May 26 2008
The Movie 21, Variable Change, and Monty Hall
During the movie 21 (read my review here), we are introduced to a seemingly simple example of probability. The scenario is this:
A game show host tells a contestant to choose between three doors. Two doors have a goat behind them…and the other has a brand new car. He tells the contestant to choose a door. The contestant chooses door #1. Then, the host, knowing what is behind each door, reveals what is behind door #3-a goat. He then asks the contestant if he would like to change his choice or stick with door #1.
What would you do?
This mathematical scenario is known as The Monty Hall Problem and is an example of the principal of variable change and conditional probability.
My initial thought would be to stick with door #1 as it was the original hunch and the probability is 50/50. But this is the wrong answer as the probability is not 50/50, which surprised me and has confounded many on the Internet. On the IMDB board for 21, there are more than 10 pages of discussion regarding this very point. Who says that everyone these days is intellectually apathetic?
The right answer? You should change to door #2 because the probability that the car is behind door #2 is greater: 66.6% to 33.3%.
Why is this?
Well, we all know at the beginning that the probability that the car is behind the door that we choose is 33.3% or 1/3. But then, then the host reveals that there is a goat behind door #3 and we have just two choices, which most of us assume have an equal chance of having a car behind them. But they do not and the graphic below illustrates why.
When you begin, here is what your odds looks like:
However, after door # 3 is eliminated, you are left with doors one and two. You can see in the diagram below why door #2 has a 66.6% chance and door #1 remains at 33.3%:
This scenario illustrates the principal of variable change.
Of course, you still might get it wrong, but at least your odds are better.
I’m not a statistician, so if you agree or disagree, feel free to comment below.
Links Related to This Scenario:
Monty Hall Problem: Read a history of the problem and solution on Wikipedia.
Monty Hall Dilemma (aka 3 door game) : This is a great link. The delicate genius provides an easy simulation that you can play!
The Monty Hall Problem: John Todd writes about it in his blog, providing some more links that you may enjoy.
Wednesday Math, Vol. 23: The Monty Hall Problem: Matty Boy also discusses the issue on his blog after seeing the movie 21.
Monty Hall Problem YouTube: Here is a great graphical presentation on YouTube.
The Monty Hall Problem: Discussions from a Mathematics Professor
Let’s Make a Deal: Here, you can play a simulation of the game.
Related Posts
The explanation is wrong in the sense that you have added up the probabilities of having the car behind door 2 or 3. The very fact that door 3 does not have a car behind it does not allow you to simply add the probability of success to door 2.
Let me ask a question: If I chose door 2 initially and later it was reveiled that door 3 had a goat, then by your logic finally door 1 should be ending up with 67% probability of having a car. Now it seems that my initial choice of door is affecting the odds of having the car behind a door.
Thats why probability is finally 50%. Whats shown in movie is wrong.
I just read the theoretical proof which I should have looked for. The result is indeed correct. But again the method for proof in this blog is not at all trustworthy. for proof refer http://en.wikipedia.org/wiki/Monty_Hall_problem#Bayesian_analysis
To analyze this problem with much more accuracy
Lets extend this problem to 10 doors where car is behind one of them and remaining are empty.
You are asked to pick a door and suppose you pick door number one(1) or which ever you want.
Now see carefully,
There is
10% chance that car is behind the door you picked and
90% chance that car is behind one of the remaning doors you did not pick
Now remember that host knows which door has car behind it and he opens 8 doors from those which didn’t have car behind them and the door with car is kept close no matter that door is door number 2 or 3 or 4 or 5 or 6……..
Now host asks you if you want to change your option,
if you stick on your option, then there is 10% chance of winning
if you change your option, then according to statement 1,there is 90% chance of your win.
If you are clear then its ok other wise see this table
here 1 represents car, zero represents empty door
*Note: These are all 10 possible configurations for car placement
Each of these configuarations is equally probable
door1 door 2 door3 door4 door5 door6 door7 door8 door9 door10
1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 1
suppose u pick door 1, and stick to your choice then only for first configuration you will win and for remaining 9 configurations host will open 8 doors with no car behind them you will lose. your winnig chance is 10%
If you choose to change your options then you will win for 9 configuration and loose for only one.your winnig chance is 90%
I think the simplest way to understand this is the following:
You have a 66.67% of NOT picking the door with the car at the beginning.
