The easiest way to convince your friend of his error is to imagine the odds when there are n doors, and Monty opens n-2 of them, the ones which don't contain the prize and aren't your first pick.
How could Bayesian statistics be applied in this case? I’m wondering if the situation is just too “simple” to make it applicable.
Before Monty opens a door, so the prior probability, is 1/3 for each door.
After he opens a door, that door will have zero probability, as we know he opens a door without a car, then how to update the probabilities afterwards?
Seems the simple way to look at it is to not partition by door, but by chosen vs not chosen. Chosen is 1/3 and not chosen is 2/3 before and after Monty opens a door, so perhaps there is no “Bayesian information” revealed by opening the door anyway.
> Seems the simple way to look at it is to not partition by door, but by chosen vs not chosen.
Right. That might be the easiest way for this problem.
More straightforwardly, without that shortcut:
- Call the doors a, b, and c. Assume, without loss of generality, that we choose door a initially. Let the random variable X ∈ {a, b, c} be the door with the car.
- Our prior probability is uniformly distributed: Pr(X = a) = Pr(X = b) = Pr(X = c) = 1/3.
- The data Y that we collect is our observation of which door gets opened by the host. The likelihood function Pr(Y = y | X = x) is the probability of the observation being y (i.e., that Y = y), given that the underlying state is x (i.e., that X = x). The only non-zero likelihoods are Pr(Y = b | X = a) = Pr(Y = c | X = a) = 1/2 and Pr(Y = c | X = b) = Pr(Y = b | X = c) = 1.
- Bayes' theorem, Pr(X = x | Y = y) = Pr(Y = y | X = x)·Pr(X = x)/Pr(Y = y), gives the answer, the posterior probability, which should be seen as a function of x. The denominator Pr(Y = y) = ∑ Pr(Y = y | X = x), sum over x ∈ {a, b, c}, is a normalising factor that makes the posterior probability distribution sum to 1. It is also the probability we, at the start of the game, assign to Y = y. In our problem, Pr(Y = a) = 0 and Pr(Y = b) = Pr(Y = c) = 1/2.
How about you put in the numbers and see if it comes out right or if I have made a mistake? :-)
Edit: Sorry, I just realised that we could have made it simpler by assuming that the host opens, say, door b.
