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ZeJayMan
the farthammer

Registered: Aug 2005
Location: Glasgow
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Oct-24-2006 00:17
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Psy-T
Melody Klein

Registered: Jan 2003
Location: Haifa
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| quote: | Originally posted by pkcRAISTLIN
im not understanding the logic youve used, with the 3/5 here. why arent the odds changed to each door being 1/3 chance? if you had five doors, and five contestants got to choose a door each, the chances of a correct guess will increase that corresponding amount, but no unopened door is ever going to have more of a chance than any other unopened door. |
the game starts with 5 possible locations for the car
you bet on two of those, 2 in 5 chance for it to be there
the three you didn't bet on have a 3 in 5 chance for the car to be there
the host opens two empty doors. the car's location does not change as a result, obviously, but, the probability of the car being in one of those 3 doors has not changed, the fact that you know it's not in the third or the fourth door tells you there's a higher probability of it being in the fifth door.
i can try to rephrase myself endlessly, but if you can't understand it from my words,, just run the damn simulations lol.
___________________
People who own my ass: Citric Acid, Boomer187, Tribu, Sand Leaper,
Jackson, venomX, jamie, Renegade, Konjin, Akridrot, Miss Bliss.
Psy-T - Down The Rabbit Hole (400minute long acid set)
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Oct-24-2006 00:17
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pkcRAISTLIN
arbiter's chief minion

Registered: Jul 2002
Location:
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| quote: | Originally posted by Psy-T
the game starts with 5 possible locations for the car
you bet on two of those, 2 in 5 chance for it to be there
the three you didn't bet on have a 3 in 5 chance for the car to be there
the host opens two empty doors. the car's location does not change as a result, obviously, but, the probability of the car being in one of those 3 doors has not changed, the fact that you know it's not in the third or the fourth door tells you there's a higher probability of it being in the fifth door.
i can try to rephrase myself endlessly, but if you can't understand it from my words,, just run the damn simulations lol. |
i understand your words, i just dont think youre right
each door has 1/5 chance. take two doors away, each door has 1/3 chance. by eliminating the two other doors, the probability isnt "transferred" over to the fifth door, each change in the doors equates to a change in ALL the doors, not just the unopened unselected one.
thus it starts 1/5, then becomes 1/4 then 1/3. since you have a free choice at the beginning, the probability for all doors remains the same.
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Oct-24-2006 00:45
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ZeJayMan
the farthammer

Registered: Aug 2005
Location: Glasgow
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Oct-24-2006 00:51
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