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patriota comments on XKCD - Frequentist vs. Bayesians - Less Wrong Discussion
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The p-value for this problem is not 1/36. Notice that, we have the following two hypotheses, namely
H0: The Sun didn't explode, H1: The Sun exploded.
Then,
p-value = P("the machine returns yes", when the Sun didn't explode).
Now, note that the event
"the machine returns yes"
is equivalent to
"the neutrino detector measures the Sun exploding AND tells the true result" OR "the neutrino detector does not measure the Sun exploding AND lies to us".
Assuming that the dice throwing is independent of the neutrino detector measurement, we can compute the p-value. First define:
p0 = P("the neutrino detector measures the Sun exploding", when the Sun didn't explode),
then the p-value is
p-value = p035/36 + (1-p0)1/36
=> p-value = (1/36)(35p0 + 1 - p0)
=> p-value = (1/36)(1+34p0).
If p0 = 0, then we are considering that the detector machine will never measure that "the Sun just exploded". The value p0 is obviously incomputable, therefore, a classical statistician that knows how to compute a p-value would never say that the Sun just exploded. By the way, the cartoon is funny.
Best regards, Alexandre Patriota.
(1/36)(1+34p0) is bounded by 1/36, I think a classical statistician would be happy to say that the evidence has a p-value of 1/36 her. Same for any test where H_0 is a composite hypothesis, you just take the supremum.
A bigger problem with your argument is that it is a fully general counter-argument against frequentists ever concluding anything. All data has to be acquired before it can be analysed statistically, all methods of acquiring data have some probability of error (in the real world) and the probability of error is always 'unknowable', at least in the same sense that p0 is in your argument.
You might as well say that a classical statistician would not say the sun had exploded because he would be in a state of total Cartesian doubt about everything.
For this problem, the p-value is bounded by 1/36 from below, that is, p-value > 1/36. The supremum of (1/36)(1+34p0) is 35/36 and the infimum is 1/36. Therefore, I'm not taking the supremum, actually the cartoon took the infimum, when you take the infimum you are assuming the neutrino detector measures without errors and this is a problem. The p-value, for this example, is a number between 1/36 and 35/36.
I did not understand "the big problem" with my argument...