the premise is that a game show contestant has 3 doors to pick from and only 1 is a winner. the player picks a door and without knowing the outcome, the host allows him to change his answer.

the movie states that is always better mathmatically to change the pick to another door. 1/3 are the beggining odds. by changing doors you now have 1/2 chance. this is where i dont follow the logic.

following this method, if you change your answer a third time you are a guarrenteed winner which is absurb.

i had a low-level college prob and stats class which i did well in and greatly enjoyed. im gonna have to call b.s. on this one.

pls advise and explain in layman's terms. tia.

No. You have a 2/3 chance of getting the prize if you switch.

See this: http://en.wikipedia.org/wiki/Monty_Hall_problem

And this: http://www.youtube.com/watch?v=mhlc7peGlGg

And here is one other link from October where a similar question was asked in this forum: http://forumserver.twoplustwo.com/25...puzzle-906129/

actually you get the same chances because no information was given.

there's an assumption that underlies the question, as wiki explains:

- The contestant selects a door (odds 1/3).
- Next, the host (who konws the correct door) selects a remaining door that is incorrect and asks the contestant if he/she would like to change his/her choice.
- B/c the door chosen by the host can be excluded, the prize must be behind one of the two doors not chosen by the host (odds 1/2).
- Because the host effectively eliminated an incorrect choice, the second time around the contestant has a higher likelihood of choosing the correct door.

OP did not state the ‘classical’ Monty Hall problem correctly. He left out the condition that the MC opens a door that he knows is a loser and then asks the contestant if he wants to switch.

Provotrout corrected that error but then incorrectly said that the two unopened doors now each have a 50/50 chance of being the winner - at least I think that is what he means by 1/2 odds.

As suggested by Sherman,check out his links to see why 2/3 is the probability for the remaining door not initially selected and not opened. This can be shown by deductive logic, by Bayes’ theorem and several other ways.

I never thought he did a very good job explaining the answer in the movie, but yeah, as others have said, it is 2/3 win if you switch, provided you make the most natural assumptions about the hosts behavior. I think it's easier to see what's going on in the 3 prisoners problem, which features a related calculation.

What about if there was a million boxes, you pick one, all the other boxes are opened apart form one, would you switch then?

My friend and I actually tested the Monty Hall problem, except we used a marble under one of 3 cups. We spun a pen to randomize which cup one of us would hide the marble under while the other one was out of the room. Then, whoever was out of the room would come back in and make their first guess. One empty cup would then be revealed and they'd be asked if they'd like to switch.

We ran ~500 trials and the final result was, by switching, we were correct 67.x% (I forget the decimal) of the time.

Yes, we were bored but it was fun, damnit!

gotcha, thanks for the clarification

I haven't seen the movie but if OP meant (not stated in original question) that the host shows a remaining door ad random (= no monty hall):
There is no benefit in changing when host shows a no winner (0.5 win // 0.5 loss). The host will show 1/3 the winner in which case you will always loose.

Here should be made a mistake in the movie because when it's better to change doors it's probably a monty hall problem and odds are 2/3 of winning by changing. If you have 1/2 odds of winning by changing it's probably what I explained.

Which do you think is more probable?

A) The movie blatantly made a mistake regarding the Monty Hall problem.

B) The OP blatantly misquoted the movie.

It was meant ironically.

hmmm seems like a 50/50 situation

If you picked A, then switch to B and you'll have a 2/3 chance of being right.

This is the correct answer.

A lie can circle the globe before the truth has even put its boots on.

Hey Provo how goes it... New user here....how might one contact you privately? Got a q for ya .

Click on his name and then click the "send message" option. Choose send a private message.

You have permission to send a PM to me.

The OP got it blatently wrong, Kevin Spacey got it significantly wrong and even Wiki gets it somewhat wrong. It's essential to the Monty Hall problem that the contestant knows the host consistently reveals a goat door every time the game is played. Wiki omits this condition and Kevin Spacey implies the condition does not hold when he says, "The game show host, who by the way knows what's behind all the other doors, decides to open another door. Let's say he chooses door #3, behind which sits a goat"

The implication is that sometimes the host decides to open another door and sometimes he doesn't. He might even decide to open a prize door some of the time and just decided to open a goat door this time. Maybe he only opens a goat door when the contestant's first choice is a winner.

Of course the OP isn't even close.


PairTheBoard

I don't agree with part of this argument. It is obvious without being stated that the host will never eliminate the one door containing the prize, or the game itself breaks down. It's also known that on the actual game that started this problem, the host always opened a door (goat) and gave the choice. I agree that point does need to be explicit in the problem, but when you call the problem "Monty Hall" then we know those conditions are true.

Maybe. Howver , for someone who knows nothing about the game except what they are told I don't think it's obvious to them that the host doesn't show the goat only when they've already picked the prize, just to mess with them. In fact, that's the first thing I would worry about.


Not everyone watched "Let's Make a Deal" nor has any idea how Monty Hall played the game other than what they are being told. At any rate, "Monty Hall" was not mentioned in the movie "21".


PairTheBoard

Yes, of course you're right that this problem is often poorly posed: there needs to be the explicit requirement that the "host" will always open a second door without the "prize". I've seen this "mistake" almost ad nauseum ad infinitum, but there was at least one poker author that mentioned this "mistake".

I decided to watch the movie clip, and the often missed point is that the game show "host" may decide to only open a second door if the contestant's original choice is correct ( in which case, switching now gives a probability of zero ); at the other extreme, it's possible that the "host" only shows a second door when the "contestant's" original choice is incorrect ( perhaps the contestant is well-liked by the audience or a blood relative of the host ; if the original choice is correct, the "host" just opens the door that was chosen and reveals the "prize" ). It's also true that for any number p in [0,1] that switching doors results in obtaining the "prize" with probability p, given that there is flexibility in the "host" choosing whether or not to reveal a second door.

Here's an article where Monty Hall himself turns the tables on those who "autoswitch" ( from the middle of page 2 to the top of page 3 in the following link ):

http://query.nytimes.com/gst/fullpag...0&pagewanted=1

Hey provo..sorry for delay.. and also I'm not seeing the send message link.. would you mind sending me an email or give me a time that you will be on here and figure out a way so we can chit chat ? Talk soon...