I was just told in comments that probability p=1/n does not mean that I have 1 occurrence in n experiments in average, for large number of experiments because observed frequency has nothing to do with probability. Am I in frequentist sin? What is the meaning of probability if not frequency?

  • $\begingroup$ Is it forbidden to clarify this issue? $\endgroup$
    – Val
    Aug 22, 2013 at 15:36
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    $\begingroup$ You were told nothing of the sort: you were told that when the probability of observing an occurence in a single trial is $\frac{1}{n}$, the probability of observing zero occurrences in $n$ independent trials tends to $\frac{1}{\mathrm{e}}$ as $n$ gets larger; & that an observed frequency is not the same thing as an underlying probability. $\endgroup$ Aug 22, 2013 at 15:45
  • $\begingroup$ @Scortchi, this looks like the answer $\endgroup$
    – Val
    Aug 22, 2013 at 15:50
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    $\begingroup$ Pedantry corner: Frequency often means a count of occurrences, as in frequency distribution. Sometimes nothing is lost and clarity is gained by stressing that frequency is short-hand for relative frequency whenever that is so. $\endgroup$
    – Nick Cox
    Aug 22, 2013 at 17:00

1 Answer 1


Imagine we have a fair coin. Everyone would agree that the probability of it landing on heads is 0.5, as is the probability of it landing on tails. Suppose we then flip the coin ten times. We might not necessarily see exactly five heads. You shouldn't be particularly surprised to if we observe four or six heads instead and even observing all heads or tails isn't completely out of the realm of possibility, so the observed frequency can differ from the underlying probability. As the number of trials increases, we'd expect the observed frequency to approach the underlying probability, but it's important to keep the difference in mind.

Now, suppose you have an event that occurs with probability $\frac{1}{n}$. For example, you might be rolling a die and counting the number of times it comes up as a 1, which (for a fair die) occurs with probability $\frac{1}{6}$. Suppose you roll this die six times. Although you would expect to see one success in those six trials, you're obviously not guaranteed to see any at all (how would that even work?!).

@Whuber added in an interesting generalization of this. Suppose you have an event that occurs with probability $\frac{1}{n}$. The probability of not seeing this even approachs $\frac{1}{e}$ as $n$ gets larger. Let's go back to the die. It has six sides, one of which is a success. If we roll it six times, then the probability of seeing a number other than 1 each time is $\big(1 - \frac{1}{6}\big)^6\approx0.335$, because there are six outcomes, only one of which succeeds, and six independent trials. If we used a ten sided die instead, and rolled it ten times, then $\big(1 -\frac{1}{10}\big)^{10} \approx 0.349$, and if we used a 1000-sided die, then the probabilities of no successes (reminder: not a single one on any roll) is $\big(1-\frac{1}{1000}\big)^{1000} \approx 0.368$. For reference $\frac{1}{e} \approx 0.3679$

Now, keep in mind that these are probabilities, not observed frequencies! I think people took exception to the implication that rolling the 1000-sided die 1000 times counts as a "large number" of trials; it's not, since that's all part of a single experiment. You'd have to repeat the entire procedure many times for the observed frequencies to converge towards the probabilities.

  • $\begingroup$ Thanks for the clarification. I could not imagine that people can interpret my words so naively, especially after I have hinted that you can do more much more than $n$ experiments → you can test $p=1/n$ it experimentally. $\endgroup$
    – Val
    Aug 22, 2013 at 17:19
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    $\begingroup$ We interpret them as you write them, Val. Consider for instance your first comment: "I am more interested to know why is the probability of having success in n trials is not 1?" Call me naive if you like, but I read this as referring to "$n$ trials" and I cannot find any unambiguous reference to "more than $n$" of them. Successful communication requires both parties--author and reader--to share an understanding. When you find that readers have different interpretations of what you wrote, you should not be so quick to call them "naive," but should instead consider how to clarify your words. $\endgroup$
    – whuber
    Aug 22, 2013 at 20:04

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