by comparing the continuous integral with the discrete sum, represented with trapezia
and rectangles as shown in the sketch.

The indefinite integral can be expressed through the Polylogarithm as
$$
\int {f(x)dx}  =  - K\int {\ln \left( {1 - e^{\, - \,x/K} } \right)d\left( {x/K} \right)}
  =  - K\,{\rm Li}_2 \left( {e^{\, - \,x/K} } \right)
$$
Therefore the inequality becomes
$$
\eqalign{
  & {1 \over {2K}}\left( { - \ln \left( {1 - e^{\, - \,a/K} } \right)
 - \ln \left( {1 - e^{\, - \,\left( {a + n} \right)/K} } \right)} \right)
 - {\rm Li}_2 \left( {e^{\, - \,\left( {a + n} \right)/K} } \right)
 + \,{\rm Li}_2 \left( {e^{\, - \,a/K} } \right)  \cr 
  &  <  - {{\ln \overline P \left( n \right)} \over K} <   \cr 
  &  - {\rm Li}_2 \left( {e^{\, - \,\left( {n + a + 1/2} \right)/K} } \right)
 + {\rm Li}_2 \left( {e^{\, - \,\left( {a-1/2} \right)/K} } \right) \cr} 
$$
an example of which is sketched below.

In the limit for $n \to \infty$ we get
$$
K\,{\rm Li}_2 \left( {e^{\, - \,a/K} } \right) - {1 \over 2}\ln \left( {1 - e^{\, - \,a/K} } \right)
 <  - \mathop {\lim }\limits_{n\, \to \,\infty } \ln \overline P \left( n \right)
 < K\,{\rm Li}_2 \left( {e^{\, - \,\left( {a - 1/2} \right)/K} } \right)
$$
which is graphed below

