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...
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.
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.