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so boring

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¥u¬Ý¸Ó§@ªÌ 109  µoªí¤_: 2007-07-21
A(goat) B(Car) C(goat)


§A¿ïA,B&C ¤¤Car o¬J probability ³£«Y1/3
¦pªG§Achoose A
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A ¤¤Car o¬J probability ³£«Y1/3
¦ÓC ¤¤Car o¬J probability ´N¥[¥ª¥hB«×
so B ¤¤Car o¬J probability «Y2/3

­ø©úo¬J¸Ü

I have another example

100 boxes,99 ­Ó¦³bomb,1 ­Ó¦³10K cash
§Achoose ¨ä¤¤1­Ó(e.g. No. 7)
¤§«á¶}¦³bomb o­Ó98­Ó(e.g. ¶}²bNo.7 ¦PNo.8) boxes
box 8 ¤¤10K cash o¬J probability (99/100)´N¤ñ box 7(1/100) ¦h99­¿

¥Î3­Óbox ¦n©ö²V²clol

get it?
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¤Þ¥Î²Ä35¼Ókylau©ó2007-07-18 18:54µoªíªº¡§¡¨:
This question is quite similar to a prob question I've seen in University
http://hkumath.hku.hk/~wkc/prob/prob.html
In a TV program, a player is asked to choose a key out of three indistinguishable keys A,B, and C. Only one of them is matched with the car, the prize of the winner Without the help of the TV program host, clearly the player only has a chance of 1/3 to get the car. The host is allowed to help the player in the following two ways: (A) The host can eliminate one wrong key from the three keys first before the player chooses one from the remaining two; or (B) the player can choose one key first without using it, then the host eliminates one wrong key from the remaining two keys. At this moment, the player is allowed to change his chosen key with the remaining one. Which option is more favorable? Why?
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¥u¬Ý¸Ó§@ªÌ 111  µoªí¤_: 2007-07-29
¤Þ¥Î²Ä102¼Óyipsun©ó2007-07-21 10:05µoªíªº¡§¡¨:
just like this question
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2/3 probability if you change

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