- 帖子
- 418
- 精華
- 0
- 威望
- 300
- 魅力
- 0
- 讚好
- 0
- 性別
- 男
|
145#
發表於 2007-3-17 03:32 PM
| 只看該作者
Originally posted by ronja at 2007-3-17 01:00 PM:
thanks for telling me that you do not understand the first question
but please don't make the 1st question become the 2nd question,that's the reason why i ask the 1st question! Would you please telling me what is your reason why you ask the first question?
Can you explain why the ratio of switching to not switching must be 1:1?
Can you explain why the probability of winning the game must be 50%?
Do you have any evidence for supporting your assumption that the ratio of switching to not switching is 1:1?
thanks for telling me that you do not know how to read other's replies and superficially
blame others for misunderstanding the question instead of answering other's questions with expanaton.
I have not ever mixed up the two question.Please read the following quoted statement carefully.Originally posted by kaichun88 at 2007-3-17 12:32 AM:
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2/3
2.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但換門case玩家在捒門後,主持人有擇開及任開兩情況,雖說,擇開及任開均有兩個組合,但兩者比重不同(任開只佔1/3),固此,門的性質一樣,但擇開及任開性質不同,因此,對換錯與錯換對之比並不一樣 Originally posted by kaichun88 at 2007-3-16 10:29 PM:
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%. Originally posted by kaichun88 at 2007-3-17 06:46 AM:
#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? [ Last edited by kaichun88 on 2007-3-17 at 03:53 PM ] |
|