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Posers and Puzzles

Posers and Puzzles

  1. 05 Jan '07 13:14
    You're on a game show in front of three doors, behind two of them there are goats and behind the other one is a car. You must pick a door and win whatever is behind. After you pick a door (say #1) the host opens another door (say #3) and shows a goat, he then offers you the chance to switch doors (to #2 instead of #1)... should you switch? (demonstrate why or why not)

    If you want to get technical here are the rules for the game show:
    - Always two goats and 1 car.
    - Host MUST always open a door containing a goat after the player has picked a door.
    - If the player initially chose the car door, the door uncovered by the host is random.
  2. Standard member Palynka
    Upward Spiral
    05 Jan '07 13:59
    The car is either there or it's not, so it's 50/50 (or roughly the same probability of God existing).
  3. Standard member HandyAndy
    Non sum qualis eram
    05 Jan '07 14:30
    It behooves you to switch. Your initial chances of selecting a door with a goat are 2 in 3.. you are twice as likely to have a goat than the car. Then, the host shows you the other goat. Therefore, switching brings you to the car. If you initially chose the door with the car (a 1 in 3 chance) you lose by switching. Changing doors gives you a 66.6% chance of winning the car.
  4. Standard member celticcountry
    Copyright ©2001-2006
    05 Jan '07 14:41
    You MUST switch every time.

    your first choice is a 3:1 ratio.

    your "switched" choice is 2:1 ratio.

    This is prooved if you think of 100 doors , 99 containing goats and 1 containing a car.

    If you choose 1 door and the other 98 doors are opened to show goats. A switch is in your favour (favor) . you win 99 times and lose one time.
  5. Standard member celticcountry
    Copyright ©2001-2006
    05 Jan '07 20:01
    switch


    I can prove it.....
  6. Standard member EAPOE
    Earl of Rochester
    06 Jan '07 00:28 / 2 edits
    Originally posted by trad
    You're on a game show in front of three doors, behind two of them there are goats and behind the other one is a car. You must pick a door and win whatever is behind. After you pick a door (say #1) the host opens another door (say #3) and shows a goat, he then offers you the chance to switch doors (to #2 instead of #1)... should you switch? (demonstrate why or w ...[text shortened]... door.
    - If the player initially chose the car door, the door uncovered by the host is random.
    Ah ha a nice counter intuitive problem that illustrates the hard wierd non logical way in which our minds work (or not work as is more reasonable to say).

    The first choice gives us a 1in 3 probability of choosing the car. Then through being presented with the option of switching we reason intuitivly that we now have a 50:50 probability in choosing the car.

    However the true probability(at least in a mathematical definition of the system) is 2 in 3.

    We have simply choosen 2 out of the 3 boxes.

    So we should always switch.

    You have all fallen into the intuitive mistake when visualising the problem. . . . .In saying that the final probability is 50:50 by breaking the problem into two distinct choices.
  7. Subscriber AThousandYoung
    It's about respect
    06 Jan '07 00:51
    Goats?
  8. Standard member EAPOE
    Earl of Rochester
    06 Jan '07 00:53
    Originally posted by AThousandYoung
    Goats?
    If you were after the goat you would be onto a winner.
  9. Standard member HandyAndy
    Non sum qualis eram
    06 Jan '07 01:05
    Originally posted by EAPOE
    You have all fallen into the intuitive mistake when visualising the problem. . . . .In saying that the final probability is 50:50 by breaking the problem into two distinct choices.
    Some, but not all... see my post above.
  10. Standard member EAPOE
    Earl of Rochester
    06 Jan '07 01:42
    Originally posted by HandyAndy
    Some, but not all... see my post above.
    Indeed you were spot on.
  11. Standard member HandyAndy
    Non sum qualis eram
    06 Jan '07 02:00
    Originally posted by EAPOE
    Indeed you were spot on.
    Thank you, Earl. This is the famous and controversial Monty Hall Problem, inspired by the TV game show Let's Make a Deal. If anyone is interested, the literature is enormous.
  12. Standard member EAPOE
    Earl of Rochester
    06 Jan '07 03:14
    Originally posted by HandyAndy
    Thank you, Earl. This is the famous and controversial Monty Hall Problem, inspired by the TV game show Let's Make a Deal. If anyone is interested, the literature is enormous.
    Nice one. . . .
  13. 06 Jan '07 14:03
    Okay I've read this before and I understand the calculations above. But still it puzzles me.

    Let's say the player picks door 1. Then door 3 with a goat is opened. So after that he has a higher chance if he switches to door 2. But just before he makes a decision, he gets replaced by another candidate who can choose from the remaining doors 1 and 2. So the new player also has a higher chance if he picks door 2? Or only if he has witnessed what happened before? Witnessing or not, won't change what is behind door 2.
  14. Standard member HandyAndy
    Non sum qualis eram
    06 Jan '07 16:48
    Originally posted by crazyblue
    Okay I've read this before and I understand the calculations above. But still it puzzles me.

    Let's say the player picks door 1. Then door 3 with a goat is opened. So after that he has a higher chance if he switches to door 2. But just before he makes a decision, he gets replaced by another candidate who can choose from the remaining doors 1 and 2. So th ...[text shortened]... he has witnessed what happened before? Witnessing or not, won't change what is behind door 2.
    A new contestant, with no knowledge of what transpired, is choosing between two doors, one with a goat and one with a car. His/her chance is 50%. If the new contestant witnessed what transpired, his/her chance to win the car by switching doors is 66.6% -- the same as the original contestant.
  15. 07 Jan '07 01:57 / 1 edit
    Say you pick door #1

    If car is behind:
    1 - switch and you lose
    2 - switch and you win (because host would open #3)
    3 - switch and you win (because host would open #2)

    The key is that the host knows where the car is, and always opens a losing door.


    If you stay with # 1, however:

    If car is behind:
    1- stay and you win
    2 - stay and you lose
    3 - stay and you lose

    If you switch, you win 2 out of 3 times.
    If you stay with door #1, you only win 1 out of 3 times.


    So yes, you should switch.