playbr2

Originally posted by kantang4910 at 2007-3-15 11:58 PM:


又來鬼辯
機會率唔係  事件出現次數/所有情況出現次數  又係乜呢?

係喎你岩

老師咁我有野問

我有一粒''正常''既骰子 ....它有六個面 ,為 1 至 六

抽中''任何''一面o既 機會率唔係 六分一呢

咁我而加擲 骰子 6次 ..... ''結果'' 全部6次 都擲出 1

係咁我應用 ----> 事件出現次數/所有情況出現次數

6/6 = 1  , 『 結論 :  一粒''正常''骰子有六個面 , 但擲出 1 的機會率係100%, 不會有其他結果 』

..................咁又岩吾岩呢......如果吾岩.....係咪因為 機會率 與 事件出現次數 不會成一定比例呢

lilirayhk

Originally posted by hold_find at 2007-3-15 08:34 PM:

這是對的，因為它說"其中一個...

試想想 一個有兩個波的袋, 波一係就白色, 一係就黑色, 你而家攞其中一個波出黎睇, 發現係黑色wo, 咁另一個波係黑色的機會率又會唔會係1/3 呢?

PG-13

Originally posted by netharmon at 2007-3-15 11:38 PM:
如果真係 2/3 同 1/2 咁大差異, 不如寫個 Program 來 Simulate 10000000 times 睇吓 如果轉的勝出 是近 1/2 定 2/3.

Stimulator... http://www.userpages.de/monty_hall_problem/
OF COZ THERE IS STIMULATOR! its a world known mathmatical problem, and there are of coz more than 1 stimulator on the internet.. search it for more if u think this ones fake

PG-13

Originally posted by playbr2 at 2007-3-15 04:25 PM:
引文一篇[/colo...

The comment is wrong, without another knowingly choose it for u, the probablity will be the same.

first, Player 3 choose 1 ( win, lose, lose),
*before opening the door, the porbably of each door is 1/3 for winning

then, host 2 choose 1 (lose, lose) with the probabilty of 1/3 to pick this set or ( win,  lose) with the probability of 2/3 ,<-- this interferance is critical because the host have to choose the lose door in either situation.

The host' probability of holding a winning door would be 1/2. Because in the cases of (lose, lose) with the probability of 1/3, he has to open the lose door, which left ( lose) door. And in the ( win,  lose) case with the probability of 2/3, the host has to open the lose door, which left ( win).
As a result, with knowingly open ( lose) door, there are only 2 outcomes:
( win)or ( lose). However, because the chance of picking a (lose, lose) is only 1/3, there are 2/3 of chance that the host will have ( win)
So the probability is 2/3
--------------------------------------------------
In case the host DOES NOT knowingly open the ( lose) door
first, Player 3 choose 1 ( win, lose, lose),
*before opening the door, the porbably of each door is 1/3 for winning

then, host 2 choose 1 (lose, lose) or ( win, lose),

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
am not expert in stats and i dont think my ans is right anyways

[ Last edited by PG-13 on 2007-3-16 at 02:10 AM ]

小小康

黃興桂+田雞o既理論，答案是1/2。
原因是一係中、一係唔中，永冇中間!!!

[ Last edited by 小小康 on 2007-3-16 at 01:45 AM ]

奇

Originally posted by 小小康 at 2007-3-16 01:37 AM:
黃興桂+田雞o既理...

田雞?最後唔係掉進gap中嗎?

guswan

Originally posted by 小小康 at 2007-3-16 01:37:
黃興桂+田雞o既理...

但係黃興桂有半越位同半單刀bor..............................

playbr2

Originally posted by 奇 at 2007-3-16 01:48 AM:

田雞?最後唔係掉進gap中嗎?

吾明

playbr2

Originally posted by PG-13 at 2007-3-16 01:17 AM:


The comment is wrong, without a...

.................死仔!!!! 改內容

你頭先篇野睇到我一頭霧水, 正想回覆比你知你計漏左野......好彩再睇多次咋

奇

Originally posted by playbr2 at 2007-3-16 02:07 AM:


吾明

少林足球,田雞話唔係公就字,結果個硬幣掉進gap中