The odds only change in the event Monty were to open the winning door, which he never does.
As Kenny Rogers said ..... [MEDIA=youtube]Jj4nJ1YEAp4[/MEDIA]
Holy Cow, post: 449617, member: 50 wrote: While Monty can show you a non-winning door, that has no effect on whether or not your door is the winner.
Assuming the staff doesn't move things around behind the doors, no it doesn't change whether your door is a winner. But it DOES change what we know about the other two doors. More information given requires a new set of probabilities. Having more information about some doors than others results in unequal probabilities.
Holy Cow, post: 449617, member: 50 wrote: No matter what door you select he always has a non-winning door to show you. Once he eliminates one, there are two left. One is a winner while the other is not, In effect you always have a fifty-fifty chance
If the positions behind the doors were re-randomized, then it would be fifty-fifty, but the problem assumes things stay put.
If you didn't pick the winning door the first time, then Monte only has only one non-winner to show you and the remaining door has to be the winner. That's what makes the probabilities improve by switching.
I like my first explanation better, but this alternate view may help see why you should switch.
But if you picked the winning door he is still going to show you a losing door because that's all he has left. Thus it's fifty-fifty.
I know what you are saying and agree, but Monty doesn't play by the standard rules because he always shows a losing door. If he played by the rules, one third of the time he would open the winning door, proving yours must be a losing door.
When in doubt in probability problems, itemize the equally likely cases. After he opens a door, the remaining cases are not equally likely, as shown by itemizing.
Let??s say you picked door number 1. (At this point it??s symmetrical so it doesn??t matter what numbers are on the doors. You could re-order the numbers and get the same probabilities).
The possibilities are
X O O (1/3)
O X O (1/3)
O O X (1/3)
In the first case, Monty has two options
X O _ (1/2*1/3 =1/6 )
X _ O (1/2*1/3 =1/6)
You win 2 out of 6 by not switching and 0 of 6 by switching.
In the second case, Monty has one remaining losing door and must show it
O X _ (1/3)
You win 0 of 3 by keeping and 1 of 3 by switching
In the third case, Monty has one remaining door
O _ X (1/3)
You win 0 of 3 by keeping and 1 of 3 by switching
The end result is you win 1/3 by keeping and 2/3 by switching.
On top of that, I would suspect that it is more "human nature" to hold on to your current door. The mindset being that you would hate to have picked the winning door and then give it up in addition to the general feel that your chances have improved since they revealed one of the doors to not be it
Monty always opens a losing door. That means there are really only two doors from the start, not three as it would appear. There is the door you picked and either a winner or loser. Yours is also a winner or loser. Same odds with either door.
Now, if he chooses his door purely at random, thus frequently opening a winner, then the odds are as Bill 93 states.
It doesn??t make me an expert in probabilistic analysis by any stretch, but I did give a presentation a couple of weeks ago on Bayesian inference applied to GIS. Thomas Bayes, an 18th-century English cleric and heretic mathematician, would have quickly made a choice to switch doors with Monty based on his own Bayes Theorem, which as Bill93 so aptly re-states that as ??more information [is] given [it] requires a new set of probabilities?. Using Bayes?? Theorem, once Monty opens his door (always revealing a loser, i.e., a certainty of 100%), then the probability that the winner lies behind the door that you didn??t pick is 2/3. Bill93??s alternate analysis is one that I hadn??t seen before and darned clever. Holy Cow isn??t alone in his belief that the odds are 50-50, many renowned mathematicians have come to the same erroneous conclusion because their reasoning is based on solely on a priori circumstances, not on the a posteriori conditions that comes about when Monty opens a loser door.
The intuitive example I gave in my presentation was that if you picked one door out of 100, and Monty opened up 98 loser doors, do you still really feel, based on those 98 losers opened up in your face, that your odds are still 50-50? When new data is presented, it should be factored in, and that??s the basis for what??s called Bayesian kriging in GIS: making better predictions about unknown locations of gold deposits, oil reserves, etc., etc., based on sparse known locations that are successively Bayes-iterated with new observations.
rudy.stricklan, post: 449735, member: 3006 wrote: The intuitive example I gave in my presentation was that if you picked one door out of 100, and Monty opened up 98 loser doors, do you still really feel, based on those 98 losers opened up in your face, that your odds are still 50-50?
Great example.
This whole thread has me bumfuzzled.
It's the surveyors DUTY to open all 3 doors, and get paid get it too!.
And, to scrape around the door frames.
It's as much our duty to get the right door, as it is to know what's behind the other doors, and the trim. Just so we can argue with those who DID pick, and settle on a wrong door.
Best Regards,
You can thank me later,
😉
Ok, Kent is out 'o hat...
N
An interesting reading for young and old alike with a math interest is a novel by a Brit called
The Curious Incident of the Dead Dog in the Night.
It has a chapter referring to the Monty Hall problem.
It is a mystery as seen through the eyes of a 15 year old autistic boy.
If you like mystery and math and have an interest in autism, you may enjoy.
Holy Cow, post: 449656, member: 50 wrote: But if you picked the winning door he is still going to show you a losing door because that's all he has left. Thus it's fifty-fifty.
By sticking to your one door you never had more than a 1/3 chance. The words ??would you like to change?? aren??t magic, and this slight pause in the game doesn??t make your chances any bettter than if Monty never stopped to ask, and just went ahead and opened his two doors.
Play the game from the other perspective. Monte offers you two doors that you have to stick with. You now know intuitively, for a fact, that the sucker with the one and only door never had better than a 1/3 chance no matter what order your two doors were opened in, and no matter how long it took you know this. You stick with the one door and you are the 1/3 sucker. By switching you are playing from this other side. You are putting aside one door (never to be opened, 1/3 chance) and you get the combined odds of the other 2 doors.
Conrad, post: 450068, member: 6642 wrote: By sticking to your one door you never had more than a 1/3 chance. The words ??would you like to change?? aren??t magic, and this slight pause in the game doesn??t make your chances any bettter than if Monty never stopped to ask, and just went ahead and opened his two doors.
Play the game from the other perspective. Monte offers you two doors that you have to stick with. You now know intuitively, for a fact, that the sucker with the one and only door never had better than a 1/3 chance no matter what order your two doors were opened in, and no matter how long it took you know this. You stick with the one door and you are the 1/3 sucker. By switching you are playing from this other side. You are putting aside one door (never to be opened, 1/3 chance) and you get the combined odds of the other 2 doors.
Conrad, that??s another good way to rationalize the logic of switching. After picking a door, the odds that that the winning door is in the group of the other two doors is two-thirds. When Monty reveals one of those doors in that group to be a loser, the remaining door??s chances of being a winner are therefore two-thirds.
In lieu of flowers, the family has asked that we give $50 to someone who has dental floss in their purse.