[An.American.Expatriate]

Why is there a 2/3 chance his choice is constrained?

If I made the correct (first) choice, he is not constrained,
If I made the wrong (first) choice, he is constrained

From “my” perspective there is a 1 in 2 chance, AFTER my selection, that Monty was “forced” to open a specific door. BEFORE my selection there is a 2 in 3 chance of “forcing” Monty - which is simply the inverse of my probablity of having chosen correctly.

[Rebel_Ace]

"...Why is there a 2/3 chance his choice is constrained?

If I made the correct (first) choice, he is not constrained, If I made the wrong (first) choice, he is constrained

From “my” perspective there is a 1 in 2 chance, AFTER my selection, that Monty was “forced” to open a specific door. BEFORE my selection there is a 2 in 3 chance of “forcing” Monty - which is simply the inverse of my probablity of having chosen correctly..."

Please allow me to explain:

The PROBABILITY of an event occuring is the fraction of:

(possible desired outcomes)
---------------------------
(all possible outcomes)


Now, we want to know the PROBABILITY that Monty's choice is CONSTRAINED. Given that there are 3 doors, and the prize is behind only one of them, we get this fraction:

(I can pick 2 doors that have NO Prize)
-----------------------------------------
(there are 3 doors total)


that is 2/3.

Now, the way the game is played, Monty will ALWAYS open an EMPTY door. If he opened the WINNING door, the game would not be "fun".

Given this ADDITIONAL bit of information, you are STATISTICALLY BETTER OFF switching your choice. Monty CANNOT open a door at RANDOM, and THAT is the addional information that changes the probability computations.