Fair die or loaded die? A Bayesian posterior

Question

A box holds two dice: a fair die with P(6) = 1/6 and a loaded die with P(6) = 1/2. You pick one at random (equal chance) and roll it twice, getting a 6 both times. Using Bayes' rule, what is the probability that you picked the loaded die?

Accepted Answer

Intuition first. The loaded die produces sixes much more readily than the fair one, so two sixes in a row point toward the loaded die. Bayes' rule makes this precise by weighing how likely the data is under each die, starting from the even 50/50 prior. The recipe. Multiply each hypothesis' prior by the chance it produces the data, then normalize. Chance of two sixes under each die: P(66\mid	ext{fair}) = \left(	frac{1}{6}ight)^2 = 	frac{1}{36}, \qquad P(66\mid	ext{loaded}) = \left(	frac{1}{2}ight)^2 = 	frac{1}{4} = 	frac{9}{36}. With equal priors the 0.5 cancels, so compare the like…

Fair die or loaded die? A Bayesian posterior

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