vos Savant's explanation is faulty when he "accidentally(?)" collapsed line [1] & [2] into one arrangement, while keeping [3] & [4] as different arrangements.
Suppose Goat-A and Goat-B are NOT twins, all arrangements are:
Please copy out the text and format it with font to be "Courier New" to see the proper table layout. Then compare the table with the one in the wiki webpage, directly under "Solution" and "Vos Savant".
The problem in vos Savant's solution is that he assumed the "thing" behind "door #1" was randomly drawn. But it was not true for the "game show". (The probability of 1/3 & 2/3 would have been true if the game was a card game, and the "thing" behind door # was randomly drawn in front of the card game player. The probability of 1/3 & 2/3 was directly due to the distribution of the winning card, rather than being caused by a "switching" of game doors.)
[When the game show started, there is only 2 possibilities behind door #1 : "a car" or "a goat". (NOT "a car" or "a goat" or "another goat".) Therefore we may artificially consider both "goats" are identical twins, in spotting if the solution has any faults.) ]
To simplify the game, we may allow the player to choose to keep the prize behind door #1, or switch and keep the prizes behind both of door #2 and #3. It would become obvious that the player could only win a "car" or a "goat" (but not both) behind door #1, or if switching, "goat+goat" or "goat+car" behind door #2 & #3. Therefore, it will give us only line [1] and [3], as [2] is only a repeat of [1], and [4] a repeat of [3]. -> Thus the probability is still 1/2.
馬年得意.
回覆刪除好一句「四蹄踐落英」！梁振英必要落台！
回覆刪除振拔鳳來儀
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下愚揚蓄念
臺宦仰秣時
快驥嘶瑞意
速頌福禔辭
莫遺如意句
遲巧失熙怡
風起迎驍驥
回覆刪除四蹄捲白梨
瀟灑英雄血
矯捷顧盼媚
祝: 新春愉快
馬年博事通
回覆刪除存異也求同
事了拂衣去
魚樂定無窮
葉蒨雯的祝福也是予愛聽之曲
Referring to http://en.wikipedia.org/wiki/Monty_Hall_problem
回覆刪除For the Monty Hall Problem, the chance should still be 1/2 [rather than 1/3 (if staying) and 2/3 (if switching)].
Explanation:
Consider Goat-A and Goat-B are twins, then Door #1 does not have to hold Goat-B.
[0].Door.#1..Door.#2.........Door.#3.........if.stay.w/#1..if.switch.door
[1].Car......Goat-A..........Opened[Goat-B]..Car...........Goat-A
[2].Car......Opened[Goat-A]..Goat-B..........Car...........Goat-B
[3].Goat-A...Car.............Opened[Goat-B]..Goat-A........Car
[4].Goat-A...Opened[Goat-B]..Car.............Goat-A........Car
Thus the chance is 1/2.
vos Savant's explanation is faulty when he "accidentally(?)" collapsed line [1] & [2] into one arrangement, while keeping [3] & [4] as different arrangements.
Suppose Goat-A and Goat-B are NOT twins, all arrangements are:
Door.#1..Door.#2.........Door.#3.........if.stay.w/#1..if.switch.door
Car......Goat-A..........Opened[Goat-B]..Car...........Goat-A
Car......Goat-B..........Opened[Goat-A]..Car...........Goat-B
Car......Opened[Goat-A]..Goat-B..........Car...........Goat-B
Car......Opened[Goat-B]..Goat-A..........Car...........Goat-A
Goat-A...Car.............Opened[Goat-B]..Goat-A........Car
Goat-A...Opened[Goat-B]..Car.............Goat-A........Car
Goat-B...Car.............Opened[Goat-A]..Goat-B........Car
Goat-B...Opened[Goat-A]..Car.............Goat-B........Car
The chance is also 1/2.
Sorry, I am not sure I understand your analysis.
刪除Please copy out the text and format it with font to be "Courier New" to see the proper table layout. Then compare the table with the one in the wiki webpage, directly under "Solution" and "Vos Savant".
刪除The problem in vos Savant's solution is that he assumed the "thing" behind "door #1" was randomly drawn. But it was not true for the "game show". (The probability of 1/3 & 2/3 would have been true if the game was a card game, and the "thing" behind door # was randomly drawn in front of the card game player. The probability of 1/3 & 2/3 was directly due to the distribution of the winning card, rather than being caused by a "switching" of game doors.)
[When the game show started, there is only 2 possibilities behind door #1 : "a car" or "a goat". (NOT "a car" or "a goat" or "another goat".) Therefore we may artificially consider both "goats" are identical twins, in spotting if the solution has any faults.) ]
To simplify the game, we may allow the player to choose to keep the prize behind door #1, or switch and keep the prizes behind both of door #2 and #3. It would become obvious that the player could only win a "car" or a "goat" (but not both) behind door #1, or if switching, "goat+goat" or "goat+car" behind door #2 & #3. Therefore, it will give us only line [1] and [3], as [2] is only a repeat of [1], and [4] a repeat of [3]. -> Thus the probability is still 1/2.
I don't think vos Savant made the mistake(s) you think she made, but I don't have time or the expertise to discuss this problem with you. You can read vos Savan't account of the story behind the Monty Hall problem and how her answer was empirically tested in this book (see pp.199-209): http://www.amazon.com/Ask-Marilyn-Letters-Published-Magazine/dp/0312081367/ref=sr_1_2?s=books&ie=UTF8&qid=1391331285&sr=1-2
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