I originally posted the following as a partial answer to a question asking why a 95% confidence interval does not imply that there is a 95% chance that the interval contains the true mean (see: Why does a 95% Confidence Interval (CI) not imply a 95% chance of containing the mean?). A commenter (thanks to John) subsequently asked me to post the comment as a separate question, so here goes.

Firstly, I'm going to assume that if I select a playing card at random from a standard deck, the probability that I've selected a club (without looking at it) is 13 / 52 = 25%.

And secondly, it's been stated many times that a 95% confidence interval should be interpreted in terms of repeating an experiment multiple times and the calculated interval will contain the true mean 95% of the time – I think this was demonstated reasonably convincingly by James Waters simulation in the question linked above. Most people seem to accept this interpretation of a 95% CI.

Now, for the thought experiment. Let's assume that we have a normally distributed variable in a large population - maybe heights of adult males or females. I have a willing and tireless assistant whom I task with performing multiple sampling processes of a given sample size from the population and calculating the sample mean and 95% confidence interval for each sample. My assistant is very keen and manages to measure all possible samples from the population. Then, for each sample, my assistant either records the resulting confidence interval as green (if the CI contains the true mean) or red (if the CI doesn't contain the true mean). Unfortunately, my assistant will not show me the results of his experiments. I need to get some information about the heights of adults in the population but I only have time, resources and patience to do the experiment once. I make a single random sample (of the same sample size used by my assistant) and calculate the confidence interval (using the same equation).

I have no way of seeing my assistant's results. So, what is the probability that the random sample I have selected will yield a green CI (i.e. the interval contains the true mean)?

In my mind, this is the same as the deck of cards situation outlined previously and can be interpreted that there is a 95% probability that the interval calculated using my sample is green (i.e. contains the true mean). And yet, the concensus seems to be that a 95% confidence interval can NOT be interpreted as there being a 95% probability that the interval contains the true mean. Why (and where) does my reasoning in the above thought experiment fall apart?

  • $\begingroup$ The difference would be that in your example, you actually know what the population mean is. Therefore, out of a hundred confidence intervals, you can estimate that 95 of them will be green and 5 will be red. Saying the mean has a 95% probability of being in the CI makes it sound (to me) as if the mean is determined by the CI, whereas in reality the mean is given (although we don't know what it is) and the CI is estimated. The problem is with using the term "probability" on something that is deterministic, but unobserved, as opposed to truly random. $\endgroup$
    – Marie. P.
    Commented Sep 5, 2017 at 13:04
  • $\begingroup$ @Marie.P. Thanks for your comment. It does help to clear the mist a little. However, to continue with the somewhat contrived example, my assistant might be able to calculate the population mean but I do not have access to that information; I only have information gathered from a single sample. I tried make sure that I considered the probability that the CI (calculated using my sample data) contains the mean rather than the probability that the mean is in a given CI (as you stated), which I think of as being slightly different assertions. But the last sentence in your comment does help. $\endgroup$ Commented Sep 5, 2017 at 16:21

3 Answers 3


The confusion comes from this sentence:

And yet, the consensus seems to be that a 95% confidence interval can NOT be interpreted as there being a 95% probability that the interval contains the true mean.

It is a partial misunderstanding of the real consensus. The confusion comes from not being specific about what probability we talk about. Not as a philosophical question but as "what exact probability we are speaking of in the context". As @ratsalad says it's all about conditioning.

Call $\theta$ your parameter, $X$ your data, $I$ an interval that is a function of $X$:

  • $I$ is a confidence interval means $P(\theta\in I\mid\theta)>0.95$ for all possible $\theta$ including the true one. Probability averages over all possible $X$ at fixed $\theta$. This is what you explain in your interpretation.
  • $I$ being a (Bayesian) credible interval says $P(\theta\in I\mid X)>0.95$. Probability averages over all possible $\theta$ at fixed $X$.

Both are probability of the same event but conditioned differently.

The reason why one discourages saying "the probability that $\theta$ is in $I$ is 0.95" for confidence intervals is because this sentence implicitly means the second point: when we say "the probability that..." the conditioning is implicitly to what has been observed before: "I have seen some $X$, now what is the probability that $\theta$ is..." is formally "what is $P(\theta...\mid X)$".

