男女問題，請見圖girlboy
為何&quo... [/quote]
我現在已明白你現在的觀點,我一直沒有把問題一變成問題二,你現在亦認同天才換的決擇是對的,在樓主playbr2的問題上,當誤會又好,當意見分歧都好,應該已經解決,言語上曾有冒犯的地方,請見諒,[color=Red][size=6]不像某人[color=White]ronja[/color]存心找荐,一直回避及捉人小錯為樂,連問題問乜都未清楚,背後假設又唔知,別人提出問題又答唔到,更犯上兩大logic錯誤而不自知(用錯can't you read?同埋唔知總勝出機率受轉換選擇機會率影響)[/size][/color]
至於男女問題,不失一般性,在女女case中,設女1是已知,"已知其中一個是女孩子,另一個是女的機率"跟"已知其中一個是女孩子,生下女女的機率"的確是一樣的,但是,"已知其中一個是女孩子,另一個是男的機率"跟"已知其中一個是女孩子,生下男女的機率"是不同,前者答案是1/2,後者答案是1/4,假設男女出生機率是1:1之比情況下(這假設已被默認),在排列的角度上,出現男女和出現女男機率相同沒有錯,但是,出現男女只佔1/4,出現女男亦佔1/4,出現女女是1/2(因為包括女1女2和女2女1),但用排列的角度解決這問題較複雜,用我一直提出的組合角度較簡單,男女=女男,女1女2=女2女1,很快就可計出相同答案(已知其中一個是女孩子,另一個是女的機率是1/2),因此,圖girlboy是有問題,因為怱視了女1女2,女2女1都是女女,至於另外兩幅圖,應該冇問題
[[i] Last edited by kaichun88 on 2007-3-18 at 07:17 PM [/i]] [quote]Originally posted by [i]hold_find[/i] at 2007-3-18 15:20:
男女問題，請見圖girlboy
為何&quo... [/quote]
你真係好心機,我相信我明你你明我講乜,咁而家惟有你教佢數學,我教佢英文同禮貌:dev: [quote]Originally posted by [i]ronja[/i] at 2007-3-18 07:22 PM:
你真係好心機,我相信我明你你明我講乜,咁而家惟有你教佢數學,我教佢英文同禮貌:dev: [/quote]
好野喎,一句我明你你明我講乜,拖人地落水紌可作回避別人問題的籍口,好,我教番你做人的道理,逃避是不能解決問題,勇於認錯,才是男子漢的表現,況且,你教人英文同禮貌?咪誤人子弟la:D:D:D(一句Can't you read已經顯示你有幾勁) [quote]Originally posted by [i]ronja[/i] at 2007-3-18 07:22 PM:
你真係好心機,我相信我明你你明我講乜,咁而家惟有你教佢數學,我教佢英文同禮貌:dev: [/quote]
你認為我計錯數,咪用數學理據去證明我錯,況且,我與hold_find只昰互相討論,冇邊個教邊個,你叫人去教我數學,不只對我不敬,也是對hold_find不敬,你還有資格教人禮貌?
更正#156
原文提到"假如問有兩人排隊,已知其中一個是女孩子,列隊女女的機率,才是你1/3的答案"其實不然,答案亦是1/2,因為男女與女女機率不同(1/4,1/2),男女與女男才是一樣(1/4)
答案是1/2,希望不會另hold_find產生誤會,特此更正 n = 開始時幾多度門選擇, 由始至終不會改變, 車亦一早放係其中一度門後, 由始至終不會改變
唔換門中車機會 = 1 / n
換門中車機會 = 1 - (1 / n)
即係:
3 度門, 唔換門中車機會 = 1/3
換門中車機會 = 2/3
4 度門, 唔換門中車機會 = 1/4
換門中車機會 = 3/4
100 度門, 唔換門中車機會 = 1/100
換門中車機會 = 99/100
10000 度門, 唔換門中車機會 = 1/10000
換門中車機會 = 9999/10000
自己唔中即係人地中, 自己中的機會愈細即係人地中的機會愈大, 點解唔換呢 ? 根本唔關開唔開門事
[[i] Last edited by netharmon on 2007-3-20 at 01:23 AM [/i]] [b][color=Blue]初步加分完畢
討論繼續
PlayPlay大師 整到敢多會員頭都痛晒，
應扣20分[/color][/b]:dev::dev::dev:
[#124，#127，#132，#133]:dev:
[[i] Last edited by Triangel on 2007-3-20 at 10:29 AM [/i]] 天才真係天才...
