What is the probability that an event will happen when the probability decreases exponentially?

Question

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?

Answer 1

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, $$ 1-\prod_{k=1}^n\left(1-p_k\right)\lt\sum_{k=1}^np_k $$

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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 $2\%$. $$ \begin{array}{c|l|l} n&1-\prod(1-p_k)&\sum p_k\\\hline 1&0.01&0.01\\ 2&0.01495&0.015\\ 3&0.017412625&0.0175\\ 4&0.018640859219&0.01875\\ 5&0.019254208682&0.019375\\ 10&0.019847903640&0.01998046875\\ 100&0.019867047111&0.02\\ \infty&0.019867047111&0.02 \end{array} $$

Answer 2

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)

Answer 3

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