the sleeping Brad DeLong problem

Brad DeLong was sound asleep for several months, but after he woke up he found an old blog post and quickly calculated the probability that Lubos is an April fool; he wrote a blog post about it, because he had no memory of all the earlier debates about this.
Of course, Lubos disagreed with this calculation, the probabilities are very different according to his argument. So who is right?

I wrote about the Sleeping B. problem almost ten years ago, when it was mostly discussed in academic papers. Meanwhile it has become an internet standard and turned into a political debate: Lubos and the reactionaries vs. Sean and the liberals. Unfortunately, all the nuances of the problem have been lost, as it happens in tribal conflicts, such as the existence of solutions which are neither 1/2 nor 1/3.

In one of the many comments somebody wrote something like "if I would be really interested in this problem, I would code a simulation and see what it does". Unfortunately, if she would actually sit down to implement a simulation, she would quickly find that it does not really answer anything; this is one of the problems which cannot be settled with an experiment or simulation, because it is about the question what we actually mean with "probability". Both physicists and economists are not well trained at thinking through what it is exactly they are talking about, therefore I predict that this debate will not go away anytime soon.

So what is really the issue with this? Well, if the "a priori probability" is 1/2 then S.B. has to bet "as if" it was 1/3 due to the setup of the problem. So if you think probabilities are defined as betting odds (as many Bayesians do) then you will prefer 1/3. If you think probabilities are objective properties of physical systems then you will probably (!) prefer 1/2. (Btw, notice that S.B. has to (implicitly) use the 1/2 "a priori" to calculate the 1/3.)

I actually prefer the 3/8 solution, because it is not so clear how one should understand it. But I realized that I am a tiny minority of one a long time ago ...


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Perhaps it is useful to repeat the 3/8 solution here:
If the outcome was Head, she will only wake up on Tuesday; but if it was Tail, she will either wake up Tuesday or Wednesday with "a priori probability" 1/2 for each, since she cannot distinguish the two days.
Therefore, the probability for the just awoken S.B. that today is Tuesday is
p(Tue) = (1/2) + (1/2)(1/2) = 3/4
where the 1st term corresponds to Head and the 2nd to Tail.

Now we can calculate the probability for Head as p(Head) = (1/2)*p(Tue) + 0*p(Wed) = (1/2)*(3/4) = 3/8,
knowing that the "a priori probability" for Head on Tuesday is 1/2.