kaichun88

Originally posted by guswan at 2007-3-16 08:20 AM:
買六合彩,一是中,一是唔中.中獎機會1/2.
今次唔中,下次一定中.

仍在沉思中............

中獎機會1/2真的代表今次唔中,下次一定中嗎?
不論今次中或唔中,下次中獎機會也是1/2
更何況買六合彩中獎機會不是1/2吧,你以為自己是黃大仙

kaichun88

Originally posted by PG-13 at 2007-3-16 01:02 AM:


Stimulator... [url]http://www.u...

個Stimulator幾易用,睇下我的結果
                games           won                   lost
switched       100         64 (64,00 %)    36 (36,00 %)
not switched   50         14 (28,00 %)     36 (72,00 %)
total            150         78 (52,00 %)     72 (48,00 %)
實驗概率與或然率差別不大

至於那些認為勝出遊戲與在遊戲落敗機會相等的人有一個假設,就是換與不換的比率是1:1,

但問題中,觀眾有權選擇換與不換,而且根據數學,觀眾可以計出換後勝出率較不換高(2/3),

(計算方法有很多,而且已有不少人已經正確計出,本人不再詳述),正因如此,換是較有利,一一

之比假設並不成立(起碼對於文中那天才而言,1:0之比),而且,問題並沒有任何暗示或明示有

換與不換的比率是1:1之假設,缺乏這假設之支持下,勝出與落敗又怎會機會相等

kaichun88

Originally posted by PG-13 at 2007-3-16 01:17 AM:
...Without knowingly open the ( lose) door the host's probability of holding a winning door would be 1/2. Because in case of (lose, lose) with the porbability of having this set is 1/3 and in case of (win, lose) the porbability is 2/3.
So,withouth knowingly pick the set, the chance shall be [1/2x1/3 +1/2x2/3]=1/2

The problem has mentioned that the host knows what the doors without car are.As a

result,the condition for the host without knowingly opening the door is not applicable.

Besides,without knowingly open the ( lose) door the host's probability of holding a winning

door would be 1/2,Are you sure?

If the player chooses the right door,the host will not hold the right door.If the player

chooses the wrong door,there is 50 percents for the host opening the correct door(with

car).

Combining the above cases,the host's probability of holding a winning door would be

1/3x0+2/3x1/2=1/3which is equal to the player's probability of holding a winning door and

also the host's probability of opening a winning door.I hope I shall not misunderstand what

you mean.

kaichun88

Originally posted by 奇 at 2007-3-15 10:06 PM:

問題是...那麼另一個"都是女孩子"的機率是多少?
就是1/2了,如果是問組合的可能性,就1/3...

問組合的可能性也是1/2,男女與女男是重覆的,因為,當已知一個是女的情況下,男女與女男都=>另一個是男,重覆的組合當一個計算,排列(permutation)與組合(combination)是有分別的

[ Last edited by kaichun88 on 2007-3-16 at 05:44 PM ]

kaichun88

Originally posted by ronja at 2007-3-16 06:06 PM:


我根據而家大家討論個問題而去設...

我認為#116的post完全可以解決你的所謂的問題一
The reason why you asked the first question is that you assume that the ratio of switching to not switching is 1:1.However,I think your assuption is not convincing.It is because the player can choose whether he switches or not according to the probability of winning the game for the player who switches the choice.Let's take the talent as an example.The probability for him to switch choice is 100% since he think switching the choice will have a bigger chance to win the game.The probability for player to switch choice depends on how they think whether switching the choice is favourable.Do you have any evidence for supporting your assumption.If you think you do not have such an assumption.Can you explain why the probability of winning the game must be 50%?
Only when the chance of switching the choice is equal to that of not switching the choice,the general probability of winning the game is 1/3(the winning probability of not switching)x1/2+2/3(the winning probability of switching)x1/2=50%.

Actually,I think I have answer your question 1 with explanation in #116.You can of course say me not to answer or even cannot answer the questions but after reading the replies carefully please.

[ Last edited by kaichun88 on 2007-3-16 at 10:30 PM ]

kaichun88

Originally posted by playbr2 at 2007-3-15 12:52 AM:
先引文一篇.... 摘自某網友
...而換的機會率卻是2/3而不換是1/3...

認同網友的觀點,不過,為免某人誤會,應說成換而中的機會率卻是2/3而不換而中是1/3

kaichun88

Originally posted by hold_find at 2007-3-16 08:59 PM:



未睇清楚題目的是你，是你一直[...

