more maths tasks :D
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20 Nov 2007, 00:16
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Journals
Hi, here one more task for u from the same maths field: kombinatorik:
behind one of 3 doors is standing an auto. To win this u have to open the right door. After ur decision to open one of the doors, where u think the auto is standig, will be opened another door, where is no auto. That means there ar staying 2 doors closed, so that behind one is standing a car. Whats better, now to change another of 2 left doors and open it or stay by the door u called at the beginning and open it now?
Where do u have more chances to win?
behind one of 3 doors is standing an auto. To win this u have to open the right door. After ur decision to open one of the doors, where u think the auto is standig, will be opened another door, where is no auto. That means there ar staying 2 doors closed, so that behind one is standing a car. Whats better, now to change another of 2 left doors and open it or stay by the door u called at the beginning and open it now?
Where do u have more chances to win?
"behind one of 3 doors is standing an auto." - he simply replaced German words with English without formatting order of verb, noun etc. you know, this sentence would be good in Polish language when you only replace English words with Polish ones. "za jednym z trzech drzwi[behind ...] stoi[verb] samochod[car]. While in English, such order is inacceptable.
Dont look at him through a very critic eye, he tries his best.
besides i do understand, his English is better then my French/German :D
- You have three doors, only one has a car behind it.
- You get to choose one door, but you dont' open it.
- From the remaining two doors, one of them opens. It is empty.
- You now have two doors: the one you have chosen initially, and another one that stayed closed.
- You can keep your choice, or change it.
- What do you do? (You want the car)
Furthermore it might be more obvious when thinking of f.e. 1000 doors and eliminating 999 (one after another) instead of 3 doors and eliminating 2.
So, in the end you have 2 doors, one with 1/1000 and one with 999/1000.
- thus, the other two doors have a probability of 2/3, combined. B+C = 2/3
- one of them opens, so its probability to be the good door goes to zero (it's bad, you see it) B = 0
- but the total probability of the other two doors still has to be 2/3, thus the remaining door gets the whole 2/3. C = 2/3
So you choose C, that is, change your pick.
But the car can also be at the other door. 66% chances.
If you would agree on the numbers, would you choose to keep the lower changes?
now, ok, im realising that B & C have 66% chances. so i bet on B(or C) but then, i realise that i have 66% chances @ A and C ! so i change it again, and again... even if one door would be eliminated, i could still changing it for ages :P
masz do wyboru 3 drzwi. wybierasz A. szansa ze tam jest auto wynosi 33%. wiec szansa, ze tam auta nie ma wynosi 66%. Zygmunt Heizer otwiera drzwi C. W nich jest koza, czyli ze nie wygrales. I teraz zostaja ci drzwi A i B. Heizer pozwala ci zmienic zdanie ale nie musisz. Zostajesz przy A czy zmieniasz na B?
I teraz, wracamy do poczatku: jaka szansa byla ze w A jest auto? 33%. Bardziej prawdopobone wiec ze auta nie ma w A, tylko jest w B lub C. Ale wiesz juz ze w C auta nie ma bo zostaly te drzwi otwarte. Tak wiec szansa ze w B jest auto wynosi 66%.
I powtarzam to co powiedzial quad. To ze szanse ze w A jest auto wynosi 33%, a w B 66% nie oznacza, ze w A nie ma auta. Ono moze byc w obu drzwiach, tyle ze wedlug logiki wieksza szansa jest ze auto jest w drzwiach B. Wiec lepiej zmienic drzwi.
Ale oznacza to tyle ze jak masz pecha to i tak czy zmienisz czy nie, pewnie auto bedzie w tych drugich drzwiach :). To tak jak lotto, mozesz kupic 2 kupony, zwiekszasz wec swoje szanse dwukrotnie, ale co z tego skoro i tak nie wygrasz^^
can you get it now? :)
If the first door you picked was no car, would they tell you?
Otherwise the first choise was quite useless, in the real end it is a 50/50 question then.
A B C the doors.
You pick A. 33% Probability.
The probability of the car being at B or C is 66%.
B opens, no car there.
The probability of the car being at B or C is still 66%.
No car at B.
Thus the probability of the car to be at C is 66%.
You eliminate a door that has no car(you wrote) and therefore there is only 2 doors to chose from at the end.
its not a theorie, its praxis,
it was tested over 10 years long on computers in simulationprogramms to show that u win with 2/3 by changing the door, what u can simply see by viewing all the possible cases
many ppl here claim it's correct, but only coz they read wikipedia, in fact, this is all just bullshit, the probabilities stay 50/50, since it's a game of luck
At first the probability is 1/3 that you have a good 1, then 1 gets eliminated, so the probability is 1/2, not 2/3, since there aren't 3 doors left ;)
imagine the car is behind the door A, so:
u change the door A and change it, so u lose!
u change the door B and change it, sou win!
u change the door C and change it, so u win!
u see in 2 of 3 situations u win if u change ur door, thats why the chances of changing the door ar 2/3 and staying at the door 1/3
Maths is always right, btw ;) even in practice.