If the door that you pick at the start does not contain the car then the car MUST be behind the door that the host does not open.
Therefore if you always trade doors then your give yourself a 66.67% chance of winning the car.
It is misleading. It is accurate ONLY IF, but only if you assume Monty HAS to open one door. The problem with the 21 movie is that you do not know if Monty has to do so. If he doesn’t, and knows that you picked the car the equations are inaccurate because the above equations presume he has to pick one door to open. If you don’t know if he has to open one door, the probabilities under this scenario reset to 50/50.
The way I see it, there are 3 rules the game show host must follow for the 2/3 chance to be accurate and none of them are explicit from the way the problem is stated in the movie.
1) Whether the game show host opens a door is not dependant on the choice of the first door (this is basically what svenn said except it would still work if the game show host flipped a coin to see whether he opened the door, but not if he opened the door only if you picked the door with the car)
2) The game show host will not open the door you chose
3) The game show host will not open the door with the car behind it
These may seem obvious to some people based on the game show premise and the fact that the host knows where the car is, but if you don’t know the system that’s being followed then you are back to 50%.
I used to work at shodor, they’ve got a lot of different math “activities” that illustrate things like the Monty Hall Problem. You should note that the higher the number of tries the closer you get to the 67% win percentage.
http://www.shodor.org/interactivate/activities/AdvancedMontyHall/
[…] when you do the math, but it hard to let settle because you mind is telling you 50/50. Here is a pretty good site that breaks it down and here is a site that even let’s you try it out […]
There are many ways to explain why your probability of winning the car is in fact 66.6%. I find this easies to understand:
Here is the key: Given the rules, if you always switch your pick, your first choice will always determine if you win or not.
(1) If your 1st choice is a goat, you will always win the car. This is because the host will reveal the other goat so when you change, it will ALWAYS be to the car.
(2) If your 1st choice is the car, you will always lose. This is because the host will reveal a goat so when you change, it will ALWAYS be a goat.
Therefore: Since your odds of initially picking a goat are 66.6%, your odds of winning the car accounting for variable chnage is the same - 66%.
Not the simplest concept to understand but, it is obviously true.
most of the comments on here are correct. I did a search on it, primarily because in watching the way it was portrayed in the movie, that it couldnt be correct.
The only way it can be correct is to assume that the host will point out a goat to begin with, which wasn’t stated in the film. Once you can safely assume that, you can increase your odds by switching.
Here’s an explanation in a different way that I wrote up to try and explain this problem to people from a different angle so they could grasp it better. The problem is human understanding in time in terms of seconds rather than metres, so just put that out of your mind for a second and bear with me:
======================================
Answer to the question about variable change introduced by the movie “21″:
———-
The professor tells the main character that he is on a game-show and the game-show host tells you he has 3 doors: Door 1, Door 2 and Door 3. Behind one of the doors is a sportscar. Behind the other two doors there are goats and he asks you to pick a door at random. The person chooses Door 1. The game-show host then opens Door 3 and reveals that is has a goat in it. He then asks if they want to keep their initial answer of Door 1, or switch their choice to Door 2?
The person explains that it would double your chances of winning if you switched the cases. Why?
———-
Wow… so many confused people… I love it! Let me explain it a different way that will clarify any doubt. The REASON that you are all confused is because of the temporal variables that are irrelevant here.
Consider an equivalent scenario mathematically, but COMPLETELY different, temporarlly and the answer will become blazingly obvious and you will slap yourself in the forehead and say D’oh! Of course!
Imagine if the host did NOT reveal the goat FIRST, but AFTER you decide whether to switch or not. It doesn’t matter WHEN the host reveals the goat, the point is that he’s going to reveal a goat, and regardless of that, you get to pick which “set” of doors to go with.
Consider what’s really happening here. You are dividing the doors {1,2,3} into two sets. The first set is {1}, and the second set is {2,3}. Now, given this division, which set of doors would you choose?
Well obviously you’d choose set {2,3} because it is more likely that the car is in one of those two than if it is in door {1}. The host could just have easily said:
Host: Pick a door
Contestant: I pick door 1
Host: Ok, now given that you have picked door 1, which do you think contains the goat? Door 1? Or one of the other doors?