我好好奇為何2/3中獎人士有改變選擇,假如真的有2/3中獎人士有改變選擇,我想問你,玩家轉換選擇機會率(留意,不是轉換選擇而中機會率)又是幾多?又假如改變選擇不會提高中獎機會,轉換選擇機會率(留意,不是轉換選擇而中機會率),不是應是2/3,但為甚麼轉換選擇機會率(留意,不是轉換選擇而中機會率)應是2/3?你是如何得出2/3的結果?
其實,我已答了樓主#1的3個問题多次,但你好像並沒有認真看我的回覆
我不厭其煩地在此答多一次,
1.那依照機會率來說，你應否改變你的選擇呢??? 應該,因為改變選擇後勝出的機會是2/32.如果我說改變選擇,會提高中獎機會,那你會認同嗎??? 認同,不改變選擇的中獎機會一直是1/3,改變選擇後勝出的機會卻是2/3
3. 你認同A網友的那一些說法???為甚麽 不認同,A網友的揀波case引諭失當,假如,A網友問天才揀中白波的機會,天才會答1/2,而非A網友的1/3,理由好簡單,在揀波case中,玩者不會在抽黑波之前選波,自然地,沒有換與不換的抉擇,抽黑波後,可獨善其身在餘下的兩個波中抽出白波

假如,你仔細閱讀我這篇回覆及一直以來的回覆後,仍堅持己見,歡迎以理據指正本人的謬誤,而非用你的權威的程式實驗(Stimulator)去否定他人的想法,因為我並不知道你如何設定你的程式,我提議你不如參考一下PG-13的#93中的Stimulator.至於你的計萛結果,本人仔細睇過,你認為#49的網友的"對換錯"是有問題的,因有兩個組合,但其實不然,雖有兩個組合,而且兩個組合都會導到對換錯,但兩個組合要在本身選對的情況下才能成立,所以,對換錯仍是1/3,沒有問題的,你說的四個中獎conditions頭兩個加埋(不換而中)先得1/3,而非你所說的1/2。因此,你認為不管換或不換，結果中獎機會都是1/2,本人覺得,是非常有問題,我的計算結果是:換而中是2/3,不換而中是1/3,另外,你的中獎機會總式是立足於你的結論不管換或不換，結果中獎機會都是1/2,只要你的立足點不穩,你的計算結果以及中獎機會總式還會成立嗎?
而且,換和不換並不可以同時存在,換和不換的中獎機會總式要分開計算。
換的中獎機會總式=換而中獎+換而不中獎=2/3+1/3=1
不換的中獎機會總式=不換而中獎+不換而不中獎=1/3+2/3=1
兩者不可混為一談,借機幽一幽默,引用網友kantang4910的說話"鬼才們卻硬要將換和不換中獎機會總式混為一談,分清盲點後,答案其實非常簡單,祗是有人要標奇立異吧了"

kaichun88

Originally posted by hold_find at 2007-3-15 08:34 PM:
這是對的，因為它說"其中一個係女"，冇話"第一個係女"，所以目前餘下的可能性有男女 女男 女女，而第2個條件是"都是女"，所以只有女女符合，所以是1/3
注意題目問的不是"生下2小孩，2個都是女的機會"

你不是天才,但你是比天才還強的鬼才,我不得不佩服你的鬼才,你的多

次回覆令我回想經常在學校物理卷出現的Student A,Someone兄以及小明

奉承完,入正題吧!已知其中一個係女的情況下,請問男女與女男有何分別?怎會是1/3?

難道我擲毫連擲10次公後,第十一次擲公的機會不是1/2?

注意題目問的不是"連擲11次公的機會?"

[ Last edited by kaichun88 on 2007-3-17 at 12:54 AM ]

kaichun88

Originally posted by 2000 at 2007-3-17 01:30 AM:
我之前個part修左少少

不過我重想討論既係當參賽者在第一輪三選一後,它所持有那門的中獎機會率是三分之一....之後主持開出空門並不會改變它的機率.......那為何說參賽者的中獎機會率仍是1/2呢是因為我們假設轉選擇的比率是1/2.....但如果個個參賽者都轉選擇的話....中獎比率應是2/3wor........哪麼所謂1/2的參賽者中獎比率是反映参賽者轉選擇的比率???????

還是因為只有中與不中兩種所以是1/2←不認同這個......例子上面reply都好多了

#122已經提過,假設轉選擇的比率是1/2並不合理,在此不再詳述,其實,你應看出了參賽者的

中獎機會率仍是1/2是基於假設轉選擇的比率是1/2的關鍵,但假設在換車题目中並不成立,所

所以,我才會說單單計算勝出機率是沒有意義,因為,勝出機率與轉選擇的比率有關,題目又何

來反映参賽者轉選擇的比率?反而,鬼才hold_find 認為參賽者的中獎機會率仍是1/2是由於假

設轉選擇的比率是2/3,恕我愚昧,我真不知道這2/3從那裏拱出來?望鬼才hold_find可否賜

教

kaichun88

Originally posted by kantang4910 at 2007-3-16 08:35 PM:



連或然率計算方法中最膚淺的謬誤也會犯,還敢稱智力過人?未兔貽笑大方

難怪你會話人是鬼辯者,列卡特曾經說過,那些認為甚麼都知暁的人卻是一竅不通(大意),英

文又有句諺語繹作"來說是非者,便是是非人"自己犯了或然率計算方法中最膚淺的謬誤卻說

成別人犯了或然率計算方法中最膚淺的謬誤,不單止如此,還讚美別人自稱智力過人,假如別

人是鬼辯者,你豈不是神辯者,假如別人智力過人,你豈不是智力高過神(我正在讚你,並

無冒犯及人身攻擊之意)

[ Last edited by kaichun88 on 2007-3-17 at 07:40 AM ]