This implicit is reinforced by the (again implicit) suggestion you experience when reading "probability that $\theta$ is in $I$" that $\theta$ is the variable and $I$ the fixed object, while in frequentist analysis it is the opposite.

Finally this is made even worse when you replace $I$ by your calculated interval. If you write: "The probability that $\theta$ is in $[4;5]$ is 0.95" then this is simply false. In frequentist analysis "$\theta$ is in $[4;5]$" is either true or false but is not a random event thus it does not have a probability (other than 0 or 1). Thus the sentence could only be meaningfully interpreted as the Bayesian one.

  • 1
    $\begingroup$ 1.) Shouldn't it be P(..) = 0.95 instead of P(..) > 0.95? 2.) Does your last paragraph not contradict your formal description? Should the frequentist CI not be P(\theta \in I | \theta) \in \{0,1}? You yourself say " In frequentist analysis "θ is in [4;5]" is either true or false", but your formal description says it is "> 0.95"? $\endgroup$
    – HOSS_JFL
    Commented Apr 29, 2020 at 15:49

Part of the difference comes down to conditioning, the difference between pre-data probabilities and post-data probabilities. Before you do your single experiment (before you obtain your sample), you know that there is a 95% chance that the 95% CI will contain the true mean (this is the definition of a 95% CI). However, after you obtain your sample, you are in a different state of knowledge: you have not learned the true mean, but you have seen a particular sample of data, which may give you some new knowledge and which can affect your probability calculations.

Analogously, before you draw a card, you know that there is a 25% chance that the card will be a club. Now to make the analogy work, you cannot learn the true suit of the card when you draw it (because likewise the true mean is always hidden from you). But you may learn something new from drawing the card, for instance the color of the suit.

Let's say that you draw the card, and through some mechanism (it doesn't matter for the point), you learn that the card is from a black suit. This changes your probability: from prior information, you know that clubs are black, and that half the cards are from black suits, so now you know that the card has a 50% chance of being a club. If, on the other hand, you discovered a red card, from your prior information you know that clubs are not red, so you would now know that there is a 0% chance of your card being a club. Both these probabilities are consistent with a 25% chance of a club before drawing the card.

If you were to ignore your prior information, or if you were not told that the card was black, you would still have a 25% chance of being correct. However, you can do better if you take advantage of your prior information.

There are many examples of this with real CIs, where seeing the data gives a coverage probability that is different from the CI %. This classic example (halfway down the post) of a "misleading" CI from David McKay may help. A similar example is given by Berger.

To continue with your example of heights of people: lets say that you know that your population under study is from the Netherlands, which has the tallest average height of any country in the world (about $1.84 \pm 0.02$ m). However, lets say your sample has a 95% CI of $1.7 \pm 0.02$ m. Do you still think there is a 95% probability that the true population mean lies in that interval? I would say that, based on the prior knowledge, your specific sample was a stochastic fluke and anomalously low. In other words, the probability is much less than 95% that the true mean lies in your calculated CI.

Note, before you obtained your sample, and calculated your specific CI, your chance of obtaining a CI that encompassed the true mean was 95%. Afterwards, if you use no prior information, and assume that all mean heights are equally probable a priori, then you could, if you wanted, make a Bayesian statement that there is 95% probability that your interval contains the true mean. But realize that such a statement does not follow from the definition of a CI, and that it crucially depends on a particular assumed prior for the mean. It also depends on your normality assumption, as most frequentist CIs cannot be re-interpreted in a Bayesian manner so easily.

  • $\begingroup$ Thanks for answering, that does help to clear things in my mind. I think. $\endgroup$ Commented Mar 30, 2018 at 18:35

Your question is more philosophy than statistics. It has been discussed ad nauseam in the form of a cat in a box.


I will add, regarding

95% confidence interval should be interpreted in terms of repeating an experiment multiple times and the calculated interval will contain the true mean 95% of the time

This is one interpretation. You could also say that, before you create the interval, there is a 95% chance that the process will result in an interval that captures the true mean.


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