Contestant: Well it’s probably in one of the other doors
Host: Ok, I will let you pick EITHER of the other two doors and if it’s in EITHER one of them, you win (because I’m going to show you LATER which one has the goat and you can have the other door which has the car no matter whether the goat is in 2 or 3)
Contestant: RIGHT ON! I’m going with the set of doors {2,3}, thanks for the extra 33% chance of winning host!
Yes, I know this problem was tricky… it was set up with all those temporal variables to purposely try and confuse a human mind which has an erroneous concept of time to begin with.
Were time measured in metres, nobody would find this confusing at all.
The solution that is spoken of in the movie is wrong to anyone versed in statistical probability. There is something referred to as the law of replacement. Anecdotally, the movie is correct, scientifically, the solution is not sound due to the law of replacement regarding probability. If you remove one of the probabilities of an equation, you automatically reconfigure the statistical odds as you did not replace the probability. By the sheer fact that there are only two choices left, and not three, by definition, the odds are now 50%.
I CONCUR WITH IMANI…
REGARDLESS OF WHAT THE PROBABILITY STATS ARE AT THE BEGINNING DOESNT MATTER. AFTER HE SHOWS THE GOAT BEHIND WHATEVER DOOR. THAT DOOR IS TAKEN OUT OF THE EQAUTION. SURELY YOU DONT COUNT THIS IN YOU CHOICE ANYMORE. LET ME SEE…HMMMM…DOOR 1 HAS A GOAT SO I SHOULD TAKE THAT INTO CONSIDERATION FOR MY OTHER TWO CHOICES. NO!!! NOW THE PROBLEM BECOMES CLEARY A CHOICE BETWEEN TWO DOORS. WHO CARES ABOUT DOOR 1. WHAT IF YOU WALKED IN AFTER THE TEACHER SHOWED THE DOOR. IS IT STILL 66.7%. NONSENSE. YOU WALK IN TO THAT PROLEM, YOU SEE…TWO CLOSED DOORS, ITS A 50/50 CHANCE TO GUESS THE CAR. YOU DONT LOOK AT AN OPEN DOOR AND SAY…WELL IF I THROW THAT ONE IN. WHEN THE TEACHER SHOWS THE DOOR IT MAKES IT A TOTALLY DIFFERRENT PROBLEM. 50/50. AM I MAKING SENSE…DESTROY ME IF YOU WILL. I WANT TO THINK LIKE THE REST OF YOU…ESPECIALLY IF IM WRONG. BUT I DONT SEE IT….AAAAHHHH I ONLY SEE 50/50 HELP
Dr. Jerky - you are not taking something into account. If you walked in after he had opened one of the doors, the probabilities are still going to be 33/66, but you just won’t know this. Being oblivious of the actual probabilities does not change the probabilities. There is a 66 percent change of the car being behind door #2 because there was a 66% chance of it being behind the set {2,3} to begin with. Stop thinking about it in terms of two independent probability problems.
The problems are NOT INDEPENDENT.
For instance, if they were independent, then the host could open the door with the car by accident, but he can’t. Why not? Because the two problems are really one problem with a confusing sequence of discoveries.
Think of it like this. The probability that the car is behind door #1 is 33% at the beginning, right? RIGHT! How can that probability suddenly change, just magically, mysteriously just because you peeked behind one of the doors? It can’t. The probability that the car is behind door #1 is ALWAYS 33% throughout the whole sequence of operations. It doesn’t just suddenly change to 50%, how would that happen? Your probability background should tell you at least that much. It would be like being clairvoyant somehow, and statistics does not involve clairvoyance. That’s actually your whole argument… that the probability of door #2 can’t suddenly change to 66%, right? So why would you say that door #1 can suddenly change from 33% to 50%??
Now… think about what is actually happening… the probability that the car is behind door #2 is only changing because door #3 has been taken out of the equation. That 33% chance that the car was behind door#3 has to go somewhere because you can’t have 33/33 - it doesn’t add up to 100.
Think of it this way, as some other poster mentioned:
Imagine 1000 doors.
Pick one of those doors. Door #1.
Now we know that there are goats behind at least 998 of the other 999 doors. So open all those goats.
You’re left with Door #1 and Door #237. Do you think you now have a 50% chance of having picked the EXACT right door right off the bat? I doubt it. You probably picked wrong, which means the car is probably behind door #237 and the host is basically giving you that for free because he got rid of all the goats for you.