100萬度門,你無可能一下就中,
假設佢唔中吧,e個機會相當大,
之後佢幫你開左999998度空門,
咁最後個度門有車ge機會亦相當大.
我覺得如果,
你就咁叫主持將開剩一度空門同一度有車ge門,
比個5050ge機會你,你揀到車ge機會細好多 支持天才，不過兩種想法也有道理.............. 我覺得係1/2
因為你第一次選擇根本無意義...
第二次先係重點...
要慢慢爬格子看 我唔認同.......我覺得點都係1/2 機會..... **** 作者被禁止或刪除 內容自動屏蔽 **** the answer should be 1/2. It is because no matter what, the empty will be opened after your choice. That means one of the empty door has been excluded. The question is that the probablity of your first choice and second choice is the same that is 1/2 and they are independ. The probability will not increase as you will not be able to choose the one to be opened with car !
I want to change me answer. It should be 1/3. It is because of in the first choice, the probability of selecting the door with car is 1/3. After opened a empty door, the probability seems shifted to the rest of the two door to become 1/2 but the probability of your original choice is still 1/3. So change of selection is a good decision.
To illustrate the answer more clearly. Say there are 10 doors. You have made your selection and the probability to choose the one with car is 1/10. The host opened the rest of 8 empty door. If the selection remain unchange, the probability is still 1/10 as it does not change even some empty doors are opened. So, change of selection is correct. I support the brillant with IQ228.
[[i] Last edited by hkcchk on 2007-3-20 at 11:19 PM [/i]] [quote]Originally posted by [i]kaichun88[/i] at 2007-3-16 05:26 PM:
The problem has mentioned that ... [/quote]
Right, but i am saying "AFTER" the host opening the door but "WITHOUT" knowingly which door is the winning door [quote]Originally posted by [i]sakura310[/i] at 2007-3-14 07:42 PM:
playplay,正因為這樣,我才不要將正確...話唔定佢會話唔識計
[color=Red]例如3x=6,求2x=?[/color]
佢可能會將2=x代入去2x,得到xx,佢會話解唔到答案都唔定 XD [/quote]
2x點都係4, 唔會錯 :D I give up :cry: the more i read the more confuse i get..... :help:
i am juz a dumb one:) [quote]Originally posted by [i]lilirayhk[/i] at 2007-3-14 01:15 PM:
答案一定係1/3。我想問題重點係在於 ... [/quote]
唔係呀，你講得好清楚呢。如果玩遊戲0既人一開始就決定會揀主持開剩的門，咁只要佢一開始揀中空的門就會中奬，而空的門有2所以機會係2/3，如果佢堅持要自己開始時所揀的門，咁機會只有1/3。
[[i] Last edited by 279x on 2007-3-21 at 01:49 AM [/i]] [quote]Originally posted by [i]netharmon[/i] at 2007-3-20 01:15 AM:
n = 開始時幾多度門選擇, 由始至終不... [/quote]
要係無論有幾多度門，主持都係開淨兩度門，你的答案先至正確。 改唔改變選擇, 都改唔到機率啦~
已經由一開始1/3, 開左1道門之後, 就變左1/2~
所以中獎機會提高,
並唔係因為改變選擇,
係因為開左道門~ ﹙肯定左1道門無） 越講越複雜
我越睇越唔明
個選門遊戲係1/3中
如果唔換，即係第2次的take走一個唔中的門變得無意義
因為如果唔換(保持第1選擇)，中既機會會同未take走之前一樣，都係1/3
但如果玩者存心會換
o甘第1次選門的時候，就不是要選一個會中的門了，而是去選一個不會中的門(2/3機會)
因為選了個不中的，在第2 round主持take走一個唔中的門後
就可以去換剩下的中的那一個
所以我覺得換，就有2/3機會中架車
如此類推，如果有100個門，第1 round選個不中的就是99/100了
絕對比選中的1/100容易。
我是從玩者(選門的人)的角度去看
如果唔係一早決定第2 round換唔換
等到第2 round先再諗換唔換，感覺上真係好似變左1/2