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My friend flipped (past tense) a fair coin and does not tell me the result. From a frequentist perspective, what is the probability of heads for the flip?

The strict frequentist says, "The flip already occurred. There is no uncertainty to measure. Therefore, probability for the flip does not apply. The coin is either heads or tails."

Under what statistical paradigm could I say, "Even though the flip occurred, I do not know the result of the flip. Hence, until I know the result, there is a multiverse of outcomes. Based on the long-run frequency of those outcomes, about half would be heads. Hence the probability of heads for this particular flip is 0.5."

It appears pedantic that events that occurred (past tense) cannot be described in probability terms. Imagine if my someone asked me, "I flipped a fair coin 100 times last night, what's the probability that there were exactly 48 heads?". And then I say something snarky like, "The flips already occurred. You either had 48 heads or not. There's no uncertainty to measure."

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    $\begingroup$ The first two snarky comments are correct. The final one--"there's no uncertainty to measure"--is frankly false. If it were true, you could be absolutely confident in the outcome. But all this is irrelevant to statistical reasoning. We adopt probability models to make principled arguments from various quantitative assumptions and then compare the conclusions of those arguments to reality. This logic applies even (and especially!) to completed events like a coin toss, no matter what your philosophical stance might be. $\endgroup$
    – whuber ♦
    Aug 16, 2022 at 13:19
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    $\begingroup$ I wouldn't provide a snarky comment -- they tend not to be constructive. But 7.35% is a valid and useful answer to that question. For instance, it helps us gauge how surprised we should be at your statement. If you were to replace "48" by, say, "20," then anyone would be extremely confident that there is something inconsistent about your description of these circumstances. Maybe the coin flipping procedure wasn't fair; maybe you mis-counted; maybe you are reporting only one result out of a billion (computer) trials; maybe you are lying; etc. $\endgroup$
    – whuber ♦
    Aug 16, 2022 at 19:10
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    $\begingroup$ Such an interpretation misunderstands confidence intervals. Once a coin has been flipped, the randomness has "been removed," too. The whole point is that the analysis is based on the process of flipping the coin rather than the outcome. Therein lies the value of probabilistic reasoning. $\endgroup$
    – whuber ♦
    Aug 16, 2022 at 21:24
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    $\begingroup$ The value of math variables doesn't change over time. You can formulate your story in the past tense, the present tense or the future tense; the math will remain the same. Just be consistent in your notations. Call H the event "the flip result is head". Call F the event "your friend observes head". Then P(H) = 0.5, P(H | F) = 1, and P(H | not F) = 0. Your statement "Therefore, probability for the flip does not apply. The coin is either heads or tails." is correct in the sense that P(H | F) and P(H | not F) cannot have values other than 0 or 1. $\endgroup$
    – Stef
    Sep 14 at 16:52
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    $\begingroup$ If time plays an important part in a probability problem, we typically introduce a sequence of random variables ${(X_n)}_{n \in \mathbb{N}}$, and we are careful not to confuse $P(X_n = x)$ and $P(X_n = x | X_0, X_1, ..., X_{n-1})$. $\endgroup$
    – Stef
    Sep 14 at 16:56

2 Answers 2

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All data is in the past. Even with a live feed, information does not pass instantaneously. All data is historical data. The amount of time passing between now and the collection is irrelevant, with the possible exception of the Old Evidence problem in Bayesian Epistemology. It is not relevant here.

It is important to make a couple of distinctions as well. Frequentist methods distinguish data and parameters. In this story, you are making a probability statement about data. That is a very Frequentist thing to do.

However, it appears that your problem is not well posed. You seem to be conditioning on it being a fair coin. This is a classical statistic problem if it is a fair coin. What produces the difficulty is that we know about the coin and not the system.

Your friend, Fast Eddy, thrice convicted of bunko-related crimes, has a reputation of being able to toss a coin either heads or tails one thousand times in a row. Your other friend, a four-year-old named Aloysius, loves to toss the coin in the air and catch it. Which friend is tossing the coin?

Let us change your question. A coin was tossed five hours ago on Pluto. It is currently five and a half light hours away. It is a historical event as it has clearly happened. If it is a fair coin, what is the probability that you will observe it landing on heads? You will see the coin toss in half an hour. Does it matter that it is a historical event?

A Frequentist conditioning on a parameter value is really a classical statistician.

Now let us consider the Bayesian statistician. If the person tossing the coin does not impact the answer, I will place infinite mass on .5 and zero mass elsewhere. My Bayesian predictive distribution will be 50-50. That is trivial, though, as the Bayesian statistician is now a classical statistician.

As to your final example, the 48 heads depends on whether the tosser observed the results. If they saw them, then the answer is either 0 or 1. If they tossed the coin over a 1000 meter tall bridge with fog to prevent any viewing, then there is the same probability statement issue as above.

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  • $\begingroup$ Take the frequentist confidence interval: I collect an infinite a number of samples (each of size n from the same population) and I calculate an infinite number of 95% confidence intervals. Most of the intervals would contain the true value. But, in reality, I only have one sample of size n and only one 95% confidence interval. Under the frequentist paradigm, I cannot say that my interval has a 95% chance of containing the true value. This appears counter-intuitive to me. The probability should not depend on the order of events. $\endgroup$
    – William Chiu
    Aug 16, 2022 at 4:41
  • $\begingroup$ In the case of one sample and one confidence interval, no one knows whether the interval contains the true value. So I think it’s legitimate to say (under the frequentist definition of probability) that my interval has a 95% chance of capturing the true value. $\endgroup$
    – William Chiu
    Aug 16, 2022 at 4:50
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    $\begingroup$ In the confidence interval method, you first need to assume a (admittedly idealised) statistical model of the data generating process . Once a random sample has been selected then there is no randomness left, irrespective of whether the sample result has been seen or not seen. The genius of Neyman was to use the word confidence rather than probability to indicate this reality. $\endgroup$
    – Graham Bornholt
    Aug 16, 2022 at 6:01
  • $\begingroup$ I am sure someone on this site will provide links to past questions in which this issue is discussed in great detail. $\endgroup$
    – Graham Bornholt
    Aug 16, 2022 at 6:03
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As a frequentist, once the coin has been flipped but with the outcome still unknown, I could state "we are 50% confident that the outcome is heads" (the "we" here being the communal we). Post-sample confidence is based on the pre-sample probability. However, note that confidence levels are not the same as probabilities, and cannot be manipulated in the way probabilities can be (Cox and Hinkley (1974) p. 227). Regarding your explicit question, after the flip, the frequentist can say the probability of the outcome being heads is 0 or 1.

I am not sure that all strict frequentists would say that, once the coin has been flipped, "There is no uncertainty to measure". Instead it is more that frequentists would say that there is no uncertainty worth modelling. Frequentists tend to be interested in modelling the data-generating process itself whereas the Bayesian approach is what you would need if you wish to model your own uncertainty.

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