Solving the Monty Hall Problem
In my previous post, [Introducing the Monty Hall Problem](https://www.dhirubhai.net/pulse/introducing-monty-hall-problem-larry-jones-katcc/), I gave you the background for the Monty Hall problem. I explained contest on the television show for which it is named and introduced this table which "solves" the problem using Bayes' theorem.
As a reminder, Bayes' theorem can be expressed mathematically in terms of a hypothesis and evidence as follows.
But how do we apply this to our problem?
The term, P(H), is the prior probability; that is, the probability that a (particular) hypothesis is true before seeing any data. The term, P(E | H), is the likelihood that I would see the evidence if (or given) that a specific hypothesis is true. the term, P(E), is called by a number of names:
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Let's return to our problem. Before we have any evidence, we have no reason to believe that any of the three doors is more likely to have a prize than any other door. Consequently, the prior probability for each door is 1/3.
One of the two difficult parts of Bayesian inference is determining the likelihood. In our case, our reasoning is as follows
The most difficult part of Bayesian inference is the calculation of P(E). Because of the specifics of our problem, we can calculate this value as the sum of the product of each prior probability and its corresponding likelihood; that is, we multiply each prior times the likelihood that the hypothesis of the prior is true and then sum all the products. This action results in the three values 1/6, 1/3 and 0 which, summed together is 1/2.
Now that we've filled out the hypothesis, prior, likelihood, and product columns of our table, we simply, for each hypothesis, multiply the prior times the corresponding likelihood and divide by the sum of all the likelihoods.
In other words, it always pays to switch. Surprised? You're not alone.