zengrifter
Banned
If its valid and correct there are a lot of applications, including proposition bets.
But I still don't think its valid. Lets ask our resident MIT fan, MDLBJ. zg
Lecture Scene
- Thread starter kennybenny
- Start date
But I still don't think its valid. Lets ask our resident MIT fan, MDLBJ. zg
la_dee_daa
Well-Known Member
what dont u think is valid about it?????/
la_dee_daa
Well-Known Member
you know everything you see in the movies and tv is TRUE!!!
zengrifter
Banned
C'mon, settle this for us. Ask one of the MITs if needs be, please. zg
I haven't bothered reading this whole thread, but is zg asking whether the Monty Hall problem is legit? Of course it is. Switch doors.
An example: You're on Lets Make a Deal with Monty Hall, but this time there are ten doors to choose from. Behind one door is a new car. Behind the other nine doors are goats.
You choose a door. Monty, knowing where the prize is, then opens 8 other doors showing goats every time. Then he asks you, do you want to trade the door you initially picked with the only other remaining concealed door?
Of course you do. Your chance of winning the car by switching is 90%. Your chance of having the car already is only 10%.
An example: You're on Lets Make a Deal with Monty Hall, but this time there are ten doors to choose from. Behind one door is a new car. Behind the other nine doors are goats.
You choose a door. Monty, knowing where the prize is, then opens 8 other doors showing goats every time. Then he asks you, do you want to trade the door you initially picked with the only other remaining concealed door?
Of course you do. Your chance of winning the car by switching is 90%. Your chance of having the car already is only 10%.
Harman
Well-Known Member
Ken is right.
The chance of choosing the correct door starts as 33%
He opens one goat door so the car is still there
Since the chance of choosing the correct door is low - 33%, then it is better to switch, now you know one goat has already gone.
Geddit???
Like with the 10 doors it is really unlikely you would pick the right door (1/10) so by removing 8 other than your choice it is extremely likely that you have chosen incorrectly and should switch
The chance of choosing the correct door starts as 33%
He opens one goat door so the car is still there
Since the chance of choosing the correct door is low - 33%, then it is better to switch, now you know one goat has already gone.
Geddit???
Like with the 10 doors it is really unlikely you would pick the right door (1/10) so by removing 8 other than your choice it is extremely likely that you have chosen incorrectly and should switch
zengrifter
Banned
A few of us are slow - it starts at 33%, then one is eliminated, this means that the odds are 50-50 for either remaining door - how do my odds improve by switching between the 50-50 doors?? zg
Ps - Earlier in the thread I was not alone in not getting it.
Ps - Earlier in the thread I was not alone in not getting it.
la_dee_daa
Well-Known Member
ok lets look at it this way first way by not switching you have a 33% change becuase there are only 3 doors.... ok are u still with us?
now lets look at the choice to switch you pick a door..
first option you have a 33% change you picked the right one
66% change you picked one of the other 2 doors which doesnt have the caar behind it......
so say you picked the car the first time both the remaining doors have goats behind so the host will show you one and you pick the other you lose.......
now say you pick one of the other 2 doors initally with goats behind a 66% chace... the host has to show you a goat and there is only 2 doors remaining a goat and a car so he must show the goat leaving the car.. so you switch and will win!!!!! this is the whole point that if you don't pick the right door first which is a 66% change if you have a chance to swtich after seeing the other goat you will win 100% of the time after this.
so in the case of 3 doors the only time you lose if you pick the car for your inital pick.
do you get it now????
zengrifter
Banned
No. I'd like to see a sim. zg
My 10-door example didn't help?
zengrifter
Banned
No, because it still boils down to TWO doors - each, to my limited mental
capacity, having a 50% chance. If we start with 10 and eliminate all but
two - it is counter-intuitive to me to suggest that one of the two is more
likely than the other. z
g
How about we make it a million doors, and Monty shows us 999,998 goats. Still want to stick with your one-in-a-million door?
The crucial point here is that Monty isn't randomly choosing doors to open. He knows where the prize is and is guaranteed to not accidentally show you the prize. If he were randomly opening doors, and would therefore sometimes reveal the prize, then yes, the eventual choice would be 50/50 between the two last doors. That's not what's going on here though.
If that still doesn't help, how about Wikipedia:
http://en.wikipedia.org/wiki/Monty_Hall_problem
The crucial point here is that Monty isn't randomly choosing doors to open. He knows where the prize is and is guaranteed to not accidentally show you the prize. If he were randomly opening doors, and would therefore sometimes reveal the prize, then yes, the eventual choice would be 50/50 between the two last doors. That's not what's going on here though.
If that still doesn't help, how about Wikipedia:
http://en.wikipedia.org/wiki/Monty_Hall_problem
zengrifter
Banned
Okay, we're making some progress for my non-math/semantic-based mind.
There must be some applications here - even maybe just for prop bets? zg
zengrifter
Banned
Hey E-Town-Guy, we need not feel so alone in our ignorance, now -
Wiki: When the problem and the solution appeared in Parade, approximately 10,000 readers, including nearly 1,000 Ph.D.s many using university letterhead, wrote to the magazine claiming the published solution was wrong. ...even when given completely unambiguous problem statements, explanations, simulations, and formal mathematical proofs, many people still meet the correct answer with disbelief.
You may have missed this explanation which, for me at least, takes it out of the realm of being counter-intuitive.
zengrifter
Banned
Okay, now lets see if anyone can come up with an app for it - a good prop bet. Anyone? zg
markiij
Member
This is a very old math "trick". It's better to switch cause the chance you made the right choise the first time is only 33.333%. After you switch it'll become 66.666%. It's hard to explain why but I always use the following example:
If there were 1000 doors and you would have to choose one the change of hitting the winning door is only 1/1000. Now, the quizmaster opens all other doors exept one (which he basically does in the 3-door quiz too only there's only one other door he can open). Now, he asks you if you'd like to switch. Of course the answer is YES. The chance of winning if you switched, now increased to 99.9%!
Hope this helps you a bit
.
If there were 1000 doors and you would have to choose one the change of hitting the winning door is only 1/1000. Now, the quizmaster opens all other doors exept one (which he basically does in the 3-door quiz too only there's only one other door he can open). Now, he asks you if you'd like to switch. Of course the answer is YES. The chance of winning if you switched, now increased to 99.9%!
Hope this helps you a bit
.