Dear 3ege class students!
Here is the 2nd stimulus for your discussion of the Haddon's novel!
Do you agree with Christopher's perception of The Monty Hall Problem (page 78-82)?
What is the possibility to win the car, in your opinion? Justify it logically.
Mark Haddon, the author
15 comments:
Whit Christopher perceprion i agree, but it is very coplicated to understand.
My opinion to win the car are that if you play that game you need lock, and you must use your intuition.
Rok Dremelj, 3.EGE
As we can see is Christopher very smart boy. I have read his possibility over and over again and i could not find out how is he come on this solution. I am not so smart but i agree with him. It is possible that this is the way to get a car. and it is also another way to get him... with luck.
ANŽE PERČIČ, 3.EGE
It is not easy to understand. I should have some hours of maths may be this would be good:D However his theory of possibillity prabably has one good basic.
The Monty Hall Problem does seem to confuse people. Is a truly amazing lesson on statistic. I agree with Christopher’s perception but this is difficult to understand. I am guided
by intuition and not mathemathics.
Tanja Poklukar, 3.ege
I Agree wiht Christopher's Idea or perception. On The first look it seems very complicated, but it works so i agree with Cristopher.On the other hand I agree with Roky-baloboa that you must use your intuition if you want a win this game.
I don't agree with Christopher's theory of Monty Hall problem.
Because if you pick door one, and then you change your mind, there is just 1/3 possibility of winning the car, becouse car might be behind first door.
I really don't understand Cristopher's theory.
I agree with C. perceprion!
He is very good in mathematics, so that´s why he can rackon up how many possibility is to win a car.
But I think that this is just luck.
Irena Koblar
I think that if there are 3 door there is 1/3 posibility that you get a car but when you open one door and you get nothing than you have 2 doors left and is is possibility 50:50 that you find out behind which door is a car.
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His perception is interesting that I must addmit. But if I would play a game like this I wouldn’t rely on mathematic. I think that in such games all ou need is luck. But if I tried to do some maths I would say that possability to pick the wright door is 33.3%. If it isn’t soo…than whaat could I do, I’m not soo good at maths that I could argue about that!:)
I agree with Christopher, even i don't understand his teory...he know:)
With Christopher perception of The Monty Hall Problem i agree, beacuse it is very difficult to understand. I must say that Christopher is very intelligent and smart boy.One of the most important chances is that you need a good luck. On the other hand is that is possible to win the car if there are on example: 3 doors you have just 33,3%(or 1/3) chanses to win it. But thefirst thing is LUCK ;)
I don't agree with Christopher's theory of winning a car. In my opinion lucky is the most important factor in any kind of lucky-games. I often buy a lottery ticket and tried to find a winning numbers but in the end I find out that without luck there is no chance for winning the big prize.:)
Christopher is right in his analysis f the Monty Hall Problem. I bet my life on it.
Probability of choosing car with 1st pick = 1/3. One empty door is then removed to leave 2 doors.
Therefore probability of the other door containing car is (1-1/3)=2/3 since the car MUST be behind one of the remaining doors.
Therefore the other door is twice as likely to contain the car
Therefore change doors!
Still baffled? OK, let's assume there are 100 doors (1 car door, 99 empty doors). Pick a door. Game show host removes 98 empty doors. Change you mind? Hell yes! Since the first door you pick has probability of winning = 1/100 therefore other door has probability of winning 99/100.
Still baffled? Then there's no mathematical hope for you!
Probability of choosing car = 1/3
Therefore probability of the other door containing car is (1-1/3)=2/3
Therefore the other door is twice as likely to contain the car
Therefore change doors!
Still baffled? OK, lets assume there are 100 doors (1 car dor, 99 empty doors). Pick a door. Game show host removes 98 empty doors. Change you mind? Hell yes! Since the first door you pick has probability f winning = 1/100 therefore other door has probability of winning 99/100.
Still baffled? Then there's no mathematical hope for you!
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