The Monty Hall Three Door Puzzle I was going through the Monty Hall Three Door Puzzle in "Discrete Mathematics and its Application" by Kenneth Rosen (5th Edition). While reading the excerpt from tbe book (given below) I could not quite convince myself about the solution as there was only verbal reasoning and not quite vigorous mathematics involved. So I thought of two methods:
(i) Using classical probability

(ii) Using conditional probability

But I do not know whether I am correct, I may be at fault so, it shall be appreciable if my fault is pointed out.

The Monty Hall Three-Door Puzzle: Suppose you are a game show contestant. You have a chance to win a large prize. You are asked to select one of three doors to open; the large prize is behind one of the three doors and the other two doors are losers. Once you select a door, the game show host, who knows what is behind each door, does the following. First, whether or not you selected the winning door, he opens one of the other two doors that he knows is a losing door (selecting at random if both are losing doors). Then he asks you whether you would like to switch doors. Which strategy should you use? Should you change doors or keep your original selection, or does it not matter? 
Solution: The probability you select the correct door (before the host opens a door and asks you whether you want to change) is $\frac{1}{3}$, because the three doors are equally likely to be the correct door. The probability this is the correct door does not change once the game show host opens one of the other doors, because he will always open a door that the prize is not behind. 
The probability that you selected incorrectly is the probability the prize is behind one of the two doors you did not select. Consequently, the probability that you selected incorrectly is $\frac{2}{3}$. If you selected incorrectly, when the game show host opens a door to show you that the prize is not behind it, the prize is behind the other door. You will always win if your initial choice was incorrect and you change doors. So, by changing doors, the probability you win is $\frac{2}{3}$. In other words, you should always change doors when given the chance to do so by the game show host. This doubles the probability that you will win.

Now after the first read I had the question in my mind: "Well the total probability of winning is $1$ as it is $\frac{1}{3}$ in the first case and $\frac{2}{3}$ in the second case, so is it a sure event?"
Of course not, this is so because in the first case and in the second case our strategies are different, which means the experiment is different and so is the sample space. There are not the either,or cases of a single experiment, rather they are different experiments.
This being said let us jump into the classical probability approach , the sample space( possible out-comes) and then the outcomes favorable.

Case $1$: The situation where we stick to our initial decision.
Suppose (without the loss of generality) we have a large prize behind door $A$ and false behind doors $B$ and $C$ (the three doors being named so). Now our possible outcomes are:
1) we choose $A$ and stick to decision and win.
2) we choose $B$ and stick to decision and lose.
3) we choose $C$ and stick to decision and lose.
Now we have to find the probability of winning. The outcome favorable for winning is the option 1) (namely we choose $A$ and stick to decision and win.). So there are a total of $3$ outcomes and $1$ of them is favorable for the event of winning.
By classical probability we have,
   P(winning, sticking to choice) = $\frac{1}{3}$
Case $2$: The situation where we always switch from our initial decision.
Suppose (without the loss of generality) we have a large prize behind door $A$ and false behind doors $B$ and $C$ (the three doors being named so). Now our possible outcomes are:
1) we choose $A$ (Monty opens either of $B$ or $C$ randomly) and switch doors (to $C$ (or $B$) if $B$ (or $C$) is opened by Monty (respectively)) and loose.
2) we choose $B$ (Monty opens $C$ ,can't open $A$ as it has prize) and switch doors (to $A$) and win.
3) we choose $C$ (Monty opens $B$ ,can't open $A$ as it has prize) and switch doors (to $A$) and win.
Now we have to find the probability of winning. The outcome favorable for winning is the options 2) and 3). So there are a total of $3$ outcomes and $2$ of them are favorable for the event of winning.
By classical probability we have,
   P(winning, switching choice) = $\frac{2}{3}$

Now moving on the conditional probability method:

Case $1$: The situation where we stick to our initial decision.
Let $E_1$ be the event of choosing the door containing the prize.
So $P(E_1)= \frac{1}{3}$ as one door out of 3 has the prize.
Let $E_2$ be the event of opening a door containing false by Monty.
Given that $E_1$ has occurred the probability that $E_2$ occurs is:
$P(\frac{E_2}{E_1})= \frac{2}{2} = 1$, There are 2 doors (possible outcomes) containing false and both are favorable out comes.
Let $E_3$ be the event of answering the question "Do you want to change the door?" by me.
$P(\frac{E_3}{E_1 . E_2})= \frac{1}{1} = 1$,since our strategy is fixed we have only one possible out come , i.e. $NO$ and answer so, a sure event.
Required probability of winning 
= $P(E_1)\times P(\frac{E_2}{E_1})\times P(\frac{E_3}{E_1 . E_2})$
= $\frac{1}{3}$
Case $2$: The situation where we switch initial decision.
Let $F_1$ be the event of choosing the door not containing the prize.
So $P(F_1)= \frac{2}{3}$ as one door out of 3 has the prize.
Let $F_2$ be the event of opening a door containing false by Monty.
Given that $F_1$ has occurred the probability that $F_2$ occurs is:
$P(\frac{F_2}{F_1})= \frac{1}{1} = 1$, There is only 1 door (possible outcome) containing false and it is chosen as the favorable out come.
Let $F_3$ be the event of answering the question "Do you want to change the door?" by me.
$P(\frac{F_3}{F_1 . F_2})= \frac{1}{1} = 1$,since our strategy is fixed we have only one possible out come , i.e. $YES$ and we answer so, a sure event.
Required probability of winning 
= $P(F_1)\times P(\frac{F_2}{F_1})\times P(\frac{F_3}{F_1 . F_2})$
= $\frac{2}{3}$

 A: Your reasoning is correct in both cases. What you call a "conditional probability" tracks closer to looking at each event in the game tree separately and multiplying through, the "classical probability" actually uses symmetry.
The "total probability of winning" being $1$ in your first question actually has an interpretation: notice that you must choose exactly one of the two strategies, and regardless of the door Monty opens (which never contains the big prize), exactly one of the strategies must win. Therefore as mutually exclusive and exhaustive events, their probabilities sum to $1$.
A: I've long thought that a good way to understand the Monty Hall problem is to realize that Monty opening a door and then letting you switch is equivalent to him simply giving you the option to take what's behind both of the doors you didn't initially pick. In that interpretation, the strategy of accepting his offer means that your intial selection was simply guessing a door that did not have the prize, which clearly has probability $2/3$ of success.
A: 
Now after the first read I had the question in my mind: "Well the total probability 1 as it is 1/3 in the first case and 2/3 in the second case, so is it a sure event?

No.  1/3 is the probability that the prise is behind your chosen door so you win when you stay while 2/3 is the probability that the prise is not there so you win when you switch.  You must make the choice to stay or switch.
