Help, I don't get this one. Monty Hall Problem.

[mixedgas]

The Monty Hall problem came up in reading my boss assigned on instructional design in classrooms.

I dont get how the probability flips between 1/3 chance of picking the car to better odds if you switch doors after your initial choice. What if the game show host DOES NOT know which door the prize is behind. Does this assume you get a ANSWER after choice one?

I dont get this one.

I'm probably not even phrasing my question correctly.

Wiki factors in here too.

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

Any one who can put this in very SIMPLE terms for me?


Thanks, Steve

[Solonar]

The Monty Hall theory only works if the Host does know the empty room.


Mythbusters did an episode on this a few weeks ago.

[norty303]

If the host DOES NOT know which door the car is behind then there is a 1 in 3 chance he'll uncover the car. Do you 'win' in that situation too?

I think if the assumption is that the door opened by the host is a goat, then he must know where the car is (or at least isn't)

I might write a little Oracle program to test out a few thousand iterations to prove/disprove the theory to myself...

[mixedgas]

The Monty Hall theory only works if the Host does know the empty room.


Mythbusters did an episode on this a few weeks ago.

Thanks, there is so much obtuse bullcrap about this one that I did not get it.

I sorta needed the "Britney Speers Guide to Semiconductors" for this one.
IF you have not seen BSGTS, it does exist, it is very good, and Miss Speers did try to sue it out of existance, until some one told her publicity department it was a PR winner.

Steve

[steve-o]

Just skimming the wiki article, the host does know the contents of all 3 doors. He opens the 1 door that you have not chosen, revealing a goat, leaving now 50/50 odds. If you now choose the door that you did not choose before your odds go up of winning because of a probability factor which I could not understand. The key I guess is making sense out of those probability quantum formulae. Rots-o-Ruck Steve! Sorry, no help..

-edit- Yeah the Britney Spears guide to semiconductor physics is hilarious.

[dilbert]

Can only work if the host knows the correct door so he always offers you one incorrect and one correct so 50/50... Your first choice is always 1/3 so you get better odds if you change.... factor in the mental conditioning and you should always swap........

Sure the mythbusters covered this on one show......

[colouredmirrorball]

I remember this one from a statistics course. The host has to know which is the winning door.

Each door has a chance of 1/3 to win. Pick one at random and name it door A, the other two doors B and C.
The host now announces he will open door B or C and the candidate can choose to remain at door A, or switch to the other door.
If door A was the winning door, the host can open doors B and C at a random chance. If door A was NOT the winning door, then door B or C is the winning one. In this case, the host will of course not open the winning door. This is an extra piece of information which alters the chances.

[dsli_jon]

Heya P'fessor -

Oi, would-have posted this an hour-ago, but just got off a 1 hour-long sales-call... wow.. Ok...

I dont get how the probability flips between 1/3 chance of picking the car to better odds if you switch doors after your initial choice.

It really doesn't. Your odds 'improve', of course, when the Host eliminates one of the 3 choices, from 1/3 to 2/3, but as-to 'do your 'odds' actually improve', if you switch, this quote from the subtopic "Variants" of that article is most explicative (bold-highlight, mine..): "...in any case the probability of winning by switching is at least 1/2 (and can be as high as 1), while the overall probability of winning by switching is still exactly 2/3." - it's a matter of 'how you look at it' - including the 3rd choice, in the overall odds, yes, means that your 'chances' have improved from 1/3, to 2/3... but, in the 'actual-odds', you're gonna pick the car or not, and that's 1/2..

..the rest of it, is all 'game theory', etc - fascinating stuff... I'm intrigued by the part I 'clipped' from that sentence: "...the player can have different probabilities of winning depending on the observed choice of the host..", in other-words, 'playing' the probability-table, based on which door the host chose, as-shown in this graphic: http://upload.wikimedia.org/wikipedi...tree_door1.svg - interesting 'poker-math', in that scenario...

..but - I think I know a way that you can 'better your odds', even-further... once the Host 'narrows' the odds by revealing one of the goats... and you have to 'choose', one of the two doors left - before you speak your selection - simply 'bleat'... ...this way, you aurally ferret-out where the other goat-is, and call the car... of course, if the goat that's 'left' is not the one that responds to you, well then you're screwed...

..and, heh - 'Quantum Goat Theory', Lol...
j

[mixedgas]

-edit- Yeah the Britney Spears guide to semiconductor physics is hilarious.[/QUOTE]

Its accurate, too. VERY accurate.


Booble,

Steve

[mixedgas]

Jon, Why bless your Bleating Heart for that wonderful explantation with such great mathematical insight.
I'd laugh out loud about QGT, but I'm at work and laughter is a dis-allowed state in a medical plant.
Did you have some extracurruculars in College we need to know about? You really got my goat with that one.

PS, cant wait to just plain crack up outside the plant door,





Steve