Quantum Bayesian Networks

November 9, 2012

Did the sun just explode?

Filed under: Uncategorized — rrtucci @ 7:57 pm

From Xkcd comics



  1. Notice that 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 = p0*35/36 + (1-p0)*1/36

    => p-value = (1/36)*(35*p0 + 1 – p0)

    => p-value = (1/36)*(1+34*p0).

    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.

    Alexandre Patriota

    Comment by Alexandre — May 19, 2013 @ 4:06 am

  2. Thanks Alex. I agree, the cartoon calculates a p-value which bears no relation to the sun exploding or not,. I guess the point is that Bayesians think frequentists are utterly confused

    Comment by rrtucci — May 19, 2013 @ 7:54 am

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