In a nutshell:
Your probability of having picked the right door to begin with does not suddenly jump to 50% just because somebody “in the know” removes all the goats from the equation
The Movie is correct and is based on a true theory. Look at it the easiest way possible.
The doors are as follows C=Car G=Goat
C G G - Lose
You pick Door 1 and switch to door 2 after the Host Takes away 1 goat.
G C G - Win
You pick Door 1 and switch to door 3 after the Host Takes away 1 goat.
G G C - Win
You pick Door 1 and switch to door 2 after the Host Takes away 1 goat.
There you have it. if you switch you have a 2 in 3 (66%) chance of winning.
IF YOU DO NOT SWITCH THIS IS WHAT HAPPENS
C G G - Win
You pick door 1 and do not switch to door 2 after the host takes away door 3
G C G - Lose
You pick door 1 and do not switch to door 2 after the host takes away door 3
G G C - Lose
You pick door 1 and do not switch to door 3 when the host takes away door 2.
You only have a 1 in 3 chance (33%) if you do not switch.
Here it is Plain and Simple for those who still think you have a 50% chance no matter what. Enjoy
Correction
in the 1st set of 3 options i had the #s wrong. Heres how it should be.
C G G - Lose
You pick Door 1 and switch to door 2 after the Host Takes away 1 goat.
G C G - Win
You pick Door 1 and switch to door 2 after the Host Takes away 1 goat.
G G C - Win
You pick Door 1 and switch to door 3 after the Host Takes away 1 goat.
My Bad.
66% it is!!!
my self and my colleagues had a discussion about this subject, with different opinions, obviously. we know that when you create a test medium which happens to confirm your theory, then you have the right to think that you are right. my self, being a programmer, i wrote a small script to test the cases where a player would a) change the door, b) stick with the first choice and c) choose random every time.
result tests show that the theory is right: 66% chance of choose right if change the door, 50% if you choose random every time, 33% if you stick to the first choice; with some deviations of course, but it is pretty relevant.
should my programming be bad, hopefully someone will correct me, though i do not think it is the case.
if anyone likes to test that, just copy the content of the script from http://rafb.net/p/chIo2328.html, put in a text file with the extension *.py and just run it. obviously you will have to have the python program installed on your machine (available for free at http://www.python.org/download/)
Um I understand the groupings and the percentages and all but if you know the game show host will always reveal a goat out of the three doors then your chances at picking the correct door is still 50/50! If you actually do account for a variable of change and a variable of occurance (that being the host always reveals a goat) you know you are still stuck with a 50/50 chance of picking the car!
Might I add, if the game show host did not I repeat, DID NOT know what he would reveal as he opened one of the doors and he revealed a goat then and only then does their arise an increased propability that the car is in the door you did not choose.
for instance: some say
door 1 = 33%
door 2 & 3 = 66%
and host reveals 3 as a goat you should theoretically pick 2 as it falls in the 66% grouping increases your chances but I say that line of thinking is flawed! For if you know the game show host will reveal one of the other doors within the 66% group (the doors you did not choose) as a door it re-categorizes the percentages and gives them an equal value of 50% thus making ur chances on choosing the car at 50/50!
The whole theory rests on the possibility of what the host will reveal. If it is at all possible the host would reveal the car then this variable of change theory would apply and I would thus switch my original choice!
My brain hurts! ;P
You can use the same argument to show that door 1 to prove that door 1 has a 66% chance of holding the car. Obviously, there is something wrong with the argument since door 1 and door 2 both can’t have a 66%% probability of being correct.
For example:
When I chose door 1, I was actually choosing the set {1, 3} which has a 66% chance of having the car in it. Since door 3 had a goat, door 1 now has a 66% chance of having the car and door 2 has only a 33% change of having the car.
Clearly this line of reasoning must be can not be correct because it can be used to “prove” that both door 1 and 2 are the best choice.
Once Monty reveals that door 3 has a goat, it is removed from the problem and the contestant is faced with an entirely new choice: two doors each with a 50% chance of holding the car. The key to understanding this is that there the second choice is not a continuation of the intial problem, it is an entirely new problem.
> When I chose door 1, I was actually choosing the set {1, 3} which has a 66% chance of having the car in it. Since door 3 had a goat, door 1 now has a 66% chance of having the car and door 2 has only a 33% change of having the car.
