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桜 & 恋

Sunday, July 09, 2006

food for thought

you are on e game show on television. on this game show e idea is to win a car as a prize. e game show host shows u 3 doors, he says tt there is a car behind 1 of the doors and there are goats behind the other 2 doors. he asks u to pick a door. you pick a door but e door is not opened. then e host opens 1 of e doors you didnt pick to show a goat(because he knows wat is behind e doors).then he says tt u have 1 final chance to chane ur mind b4 e doors r opened and u get a car or a goat so he asks u if u wan to change ur mind and pick e other unopened door instead, what shld u do?

Marilyn vos Savent said tt u should always change and pick e final door because e chances are 2/3 tt there will be a car behind e door.

but if u use ur intution u think tt e chances is 50-50 because u think there is equal chance tt the car is behind ani door..

many said tt Marilyn vos Savent was wrong...

here are wat they said

-'im very concerned with e general public's lack of mathematical skills,please help by confessing your errors' ROBERTS SACHS, PH.D..., GEORGE MASON UNIVERSITY

-'you r utterly incorrect... how many irate mathematicains are needed to get you to change ur mind?'E, RAY BOBO. PH.D., GEORGETOWN UNIVERSITY

below is pictorial explaination.. u decide whether he right or wrong or both...

solution


posted by kai/kaispiral at Sunday, July 09, 2006

8 Comments:

  • First picture on your blog and you post some boring ole' mathematical picture... (-_-lll) IMHO, no matter what you choose, no matter how great or unlikely your chances are, whether you win or not really has nothing to do with your choice. If you win, it was meant to be, if not, it was also meant to be. Things NEVER go 100% the way you want it, no matter how properly you plan for it, calculate the odds for it, minimise the chances of anything happening. EVERYTHING depends on what i like to call the random variable, that which we have no control over. As such, in conclusion... there is no right and wrong... only fate!

    By Anonymous Anonymous, At 10:13 PM  

  • Since the host eliminates one door for you, doesn't it mean then, that you really only have 2 choices? Goat or car, behind the door? In other words, shouldn't one "goat" tree of your diagram be eliminated since the door you choose is already fixed and behind it is only either a goat or a car?
    All games that involve gambling are made in such a way that the chances of the dealer winning are superior to that of whoever plays it. But in this case, i would conclude that the chances here ARE in fact 50/50.

    By Anonymous Anonymous, At 10:22 PM  

  • agrees with false lies!

    By Anonymous Anonymous, At 11:51 PM  

  • I would like to offer an alternative way at looking at this...

    The two goats, being identical, makes it such that picking one or the other, both cases are to be lumped under one scenario: you picked a goat. As such, the probability of choosing a car on your first try is 1/3, whilst that of picking a goat is 2/3.

    Once the host reveals the other goat, your chances of getting either a goat or car is halved in either case. If you chose the goat on your first try, you have 2 choices, both with a 50/50 probability: change and get a car, or stay and get the goat. In the instance that you chose the car as your first choice, the chances are, once again 50/50 of getting it.

    So here's the solution via tree diagram. Your overall chances of winnning a car at the end of the game is
    2/3(if you chose a goat first) x 1/2 (the probability that you choose to change)+ 1/3(if you chose the car first) x 1/2(the probability that you choose to stay) = 1/2, ie 50%.

    There you have it, the chances of you winning either a car or a goat is 50/50 each.

    By Anonymous Anonymous, At 4:27 PM  

  • here another mathetical explaination(which i know very sure myself)

    let e door be called x y z

    let cX be e event tt e car is behind door x and so on

    let hX be e event tt e host opens door x and so on

    supposing tt u choose door x, e probability tt u win a car if u then switch ur choice is given by e following formula:

    P(hZ^cY)+(hY^cZ)= P(cY).P(hZ|cY)+P(c2).P(hY|cZ)=(1/3.1) +(1/3.1)=2/3

    By Blogger kai/kaispiral, At 8:24 PM  

  • actually don understand at all =.=

    By Blogger kai/kaispiral, At 8:26 PM  

  • As i said, two doors with two goats behind them, whether you pick one or the other, it doesn't matter: You picked a goat. As such, two doors have to be lumped together as on identical event...

    And how in tarnation do you "know very well myself" and yet "actually don undesrstand at all"??? -_-''' Kai-kun wa bakaaa...

    By Anonymous Anonymous, At 9:47 PM  

  • Oh, on a side note, fate is cruel so you'll most likely lose anyway 99% of the time.

    By Anonymous Anonymous, At 9:48 PM  

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