Is something wrong with the following reasoning? Mostly I wonder how could one derive uniformly random arrival from ignorance. But even if that derivation is invalid generally, it seems reasonable here. Could someone please explain?

As a young man Mr Gott visits Berlin in 1969. He’s surprised that he cannot cross into East Berlin since there is a wall separating the two halves of the city. He’s told that the wall was erected 8 years previously. He reasons that: The wall will have a finite lifespan; his ignorance means that he arrives uniformly at random at some time in the lifespan of the wall. Since only 5% of the time one would arrive in the first or last 2.5% of the lifespan of the wall he asserts that with 95% confidence the wall will survive between 8/0.975 ≈ 8.2 and 8/0.025 = 320 years. In 1989 the now Professor Gott is pleased to find that his prediction was correct and promotes his prediction method in prestigious journals. This ‘delta-t’ method is widely adopted and used to form predictions in a range of scenarios about which researchers are ‘totally ignorant’.

  • $\begingroup$ Brainstorming how things could go wrong: If the lifespan of the wall were a random variable and the wall had a constant risk of being torn down, then you could argue it's much more likely to observe the wall while the wall is young... $\endgroup$ Dec 16, 2015 at 12:44
  • $\begingroup$ There is a good Bayesian vs Frequentist discussion to this problem at: youtube.com/… $\endgroup$
    – VRaina
    Jul 28, 2020 at 20:41
  • $\begingroup$ What do you mean by ignorance? Mr Gott is using quite some information. He has made an observation of the age of the wall and he is using the assumptions that the wall has finite lifespan and that the probability to observe a wall at a certain time during it's the lifespan is uniform. $\endgroup$ Jul 28, 2020 at 21:31

2 Answers 2


The problem you are describing is related to doomsday argument and sunrise problem.

Estimating something from ignorance in this case relates to Bayesian estimation with uniform prior. In basically any inference problem you have some data $D$ and some parameter $\theta$ that you want to learn about using your data. There are multiple different methods that can be applied to such problems and Bayesian approach is one of them. The general idea is that you can use your prior knowledge about $\theta$, data and Bayes theorem to learn something about $\theta$ (i.e. posterior):

$$ \underbrace{P(\theta|D)}_\text{posterior} \propto \underbrace{P(D|\theta)}_\text{likelihood} \times \underbrace{P(\theta)}_\text{prior} $$

The basic idea is that you plug-in your prior into this formula and then check which of your prior expectations are likely given the data you have. Prior is some distribution for $\theta$ that is assumed a priori, that is, before seeing the data. You can make different assumptions about $\theta$ based on your actual problem and your subjective judgment.

One simple choice that can be made is to assume that you have no knowledge whatsoever about $\theta$ just that it lies in some $[a,b]$ interval (so in fact you have some knowledge and make assumptions). In such case you use uniform distribution $\mathcal{U}(a,b)$ for $\theta$ and assume a priori that all the values in this interval are equally likely. Next, you update your assumptions by confronting them with the data. In this case ignorance-prior is used to make assumptions about $\theta$ that are to be tested and verified against the data. This is helpful because it gives you method of finding candidate values of $\theta$. Rephrasing it differently, you know nothing about $\theta$, but still you start with something to plug-in in the place the unknown to learn about it.

Notice that when using uniform priors this approach is coherent with maximum likelihood estimation, but when using non-uniform priors it can lead to different results that are influenced less or more by the prior. Bayesian approach could be also helpful in situations like the one described in your quote, where we do not have much data about problem of interest, where priors help to overcome those limitations by including out-of-data information in our statistical model.

J. Richard Gotts (1993) example is pretty simple and it needs only few assumptions and some basic algebra to understand it. Imagine that you have some point $x$ that lies on the line $[a, b]$, but you do not know exactly where it lies. For a moment let's forget what value does exactly have the beginning of the line $a$ and the end $b$, but let's think of them as $0\%$ and $100\%$ of total length $b-a$. Let's assume that $x$ can lie anywhere on the line, i.e. $X$ is a random variable uniformly distributed over $[a,b]$ interval. Making this assumptions lead us to conclusion that we have $0.95$ probability that $x$ is somewhere in the $95\%$ middle part of the line (recall that for uniform distribution $P(X < x) = \frac{x-a}{b-a}$). So if we are in the $95\%$ middle region of the line, than $x$ is at least $0.025(b-a)$ or at most $0.975(b-a)$.

Gott, J. R. (1993). Implications of the Copernican principle for our future prospects. Nature, 363(6427), 315-319.

  • 1
    $\begingroup$ do you mean $0.025(b-a)$? $\endgroup$
    – Slim Shady
    Oct 3, 2021 at 16:15
  • $\begingroup$ @SlimShady thanks, good catch! $\endgroup$
    – Tim
    Oct 3, 2021 at 16:19
  • $\begingroup$ But isn't Gott's reasoning stupid for this exact case? Since if he came to the wall 1 day after the wall was built, by his reasoning, the estimate would then be $1.02<x<40$ days. So the situation that he was "correct" happened just by pure chance? $\endgroup$
    – Slim Shady
    Oct 3, 2021 at 17:23
  • $\begingroup$ @SlimShady it makes assumptions and makes an educated guess based on the assumptions. As any educated guess, it can be wrong. Notice that if the only thing you know is the age of the wall, you cannot precisely answer how long would it last. You can only guess. The guess would be only as good as the assumptions it makes. I recommend you to try coming up yourself with such guess, how would you approach the problem? $\endgroup$
    – Tim
    Oct 3, 2021 at 17:47

As a addition to Tim answer, I would note that Gelman and Robert wrote in the paper "The perceived absurdity of Bayesian inference" (http://arxiv.org/pdf/1006.5366v2.pdf) in Section 4 called "The doomsday argument and confusion between frequentist and Bayesian ideas" that the doomsday argument (and Gotts one) is essentially a frequentist and not a bayesian view:

For our purposes here, the (sociologically) interesting thing about this argument is that it’s been presented as Bayesian (see, for example, Dieks) but it isn’t a Bayesian analysis at all! The ”doomsday argument” is actually a classical frequentist confidence interval.

And explain in plain english why. Moreover, they also comment that

The doomsday argument is pretty silly and also, it’s fundamentally not Bayesian.

I am no such an expert to claim these two points but as both Gelman and Robert agree on that, I think it is good to know...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.