The premise is basically that there is an infinite amount of trials and a trial could either be a success or a failure. In this trial set, is it actually possible for there to be one and only one success?

I was thinking about it, and the only way I thought that it would be possible is if that one success had an infinite number of conditions to be fulfilled to be a success.

But if that success had an infinite amount of conditions to be fulfilled, then wouldn't it never happen?

And is there are way to do express this question, and its answer, mathematically?

  • 6
    The infinite sequence 1, 0, 0, 0, 0, 0, ... has exactly one 1 and all the rest 0's. These could represent coin flips, and there's no reason this result couldn't occur. It's unlikely, but no more unlikely than any other particular sequence. Can you frame a more precise question?
    – user4894
    Commented Jul 25, 2017 at 22:43
  • @user4894 That's great; I think I read the question as asking for help framing the concern more precisely -- would you consider maybe moving this to an answer? You could give some intuition as to the "no more unlikely" piece of it, which I think is probably counter-intuitive
    – Joseph Weissman
    Commented Jul 26, 2017 at 0:39
  • "If the test never succeeded before, succed. Else fail." Will give you exactly 1 success however many times you try.
    – armand
    Commented Apr 20, 2022 at 23:34

6 Answers 6


Honestly, this should probably be moved to Math.SE, but since it's here, I might as well answer it.

If what you are really talking about is sampling a probability distribution, then the answer is "no." If there is a finite probability to achieve success in any given trial, then success will be achieved infinitely often during infinite trials.

On the other hand, if you are willing to consider a Markov Chain (or other random process), then the answer is "yes." A sequential random process (like a Markov Chain) is an ordered series of states. Often, a state is "persistent" meaning that it is returned to infinitely often in a never-ending trial. However, sometimes there are states known as "transient" which are exactly opposite - they are returned to a finite number of times during a never-ending trial despite having finitely measurable transition probabilities (i.e. the likelihoods of going into or out of that state are not infinitesimal).

Constructing a Markov Chain that has a transient state labeled "success" is a trivial matter. However, reducing the number of successes from "finite" to "one" is probably impossible if you want to both play fair (i.e. not just construct the trivial state space) and also not have to rely entirely on luck (i.e. be able to get one success consistently during infinite trials).

  • I was talking about a sampling distribution, so thanks for the succinct answer: "no". I still have a couple of questions though: 1) What is a sequential random process? 2) Are the states in a Markov Chain binary, like the state could be either 1 or 0? 3) How can a state ever be "transient" and only return a finite number of times in a never ending trial? Commented Jul 26, 2017 at 18:21
  • So there's a lot going on here. I'll try to keep this short. 1) It's a bit of an imprecise term and not really the right turn of phrase, but I used to because I thought it would be clear what I meant - namely, a process which produces a list (or sequence) of outputs by way of a collection of probabilistic rules. 2) The states in a Markov Chain are "binary" in only the most trivial sense - the chain can only be in one state at a time. So at any given moment one state will be "on" and all others "off." But the value of each state can be anything you want, whatever you construct it to be.
    – Geoffrey
    Commented Jul 27, 2017 at 5:50
  • @IkechukwuAnude 3) The proof of that is way beyond the scope of what I can do here, but it is a well-known result that can be found in any treatment of Markov Chains. Here is one trivial example: consider a 3-state system that starts in "state 1" (henceforward, S1). S1 has a 50% proability to stay on itself and a 50% probability to transition to S2. In turn, S2 has a 50% chance to go to S3 and a 50% chance to stay on itself. Likewise, S3 is 50% to stay on itself and 50% to go to S2. A cursory examination of this system will reveal that S1 must be transient, while S2 and S3 must be persistent.
    – Geoffrey
    Commented Jul 27, 2017 at 5:58
  • Thank you for the answer. I will be sure to look that up. I've never heard of a markov chain until you brought it up. Commented Aug 3, 2017 at 23:28
  • 2
    You overlook the case where the probability is zero. Probability zero events can happen, they just do so with probability zero.
    – user6559
    Commented Nov 25, 2017 at 15:13

Depends on the probability distribution and the sampling procedure.

If your samples are IID then no, it is impossible. IID is an acronym for "independent and identically distributed". The "independent" part means that the probability of draw for a given sample doesn't depend on the outcome of any other sample. (E.g. rolls of a die or flips of a coin are independent. Picking a card from a deck is not independent, because it depends on what you drew from the deck earlier. Picking a card from the deck with constant replacement and reshuffling after each draw is independent.) "Identical" means the probability distribution from which a sample is drawn is the same for all samples. For example, two separate rolls of a 6-sided die is considered two identically distributed samples. If you roll a 6-sided die and an 8-sided die, these rolls are independent, but not identically distributed.

