Originally Posted by

**SoFlaLaw**
I have reviewed this Monty Hall Bayes Theorem problem thousands of times. Although I understand the alleged probabilities solution through the application of Bayes Theorem to is underlying facts, I always come back to wondering how this does not become 50:50 after 1 option is displayed without the prize being behind that initial option.

Let me explain my confusion:

1. We have 3 doors, behind 1 there is a prize, and behind 2 there are booby (no jokes please) prizes or no prizes. So, there exists 3 possible winning venues.

2. We select 1 of the doors (for simplicity purposes, assume I select Door 1), hoping to win the prize, at which time we have a 1 out of 3 chance of success. Remember, there are 3 possible winning choices (Door 1 - 33 1/3% chance; Door 2 - 33 1/3% chance; and Door 3 - 33 1/3% chance).

3. Game show host opens 1 of the 2 doors we did not select (either of Door 2 or Door 3, and will assume for example purposes, it was Door 2), and shows that behind it, there was no prize or a booby prize (be nice). We are offered the opportunity to switch from our choice, Door 1, to the remaining door. Should we switch doors and hope for a doubled chance from success from our 33 1/3% chance with Door 1?

Now, applying Bayes Theorem to this dilemna, as Canceler explained, there was a 66 2/3% chance that the prize was behind 1 of Door 2 or Door 3 (which I will refer to as "Door 2/3") that we did not select, and that does not change, so if given a choice to switch, we should switch from our 1/3 (33 1/3% choice of Door 1) to the 2/3 (66 2/3% choice of Door 3 - keeping Door 2's percentage).

Being a lawyer and (amateur economist - earning my B.S. Degree in Economics) I cannot accept the validity of Bayes Theorem being the solution, for the following reasons:

1. There existed three realities under which I would find the prize when making my initial door selection (i.e. - Door 1, Door 2, and Door 3).

2. Each had attached an equal probability of success (33 1/3% each).

3. When the game show host reveals 1 of either Door 2/3, and it does not have the desired prize behind it, then, the initial reality of the prize being behind it is revealed to have been a false reality.

4. The fact that it was a non-option was exposed, and its impact upon the initial decision is now revealed to include a null set (0%). It is as if it never existed.

Consequently, at that point in time, the new reality (actually the ongoing reality) is that the prize was behind either Door 1 (which I selected) or Door 2/3 (whichever being the one not exposed by the game show host, in this example, Door 2 exposed with no prize, Door 3 remaining). Consequently, Door 1 and Door 3 (the remaining door) now each have a 1:1 (50%) likelihood of having the prize behind them, as it has now been proven that the exposed door (Door 2) never existed as a legitimate option for success.

All available data has to be accounted for, and the fact that one of the doors is removed from the equation, changes the equation such that the removed door never existed for purposes of determining which door the prize is actually behind (both now, and when I made my initial selection). To NOT take this information (the nullification of one of the doors having the prize behind it), would be like not taking a third of a shoe into account when keeping a running and/or true count (forgetting unbalanced card counting systems for the moment).

Please explain why my thought process and analysis is incorrect. I would appreciate any educational value that can be imparted such that I see why I should not consider the fact of the removal of 1 option when determining my odds/chances for success, based upon my now existing options.

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