Think of it this way:
The car is behind B
You get to choose one out of three doors, you choose A
They eliminate C, and since you know there's nothing there, you can choose from A or C, wich make the chances 50/50, since now you have to pick one out of two instead of one out of three
probably from math point of view they are right, but from clearly logical point of view this is one big bullshit
imagine the car is behind the door A, so:
u change the door A and change it, so u lose!
u change the door B and change it, sou win!
u change the door C and change it, so u win!
car is on A
- if you choose A then switch to B or C - you lose.
- if you choose B then switch to A - you win (C is thrown out)
- if you choose C then switch to A - you win (B is thrown out).
pls get the real meaning of POSSIBILITY, PROBABILITY, CHANCE. Your chances are indeed bigger if you switch, but that doesnt mean you will actually win since its about a luck and you play it only once. but your chances are bigger, if you played 3 times, then you should win 2 out of 3 games - and guess what, thats not sure, if you have bad luck you can lose 3 out of 3 games, but again.... you have bigger chances to win.
dont know how about you, but i prefer having bigger chance than lower chance.
You choose A. B/C eliminated. You change. You loose.
You choose B. C eliminated. You change. You win.
You choose C. B eliminated. You change. You win.
Suppose car is at B.
You choose A. C eliminated. You change. You win.
You choose B. A/C eliminated. You change. You loose.
You choose C. A eliminated. You change. You win.
Suppose car is at C.
You choose A. B eliminated. You change. You win.
You choose B. A eliminated. You change. You win.
You choose C. A/C eliminated. You change. You loose.
Statistically, there are 9 possible situations. If you make the change, in 6 of them you win. That's a 66% probability to win!
If you don't make the change. There will still be 9 possible situations. Only 3 lucky. There's a 33% probability to win.
K-pish?
3 ways of hiding one car behind one door out of three
3 ways of picking a door.
Are you really trying to argue there are 12 ways to place a car behind one of three doors and then choose one of the three?
its all about the meaning of possibility. if you was born under the dark star and has got bad luck through all the life, you prolly wont win anyways :D
I was talking about the probabilities, not the placement nor the way you choose the doors. You just cram to probabilities of having it wrong in one choise ;)
but im off to sleep now, cya
If you pick the right case from the start, it's just one case, doesn't matter which of the other two doors opens, you have already made a choice and thus selected one of the 9 possible cases (actually 3, since there is the restriction of the car really being there).
i completely get mathemacial view of it, but slajdan showed me that problem i asked at the start of this post.
My 3 options: Sleep [x]
Figure it out [ ]
Eat [ ]
This would have to be done in real life rather than on a PC for ten years, where the equation was probably terribly flawed.
There is 2 doors to choose from, don't need much logic to understand the possibility of a win.
that means such a strategy, where u have more chances to win, u should play. And this one is from me where the chances ar 2/3. With ur strategy u have the chances 1/1.
whats difficult to understand that? :)
So I know the answer only because of that. :P
(yes I suck at maths please tell me what I'm doing wrong)
so, u pick a door with chances 1/3, but after u are given an opportunity to change it u got chances 1-1/3 = 2/3 if u make the change. Because it's like u would have selected both the other doors in the beginning (just the surely wrong one has been cut off now)
it's not easy to explain, but if u think about it a bit, it should be clear :)
Past means nothing in a case like this because you don't realy learn anything from them removing a door, if you picked the right one in the beginning or not does not matter.
2/3 - 1/1 = 1/2 because you have to remove one from both rows, simple as that.
I know that it is "correct" by some stanards counted mathematically, but practically it is not.
rätt 1/3 + fel 1/3 + fel 1/3 = 3/3 , 3/3 - 1 fel (alltså 33%), 33/2 = 17 , 33 + 17 & 33 + 17 = 50 & 50
if the host didnt know where is the car, then its 50/50. but if host knows, its 2/3.
rätt 1/3 + fel 1/3 + fel 1/3 = 3/3 , 3/3 - 1 fel (alltså 33%), 33/2 = 17 , 33 + 17 & 33 + 17 = 50 & 50
I understand you, but it is still wrong. The 3 will turn to 2.
is this what you say?
that the anser but its all about luck in real life i would stay with the door i picked =D