WRONG. When you chose door 1, you were splitting the set into {1} and {2, 3}, therefore the set {2, 3} has a 66% probability of being correct, and the set {1} only has a 33% probability of being correct.
> Clearly this line of reasoning must be can not be correct because it can be used to “prove” that both door 1 and 2 are the best choice.
The set of doors {1, 2} DOES have a 66% chance of being correct. Had the division been made in this fashion (ie: by the contestant choosing door 3 initially), then YES absolutely he should switch to the set {1, 2}, rather than sticking with his original choice of {3}.
I don’t understand how people don’t understand. Forget the math you think you know, or learned once long ago in school. This is too simple to theorize.
Try using some common sense.
If there are three doors and behind one contains a car, would you like to pick one door OR do you think your odds of winning might be better if you chose two doors?
If you think your odds are better by choosing one door, then I’d like to play this game with you for real. Any takers?
Owned.
Jack, nice try but you got it wrong.
You aren’t given that choice at the beginning. You make your first choice when it is a one in three chance. You then make another choice when it is a one in two chance.
Your argument would ONLY be right if you were offered door one or doors two and three BEFORE any have been opened. Before you make your choice there is a two-thirds probability that doors two and three have the prize but once you have removed one of those then it is 50/50 between the remaining choices.
I don’t understand how you can’t grasp this most basic concept?
I can appreciate why people think the answer is the way it is but it’s an optical illusion for the brain.
Martin, actually Jack is right. You ARE given that choice in the beginning. When the host removes the goat from the equation, he essentially turns back time and makes the probability of your next choice change - hence variable change. The ONLY way what you are saying would be true is if the host picked a random door to remove, which means he might remove the car. Since this is not an option, the host’s choice is NOT random, and hence changes the variables in the equation.
The easiest way to think of it is imagine 1000 doors. You pick one of the doors, the host removes 998 other doors with goats. Do you think you now have a 50/50 chance that you picked the car in the first place when there were 1000 other doors? I think not.
Svenn and patti got it correct from a math point of view. the problem was not stated correctly in the movie. Those that have taken probability knows that you must examine if the outcome of one trial affects the outcome of the next. For instance when you roll a dice the chance of rolling a 5 is always 1/6. Each role is independent of the other. So in this case it seems it would be 50% because you are picking 1 out of 2 doors.
I think this is what Imani and Dr. Jekyl are thinking. It is what I thought after watching the movie. But as stated by Svenn and patti, the rules of the game means that the previous outcome has an effect on the next outcome because the game show host knows where the car and will not be picking at random, therefore the trials are not independent of one another.
finally there is this site
http://www.shodor.org/interactivate/activities/SimpleMontyHall/?version=1.6.0_05&browser=MSIE&vendor=Sun_Microsystems_Inc.
play and you will see the probability of 66% when you switch doors. Notice how the game never picks your door, it always picks one of the other 2 doors which has a goat. That is the key to understanding the problem
Drew, I am an enthusiastic but rather mediocre first year maths student who very much appreciated your explanation, which makes perfect sense. I think if you combine the idea of sets with an increase in the number of goats (doors) involved, it does become quite intuitive. Thanks a lot. Also gunoieru simulation always helps too!
Ok People, I graduated from high shcool, started college and realized building houses was my gift. Needless to say I work with quit a few non graduating high school students. So sometimes I need to explaing myself clear in all languages, even when I only understand English so so. Lets go guys and girls stay with me.
When the game show host first gave us three doors to choose from I feel not one of us had a problem feeling we had a 33.3% chance of picking the right door. why did you and I say 33.3% chance of winning because 3doors X 33.3% = 100% right.
Now when door 3 is opened and is a goat that 33.3% needs to be eliminated from the starting point and needs to be divided equally in half which is 16.65% and added to door one and two evenly, which now brings doors one and two up to 50% each. Cut and dry. If I picked door one and my girlfriend picked door two and we both switched to the other door that would mean I would have a 66.6% chance and she would have a 66.6% chance of winning. 66.6% + 66.6% adds up to 133.2% now can we all see that this is not possible. Can we see that 133.2% would sound like a politician speaking. I hope you all can see this now in what ever language you read or what ever politician you are hearing. 133.2% is not what it can be. It can only be 100%. If I change my door and get 66.6% you can be rest a sure my girl friend will want her 66.6% as well. This is the problem we all want more than 100% and no one can give us more than 100%. Therefore door two can not get the 33.3% added to it and door one get nothing right?
