How likely is 1 in 10000?

If an event has a probability of 1:10,000, therefore in 100,000 trials it would then be likely to occur 10 times; in 1,000,000 trials, it would be likely to occur 100 times, but would it not be also just as likely that it occur in any given set of 1,000,000 trials any number of times, for example: 98 times, 99 times, ...

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a 1 in 10000 probability, what is the likelihood probability that in 10000 trials it will occur exactly 1 time $1/eapprox 0.3679$, as near as makes no odds. (The probability that it happens exactly 0 times is almost exactly the same.) Edit: As Mark L Stone quite rightly points out, I've taken your question as implying the trials are independent without establishing that it's the case. This is a critical assumption (and may not be reasonable in many situations). Nevertheless I'll continue answering on that basis, because I continue to think that it was your intent. The same is true for $n$ trials and a probability of $1/n$, for any sufficiently large $n$. The probabilities (for any large $n$) look much like this (showing the case for $n$=10000): If an event has a probability of 1:10,000, therefore in 100,000 trials it would then be likely to occur 10 times; in 1,000,000 trials, it would be likely to occur 100 times, but would it not be also just as likely that it occur in any given set of 1,000,000 trials any number of times, for example: 98 times, 99 times, 101 times, 96 times, 102 times, etc. Not quite: 99 and 100 have the same chance, but everything else has a lower chance:

(the probability continues to go down as you move further out).

Specifically, you're dealing with a binomial distribution with $n=1000000$ and $p=1/10000$.

Since $n$ is large and $p$ is small, it's well approximated by a Poisson distribution with mean $lambda=np=100$. how many trials must be averaged and accounted for to approach a statistical certainty that a particular result is actually 1:10000, and not 1:9999 or 1:10001 You can't be certain it's actually 1/10000, since you can be arbitrarily close to it but different from it. In $n$ trials, the expected number of successes is $np$ with sd $sqrt{np(1-p)}approx sqrt{np}$. If $p=1/10000$, and $n=10^{12}$, then the expected number of successes is $10^{8}$ with sd $10^{4}$; if $p=1/9999$ the expected number of successes would be $100,010,000$ ... about one standard deviation away -- not enough to tell them apart "reliably". But with $n=4 imes 10^{12}$, you're about $2$sd's away, and you can tell them apart more easily; that's probably about as low as most people would want to go. At $n=10^{13}$ you could tell them apart quite well (the chances of 1/10000 looking like 1/9999 or 1/10001 or anything further away by chance are pretty small by that point). Say you were happy with $10^{13}$ trials for distinguishing $p=1/10000$ from $1/9999$. If you wanted to rule out 1/9999.5 at the same confidence as you had for ruling out 1/9999, you'd need 4 times as many trials. You can see that pinning down proportions to many figures of accuracy (when $p$ is very small) requires a lot of trials; you need a sample size several times more than $(1/p)^3$ to get the estimate accurate enough that you can rule out $p=1/(kpm 1)$ when it's really $1/k$. let's say after 10,000,000,000 trials the result occured 999,982 times, would you then state the probability for the next trial to be 1:9999.82 or 1:10000 or some calculated result involving the deviation? ..(Or I guess the same could be asked after only 1 set of 10,000 trials with much less accuracy!)

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Yes, it could be asked at 10000 trials or 1000 or 100.

Let's simplify things and take 10000 trials and 98 successes. One could of course take as a point estimate of the probability of a success 98/10000 = 0.0098 but this won't actually be the underlying proportion, only an estimate of it. It might well be 0.944... or 0.997... or any number of other values. So one thing people do is construct an interval of values that would be (in some sense) reasonably consistent with the observed proportion. There are two main philosophies of statistics (Bayesian and frequentist statistics) that in large samples would usually tend to generate similar intervals but which have rather different interpretations. The most common would be a (frequentist) confidence interval; an interval for the parameter ($p$) that would (over many repetitions of the same experiment) be expected include the parameter a given proportion of the time. A typical Bayesian interval would start with a prior distribution on the parameter representing your uncertainty about its value, and use the data to update that knowledge of it to a posterior distribution and from it obtain a credible interval. Confidence intervals are very widely used (though a credible interval may come closer to your expectations about what an interval should do). In the case of binomial proportion confidence interval, as here, there are a variety of approaches, though in large samples they all give you pretty much the same interval. with dice even 6 x 10^9 trials may not result in exactly 1 x 10^9 for each of six results Correct; you would expect (with fair dice) to get between 999.94 million and 1000.06 million success almost (but not quite) every time you tried it. If actual probability is 1:10000, then increasing trials within the expected deviation would tend to confirm that It will nearly always continue to be consistent with it (and with a range of other nearby values). What happens is not that you can tell it's 1/10000, but that the interval of probability values consistent with your results will get narrower as the sample size grows.

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