The Monty Hall Problem

The Monty Hall problem is one that you may be familiar with already (and you probably do not even know it).

Some of you may remember this scene in a recent movie:

The problem has been debated by statisticians for decades.

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My take on the Monty Hall problem:

You are on a game show and the host asks you to pick from three doors (one door has a car behind it). You pick door #1. The host opens door #3 (which uncovers a goat behind it) and offers you the option to change your choice from door #1 to door #2.

At the beginning you had a 33% of guessing the door that hides a car behind it. After the host opened door #3 (and showed you the goat) only two possibilities remain: door #1 and door #2. That’s a 50% chance that either door has a car behind it. When the host asks if you would like to change your choice to door #2, a new game has officially begun.

Bayes’ theorem, as it applies to the Monty Hall problem is still appicable. Accept now, where the denominator is computed using the law of total probability as the marginal probability as seen here:

is inapplicable.

Why? Because the marginal probability cannot be measured yet in this new game (and is unnecessary because the solution will be presented by the host after affirming choosing door #1 or door #2).

…in other words, switching your choice to door #2 will give you the same odds (50%) as keeping your original choice.

In the first game there were three doors. In this new game there are only two doors.

If you consider this game as a new one, then switching your original choice from door #1 to door #2 (as encouraged by Vos Savant) now seems arbitrary.