Probability of at least one success after infinite trials [closed]

Question

I have a distribution problem where in every trial the probability of success is equal to any number between 0 and 1 (0 inclusive but 1 exclusive $\rightarrow 0 \leq P($success) < 1$). We also have a guarantee that it can not be always 0.

Now i just want to show that if i run infinite trials than the probability of having at least one success will eventually converge to 1.

In terms of logic i believe this is quiet trivial to understand but i dont know how to prove this mathematically.

Edit: I forgot to add that the trials are independent of each other. All the insights on the problems when the variables are somehow dependent was, however, really helpfull.

Answer 1

There are infinitely many non-zero probabilities of success. Since the zero probabilities are irrelevant, and you will always eventually reach all the non-zero probabilities, consider only these non-zero probabilities of success and write them as a sequence: $\ \{p_i\}_{i\in\mathbb{N}};\ 0<p_i<1\ \forall i$.

Then, the probability of at least one success equals $$\ 1- \text{P(no successes)}\ = 1- \displaystyle\lim_{n\to\infty}\prod_{i=1}^n (1-p_i);\quad (0<p_i<1\ \forall i)$$

As Peter alluded to in the comments, the limit $\ \displaystyle\lim_{n\to\infty}\prod_{i=1}^n (1-p_i)\ $ converges to a non-zero number if $\ p_i\ $ are fast decaying/converging to zero. The only number $\ \displaystyle\lim_{n\to\infty}\prod_{i=1}^n (1-p_i)\ $ cannot be is $\ 1,\ $ because $\ 0<p_i<1\ \forall i.$

Therefore, the probability of at least one success could be any number strictly greater than $\ 0\ $ and $\leq 1.$ It all depends on what the $\ p_i\ $ are.

An easy way to see that the infinite product of numbers between $0$ and $1$ could result in a number $\ \geq0\ $ but $\ <1\ $ is the following. For any $\ u>0,\ \varepsilon>0,$

$$ u-\varepsilon = u\left( 1-\frac{\varepsilon}{u} \right)$$

In other words, taking away by a small amount $\ \varepsilon\ $ is the same as multiplying by a number just less than $\ 1\ $ (there always exists such a number and this number is $\ 1-\frac{\varepsilon}{u}).$ So our infinite product is really just some decreasing sequence.

Answer 2

We need more info to answer whether your guess it's true or not. Even if we assume that $p_i$ is always greater than zero, that probability doesn't necessarily tend to 1. As others have pointed out, the probability of never having a success can be written as:

$$P(0) = \prod_i(1-p_i)$$

If we define $q_i = 1 - p_i$ and take the $\log$ of the above, we have:

$$\log (P(0)) = \sum_i \log(q_i)$$

So, it's just a matter of how the above series behaves: if it diverges, taking the $\log$ means you have a zero probability of never getting a success; but if it converges to some value, that's not the case.

If, on the other hand, your $p_i$ are iid from some stationary distribution having a greater than zero expected value, the sum above will diverge and the probability of having a success will tend to 1.

Answer 3

This seems to be false (unless there are more conditions), like some comments have pointed out. As a counterexample, take $0 < e^{-1/n^2} < 1$ the probability of failure for all natural $n$. The probability of not seeing one success is the probability of seeing all failures, in this case it's just the product of $e^{-1/n^2}$, the limit is just $e^{-\pi^2/6} \approx 0.19$, which is not $0$, as you claim it would be.

Answer 4

[Hint] Think about the complement of the outcome you desire, if you define $p$ to be the probability of success, then the probability of failure for $n$ trials is given by by $(1-p)^n$, since $0<1-p<1$ so as $n$ tends to $\infty$,$(1-p)^n$ tends to $0$, so...