It's easy to prove that infinite IID samples from a distribution, whose probability of success is p>0, cannot possibly have any number of finite successes. Simply take the limit of the Binomial distribution for any positive integer k, in the limit as the number of samples approach infinity and you'll get zero. (Sorry if this paragraph was hard to read; the first sentence is really the key point, the second is only in case you want to try and prove it yourself, but can be ignored.)

You can get one positive sample out of infinite rolls if the distribution is either not independent or not identically distributed.

If it's not independent, you need only set it up as a Markov chain (which, loosely speaking, is essentially sequential probability distribution where each draw depends on earlier draws in some manner) that's set so that the probability of drawing a "success" doesn't start at zero, but can be irreversibly set to zero at some point in the sequence. (For example, this can easily be achieved by something called a "Hidden Markov Model".)

If the samples are independent, but not identical then the answer is even simpler. Only have one sample draw from a probability distribution for which a success is possible, and all other samples drawn from a distribution from which a success is not possible. Though I suspect that this scheme would violate a certain implicit assumption in your original question.


Use an unfair dice.

For example, supposing six is success, and all the rest are are fails, 'program' a dice to choose at random the throw on which it will display a six, and on all other throws any other number.

This will then display the characteristics you want; such a dice is hard to build and program, admittedly, but one can quite simply simulate such a situation on a computer, it's child's play to do it.

More interestingly, is there a naturally occurring such situation? No, I don't think so, merely because most natural things are actually finite.


As user4894 pointed out, it's trivial to have such a set of trials. One such example would be a set of trials which included a one-time event, perhaps which consumed a resource that is no longer in the universe. Miracles might suffice as well. Such a set of trials would generate a sequence of success and failures like (1, 0, 0, 0, 0, 0,...)

If we constrain the problem to only independent identically distributed random trials, which is a typical assumption in scientific experiments you run into more interesting issues. Such random variables, by definition, must have an associated probability space which has an associated probability measure. This is the mapping from events to the probabilities of those events occurring. From the definition of a measure, we have to map our events (1 or 0) to a number on the "extended real number line," which is all real numbers, + ∞ and – ∞. This will be a conundrum because you're going to need an infinitesimal to describe the probability of event 1, and infinitesimals are not part of the extended real numbers. Basically, the concept of random variables falls apart when you try to explore such a system.

Similarly, you can construct the probability of a given set of events (such as "all infinite set of trials with only one success") by using the usual rules. This would be a way of saying "given a finite possibility of a 1, how unlucky do you need to be to get only a single 1." You can try to explore things like the p-value of such a result, but you quickly run into similar issues where the probability of a set of events is an infinitesimal.

  • I'm not sure the problem is with the concept of random variables -- equating "probability zero" with "can't happen" may be to blame.
    – user6559
    Commented Nov 25, 2017 at 15:17
  • @Hurkyl You're right that the issue is you can't equate "probability zero" with "can't happen" when dealing with an infinite number of trials. What I describe is simply the formal argument for why the mathematics of probability break down in these particular cases.
    – Cort Ammon
    Commented Nov 25, 2017 at 15:37

If the probability is non-zero, and the trial is run infinite times, then the outcome will necessarily happen, by definition. Infinity is a mathematical tool with exactly that purpose. You could run the trial for the age of the universe however, and it potentially never happen.

  • 1
    If the probability is non-zero, and the trial is run infinite times, then the outcome will necessarily happen -- No that is not true. Consider flipping a fair coin infinitely many times, one flip for each positive integer 1, 2, 3, 4, 5, 6, 7, ... Why couldn't they all be heads? It's unlikely, but so is every other specific sequence of flips. I believe I made this same point in another comment above.
    – user4894
    Commented Feb 23, 2018 at 0:48
  • Infinitely unlikely is different to finitely likely. It's like how 0.999.. recurring is provably equal to 1 en.m.wikipedia.org/wiki/0.999...
    – CriglCragl
    Commented Feb 23, 2018 at 7:41

Its completely possible if the infinite trials are not random

This is a Math SE question, but it's perfectly possible, and very easy to express mathematically

X = 2

You could try an infinite amount of numbers and only 1 will be true, just don't try 2 again.

If what you meant was

I want to create a set of rules such that an infinite number of randomized, independent trials will result in exactly 1 yes, then the answer is no

In this example, you may evaluate 2 zero times. Or 30 times.

You could encode that each number is evaluated exactly once, but now it's not random. You could also simply stop when you get the first yes, but then it's not independent.

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