stinkynuts said:
This is an extremely devious question, which is very counter intuitive. It would appear that the odds would be 50/50, but in fact, it is advantageous to switch.
This problem was successful answered by Marilyn Savant (who has the world's highest IQ). She answered the question in her column, only to be criticized and ridiculed by math professors and others.
To illustrate the point, think of the problem like this. Instead of three doors, let's say there were 100 million doors. You choose one. The host then opens all 99,999,998 remaining doors, leaving one other door unopened, along with the one you chose. Obviously, it's to your advantage to switch, since you couldn't possibly have guessed the correct door the first time (unless you were extremely lucky). The door that the host left unopened is highly likely to be the correct door.
By the way, in the original problem, it's a goat not a sheep.
First of all let's be very clear (or I have it all wrong): when you're standing there, with sweat running, there are two closed doors, one of which you've picked and the one door (or 99,999,998) the host opened, revealing no prize. Why is it obvious "it's to your advantage to switch, since you couldn't possibly have guessed the correct door the first time (unless you were extremely lucky)." It sure isn't obvious to me, because the amount of luck required never changed.
Why "extremely lucky" instead of just lucky? Surely the 'extremely lucky' person is the host who opened all those doors w/ the audience going wild and wilder and never found the car. Or could it be—No, never—fixed? How can one prove the odds changed anyway?
This is a thought problem and here's another way to think of it: any given door has a prize or has none. Either/or. 50/50. From the house's point of view the odds change by increasing the number of prizeless doors. Increasing them, then taking them out of play by opening however many is showmanship but doesn't change
the odds you face when deciding which are the same odds you face calling a coin toss: you can only call it one of two ways and have no evidence (including permutations, combinations and odds) to help you. That door either has a prize, or it has a sheep. Pick one.
The suspiciously named Ms. Savant may think that proving some number of doors had no prize demonstrates that your original choice couldn't be right, but the only 'proof' offered here is 'you couldn't be so lucky'. Given that I'm either gonna be right or wrong, I'd say the 99,998 successive wrong doors just confirm that I picked right the first time.
----------------S P O I L E R W A R N I N G-----------------------------
PS: Now that I've spoiled all the fun by Googling the answer, may I point out how vital a full, concise statement of the problem is. The element that wasn't clear here was that the host will not show you the prize when opening 'his' door(s).
------------------I D I D W A R N Y O U------------------
Though the odds are you picked a goat door, the only time switching doors fails is if you were lucky enough to pick the car door right off. But in all other instances—and when the game began the odds were stacked against picking the car—the pool of goat doors is being reduced by the host, until just the two doors are left: your original (which had only a 1 in however-many-doors chance of being right) and the last door, which for sure has a one in two chance now.
Nice puzzle, thanks for posting it
mmouse. And the Google hits are very entertaining