Hi there Rob, I see what you are saying, but you’re forgetting one thing. If your girlfriend had picked door #2, it probably would’ve been your door #1 that would’ve been opened with a goat, so you can’t think about it that way. It goes ONE person at a time, not two. The reason is because the door that the host opens depends upon where the car is. The host is tricking you. He’s gonna change his door selection so that it will always be 66.6% chance of it being the door you didn’t pick. That’s the catch.
Let me try to explain a different way.
Say you pick door #1 initially. We all agree that you have a 33% chance of winning the car. Then you ask the host to go behind the stage and find a goat and tell you which door has that goat. Do you think the act of the host looking behind the stage suddenly changes the 33% probability that you picked the car? Of course not, you said so yourself in your message. You have 1/3 chance that you picked the car, ALWAYS, no matter what happens. You can NEVER change the past and that 33% is FIXED IN STONE. So stop thinking of it like that. The 33% does not change to 66%, it does not change to 50%. It STAYS 33% and has to.
What does change is that the host was kind enough to let you pick the TWO doors (door #2 AND door #3). So when he asks if you want to change, he’s not asking, do you want to switch from door #1 to door #2. He is ACTUALLY asking, do you want to change your selection from door #1 to BOTH door #2 AND door #3. And of course you’d say yes, because picking two doors is always better than picking one.
It becomes super easy to visualize when you think about 1000 doors. You pick door #1, and then the host goes and removes goats from 998 of the other doors, leaving only door #1 and door #237. Knowing that there’s a car behind one of them, do you want to change your selection to door #237? Or do you want to stick with your original choice of door #1 and hope to heck that somehow the probability jumped from 1/1000 to 50% magically just cause somebody went behind the stage for a minute???
When in doubt, use brute force.
I created a table where I mapped out every possible permutation of the problem. There are only 24 possible outcomes, and in 12 of them staying the correct choice, and in 12 of them switching is the correct choice. 50/50.
http://www.andrewcmurphy.com/monty_hall.html
First of all, I agree with the switch, but only if the host opening one door is constant. If the host has a choice of whether to open a door or not then you don’t switch. I’ll be the host and we’ll play 1000 times. I’ll open 1 door after you choose and give you a choice to switch 100 times out of the 1000. You switch every time you have a choice. You won’t win once out of the 100 becuase I can’t afford to give away a car. You must know whether the host intended to open a door no matter what door you picked in order to gain an advantage by switching.
If the host opening a door is a constant, if you switch every time then you win if you were wrong on your first choice and lose if you were right. 33 percent you were right and 67 you were wrong, so 67 percent you win and 33 you lose.
You would still feel the fool if on game night you swapped doors but the car was behind door #1. You could discuss probability with your mates at the bar and explain to them that if you repeated this process 2 more times you would have won the car twice!
Andrew, you have made an error in your analysis. You are counting the scenario where you picked the correct door initially twice. If you look at the row of numbers in your table denoted by “Your initial pick is door number:”, you will see that you are giving a 50/50 probability for having selected the correct door initially, which is wrong. It is a 1/3 probability for having selected the correct door initially.
You are essentially asking the question “Given that I have a 50/50 chance of having selected the correct door right off the bat, what is the likelihood that switching doors will make me a winner?”.
To the player, it is unknown whether the host knows whether which door has the car as not stated in the problem statement.. In this scenario, the probability that the host knows which door has the car, to the player is 1/2. Now if the host does not know which door has the car behind, the probabilities of the player’s winning or losing by keeping the 1st door is the same i.e. 1/2 when the host reveals that the 3rd door has the ghost. In case the host knows which door has the car, behind, the player indeed has a winning probability of 2/3 on switching according to simple Bayesian analysis.. Hence the cumulative probability of winnin will be..
Probability of host knowing the door having car X 2/3 + probability of the host not knowing of it X 1/2 = 1/2*2/3 + 1/2*1/2 = 1/3 + 1/4 = 7/12
Probabilty of losing hence would be 1-7/12 = 5/12 which can also be proved in the manner stated above.