The Monty Hall Problem
The Monty Hall Problem is a famous (or rather infamous) probability puzzle. Ron Clarke takes you through the puzzle and explains the counter-intuitive answer.You can read more about this problem, and the controversy, on Marilyn Vos Savant's website www.marilynvossavant.com
Channel: Howto
Uploaded: January 21, 2007 at 3:38 pm
Author: niansenx
Length: 00:05:48
Rating: 4.82
Views: 235464
Tags: monty ron goats puzzle clarke show maths three doors mathematics math probability hall car gameshow problem game
Video Comments:
giovea (January 8, 2009 at 4:49 pm)
what you said is from a RETROSPECTIVE kind of view, set the time line and realize pre test is prospective and post test retrospective. As a player you have no point from which you can get a retrospective! the mistake is there! you circle time by crossing perspectives with no common point, in a sense... player can't tell between A or B!
giovea (January 8, 2009 at 4:45 pm)
brain rotemanacer i said no 1/4... pick car and switch to goat A, loose, pick car and switch to B, loose. Pick and switch , win, pick B and switch, win. this is considering he can only show goat A OR B. so is 50/50.
All those calculations were the showman's view point, that you don't know, that's why you are asked to calculate in the first place. If you're asked to pick a goat i say it's 2/3..yet still 1/2 if you´re showned one. picking the other goat or the car must add up, otherwise???
All those calculations were the showman's view point, that you don't know, that's why you are asked to calculate in the first place. If you're asked to pick a goat i say it's 2/3..yet still 1/2 if you´re showned one. picking the other goat or the car must add up, otherwise???
BrainRotMenacer (January 8, 2009 at 12:27 am)
Actually the host showing either door equates to 1/6 each, not 1/4. So you add them up for a total of 1/3.
giovea (January 7, 2009 at 11:38 pm)
well 66.6% of 2 doors is something like 1.33 /2...
can we be stupid enough to consider choose the revealed goat or perhaps you need a car so much you lost your mind!
and should he ask you to reconsider your reconsideration, then what????
hehehe
can we be stupid enough to consider choose the revealed goat or perhaps you need a car so much you lost your mind!
and should he ask you to reconsider your reconsideration, then what????
hehehe
studmuffin018 (January 7, 2009 at 9:32 pm)
I learend this from the movie 21
giovea (January 7, 2009 at 7:51 pm)
Anyway for those interested is fullstrikes.blogspot. I really didn't liked this mess, more when i found out it was 40 years old! incredible...
giovea (January 7, 2009 at 7:42 pm)
hei brainromenacer. Good thinking to name the goat BUT you didnt work well enough on one branch, YOU MISSED to branch the too option when you pck the car, the host can show you goat A or goat B, eitheir way you loose, so really 50/50, which makes sense since is a 1/2... !!!OBVIOUS SINCE THE START
blacky159 (January 7, 2009 at 7:23 pm)
thanks, great explenation!
giovea (January 6, 2009 at 3:52 pm)
it's all wrong, it's a dependant chain of events live taking a ball from a bag with 3 and not replacing it, and it's statisticly irrelevant if you "choose" before or after door 1 opens. "event" that matters is doors revealed.
those were the mistakes of this dogma.
so, following a time line we branch the chain of events, door1 opens, then door 2 opens.
1STCASE. Door 1 goat (2/3) x door 2 is 50/50 (goat or car )= 1/3 each , congruent!
2STCASE Door 1 car (game over) x door 2 1/1 goat=the other 1/3
those were the mistakes of this dogma.
so, following a time line we branch the chain of events, door1 opens, then door 2 opens.
1STCASE. Door 1 goat (2/3) x door 2 is 50/50 (goat or car )= 1/3 each , congruent!
2STCASE Door 1 car (game over) x door 2 1/1 goat=the other 1/3
giovea (January 6, 2009 at 3:27 pm)
No it is not determined, just a DOGNMA.you have to PROVE IT like i do , pre test and post test, figures must be the same- 1/3!!!!
2 events in chain
1st goat (2/3)x 50/50 chance (2/3x1/2= 1/3 each)
1st car (1/3) x goat (100=1) =1/3 the remaining
pre test 1/3, pos test 1/3, correct calculations . end of point
2 events in chain
1st goat (2/3)x 50/50 chance (2/3x1/2= 1/3 each)
1st car (1/3) x goat (100=1) =1/3 the remaining
pre test 1/3, pos test 1/3, correct calculations . end of point
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