考慮次序與否 1.1

這段改編自 2010 年 6 月 15 日的對話。

初學機會率的其中兩個最大難處是,要釐清「什麼時候要考慮次序」和「怎樣為之『相同情況』」,例如:

有一個袋子,內裡有十張卡紙。每張卡紙上,都寫上了一個英文字母。那十個字母分別是「AAABBBCCCC」,即是三個 A、三個 B 和 四個 C。你將要抽其中三個字母出來。被抽出來的卡紙,不會放回袋中。

假設整個過程是隨機的,即是各個可能性的機會均等。問題是,你抽中三個都是 A 的機會率是多少?

P 方法:

總共要抽三個字母:

(_)(_)(_)

抽第一個時,總共有十個字母,而你想要的 A,則有三個。所以,第一個機會率分數是十分之三(3/10)。

(3/10)(_)(_)

抽第二個時,總共餘下九個字母,而你想要的 A,則有兩個。所以,第二個機會率分數是九分之二(2/9)。

(3/10)(2/9)(_)

類似地,第三個機會率分數是八分之一(1/8)。

(3/10)(2/9)(1/8)

結論是,抽到三個 A 的機會率是 1/120。

(3/10)(2/9)(1/8)

= 1/120

在用「S 方法」驗算前,我們先考慮,我們需不需要,再額外考慮「次序問題」呢?

(HYC:你的意思是,你只考慮了,抽到「AAA」這個籠統的情況。但是「A」其實有三個,所以會形成六種可能性。

方便起見,我叫第一個 A 做「A1」、第二個 A 做「A2」和 第三個 A 做「A3」。那六種可能的結果是:

(A1)(A2)(A3)

(A1)(A3)(A2)

(A2)(A1)(A3)

(A2)(A3)(A1)

(A3)(A1)(A2)

(A3)(A2)(A1)

那樣,我們需不需要再把,以上的結果乘以 6 呢?)

不需要,因為剛才那幾個機會率分數,其實已內置了次序的考慮:

3/10)(2/9)(1/8)

正正是因為第一張被抽出來的卡紙,無論是 A1、A2 還是 A3 都可以接受,第一個機會率分數的分子才會是 3。你那種結果,正正是分子的(3 x 2 x 1)。

3/10)(2/9)(1/8)

= 6/720

— Me@2013.01.20

致讀者:如發現本文有思考漏洞,或者運算錯誤,請以電郵告知本人。謝謝!

— Me@2012.10.17

2013.01.20 Sunday (c) All rights reserved by ACHK

Godel 8

Limitations of Godel’s theorems

The conclusions of Godel’s theorems are only proven for the formal theories that satisfy the necessary hypotheses. Not all axiom systems satisfy these hypotheses, even when these systems have models that include the natural numbers as a subset. For example, there are first-order axiomatizations of Euclidean geometry, of real closed fields, and of arithmetic in which multiplication is not provably total; none of these meet the hypotheses of Godel’s theorems. The key fact is that these axiomatizations are not expressive enough to define the set of natural numbers or develop basic properties of the natural numbers. Regarding the third example, Dan E. Willard (Willard 2001) has studied many weak systems of arithmetic which do not satisfy the hypotheses of the second incompleteness theorem, and which are consistent and capable of proving their own consistency (see self-verifying theories).

Godel’s theorems only apply to effectively generated (that is, recursively enumerable) theories. If all true statements about natural numbers are taken as axioms for a theory, then this theory is a consistent, complete extension of Peano arithmetic (called true arithmetic) for which none of Godel’s theorems apply in a meaningful way, because this theory is not recursively enumerable.

The second incompleteness theorem only shows that the consistency of certain theories cannot be proved from the axioms of those theories themselves. It does not show that the consistency cannot be proved from other (consistent) axioms. For example, the consistency of the Peano arithmetic can be proved in Zermelo–Fraenkel set theory (ZFC), or in theories of arithmetic augmented with transfinite induction, as in Gentzen’s consistency proof.

— Wikipedia on Godel’s incompleteness theorems

2013.01.20 Sunday ACHK