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In today’s XKCD, a pair of (presumably) physicists are told by their neutrino detector that the sun has gone nova. Problem is, the machine rolls two dice and if they both come up six it lies, otherwise it tells the truth.

The Frequentist reasons that the probability of obtaining this result if the sun had not, infact, gone nova is 1/36 (0.027, p<0.05) and concludes that the sun is exploding. Goodbye cruel world.

The Bayesian sees things a little differently, and bets the Frequentist $50 that he is wrong.

Let’s set aside the obvious supremacy of the Bayesian’s position due to the fact that were he to turn out to be wrong, come the fast approaching sun-up-to-end-all-sun-ups, he would have very little use for that fifty bucks anyway.

What prior probability would we have to ascribe to the sun succumbing to cataclysmic nuclear explosion on any given night in order to take the Bayesian’s bet?

Not surprisingly, we’ll need to use Bayes!

Where M is the machine reading a solar explosion and S is the event of an actual solar explosion.

Assuming we are risk neutral, and we take any bet with an expected value greater than the cost, we will take the Bayesian’s bet if P(S|M)>0.5. At this cutoff, the expected value is 0.5*0+0.5*100=50 and hence the bet is worth at least the cost of entry.

The rub is that this value depends on our prior. That is, the prior probability that we ascribe to global annihilation by complete solar nuclear fusion. We can set P(S|M)=0.5 and solve for P(S) to get the threshold value for a prior that would make the bet a good one (ie not a Dutch book). This turns out to be:

Which is ~0.0277 — the Frequentist’s p-value!

So, assuming 1:1 payout odds on the bet, we should only take it if we already thought that there was at least a 2.7% chance that the sun would explode, before even asking the neutrino detector. From this, we can also see what odds we would be willing to take on the bet for any level of prior belief about the end of the world.