主場之利
學校名稱：皇仁舊生會中學 2008 至 2009 年度借調老師：陳德貞
引言
相信有留意英格蘭超級足球聯賽的同工均會發現一般球隊於主 場作賽表現會較為出色，得勝機會會較高。這種情況有沒有數據支 持? 如何解釋?
在日常生活中，我們遇到的事情有三個情況：必然發生、有可
ڇֲൄسխΔݺଚሖࠩऱࠃൣڶԿଡൣउΚؘྥ࿇سΕڶױ౨࿇سΕױ
由上表可見，榜中和榜末的球隊在主場和客場作賽時成績有顯著的 分別，「主場之利」對實力稍遜的球隊是有幫助，當中以曼城的差 距最大；反之，榜首的球隊在主場和客場作賽時成績有沒有顯著的 分別，即實力較強的球隊不論在那兒作賽也能發揮水準、保持佳 績。
在現實的情況裏，球隊能否勝出要視乎許多因素，例如天氣、
球員的狀態、教練的策略等，當中以球員的狀態和教練的策略尤其 重要。假設韋斯咸在主場作賽，60% 球員因滿心歡喜地認為自己佔 主場之利能有較大的機會勝出，作賽時掉以輕心而不在狀態；假設 韋斯咸在客場作賽，50% 球員因而激發鬥心，拚命作賽。根據以往 數據，當韋斯咸在主場作賽時，有 44% 的機會勝出，反之在客場作 賽時，有 26% 的機會勝出，假設其他情況不變，那麼該隊能否反敗 為勝？
設 a為韋斯咸在主場作賽及球員在狀態時勝出的概率；
b為韋斯咸在主場作賽及球員不在狀態時勝出的概率；
c為韋斯咸在客場作賽及球員在狀態時勝出的概率；
d為韋斯咸在客場作賽及球員在不狀態時勝出的概率。
以樹形圖（下圖）去表達現在的情況：
賽，勝出的概率會因球員在狀態下超過一半。另一方面，根據以往 數據，當韋斯咸在客場作賽時，有 26% 的機會勝出；
P (勝| 客場) = 0.26 P (勝∩ 客場)
P (客場) = 0.26 P (勝∩ 客場) = 0.26 × 0.5 0.5(0.5)c + 0.5(0.5)d = 0.26× 0.5
5c + 5d = 2.6 c = 2.6− 5d
5 . . . (ii)
按理，當球隊在客場作賽和球員不在狀態時勝出的概率 (d) 是很小，
例如若 d = 0.02，則 c = 0.50；由此可見，雖然球隊的整體勝出的 概率只是兩成多，但當球隊在自己的客場作賽時是在狀態的話，勝 出的概率會提升至一半。這樣我們看見無論韋斯咸是在主場或客場 作賽，勝出的概率均是少於一半：
P (勝| 主場) = 0.44 < 0.5 P (勝| 客場) = 0.26 < 0.5
因此我們不能說「主場之利」是必勝之因，那麽這是否代表韋斯咸 勝出的機會永遠是少於‘負或和’的機會呢？從以上的計算，我們 看見當韋斯咸球員是在狀態下作賽，無論韋斯咸是在主場或客場作
賽，勝出的概率均是大過或等於一半：
P (勝| 主場，在狀態) = 0.8 ≥ 0.5 P (勝| 客場，在狀態) = 0.5 ≥ 0.5
而
P (勝| 主場，不在狀態) = 0.2 < 0.5 P (勝| 客場，不在狀態) = 0.02 < 0.5
這種在分組比較和在合併考慮中得出不同或甚至是相反的結論的現 象可稱為辛普森悖論。以這例子來說，考慮了球隊是否在狀態下作 賽的因素，球隊在主場作賽仍比在客場作賽有較佳勝算，但是我們 不能說「主場之利」是必勝之因，反之，在狀態下作賽也可能反敗 為勝。
因此，不論是在瀏覽網路新聞或報章雜誌中的統計數據資料時，
我們應注意此現象，特別是當統計數據是以某種分類的方式作比較 時，一定要仔細計算合併後的數據，以免遭到誤導。
參考資料：
1. 2008-09 賽季英格蘭超級聯賽主勝積分排名 http://app.gooooal.com/competition.do?lid=
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哪一科老師較易在最受學生歡迎的老師選舉中獲勝？
學校名稱：保良局馬錦明夫人章馥仙中學 2008 至 2009 年度借調老師：何綺紅
引言
有些學校會於每年舉辦最受學生歡迎的老師選舉，增強師生之 間的感情。任教哪些科目的老師會較易獲獎呢? 原因何在?
學生會每年都會進行最受學生歡迎的老師選舉，讓學生以一人 一票的方式選出最受學生歡迎的老師，自然地獲得最高票數的老師 會獲得最受學生歡迎的美譽。
去年的最受學生歡迎的老師選舉的部分投票情況如下：
老師 A 老師 B 老師 C 老師 D 老師 E 老師 教授科目 中文科 英文科 數學科 音樂科 體育科 獲得票數 20 20 30 55 60
10 20 30 40 50 60 70
A 老師 B 老師 C 老師 D 老師 E 老師 最受學生歡迎的老師選舉
科任老師 獲得票數
從以上的圖表中，可見 E 老師獲得最多票數，所以 E 老師獲得 最受學生歡迎老師的美譽。
根據過往數年的選舉經驗，我發現獲得最受學生歡迎的老師多 為術科的科任老師，如體育科、音樂科的科任老師。
為何術科的科任老師較一般科目的科任老師會較多被學生選為 最受歡迎老師? 是否因為一般科目的科任老師由於要面對課程內的 家課及考試要求，所以會對學生較嚴格，經常催逼學生交功課甚至 要罰留堂溫書等，導致不受學生歡迎；另一方面，術科的課程不用 面對太大的考試壓力，上課的氣氛也較一般科目輕鬆，學生便能愉 快學習；自然地，較少學生會將神聖的一票投給一般科目的科任老 師，而大部分學生會投票給術科的科任老師。
除了上述的原因外，還有沒有其他的科學方法能夠解釋為何最 受學生歡迎的老師選舉中較多術科的科任老師獲獎？
初中數學科課程的數據處理範疇中有認識期望值的意義，我們 可從期望值的意義解釋為何最受學生歡迎的老師選舉中較多術科的 科任老師獲獎。
根據維基百科，期望值的定義為：「在機率論和統計學中，一個 離散性隨機變量的期望值（或數學期望、或均值，亦簡稱期望）是 試驗中每次可能結果的機率乘以其結果的總和。換句話說，期望值 是隨機試驗在同樣的機會下重複多次的結果計算出的等同「期望」
的平均值。需要注意的是，期望值並不一定等同於常識中的「期 望」──「期望值」也許與每一個結果都不相等。」
舉例說明，假設學生只認識教授他們的科任老師，自然地學生
只會投票給教授他們的科任老師，並且每位老師獲學生投票為最受
般科目的科任老師除非有化腐朽為神奇的魅力，否則都難以獲得最
許，當然，這個選舉制度可能令很多老師同時獲獎；另一個方法是 根據獲得票數多於期望值的百分比來判斷，百分比最高便是最受 學生歡迎的老師。當然，這選舉制度亦會有一些不公平的情況，例 如，副校長通常只教授一班高年級的班別，這樣，副校長獲得最受 學生歡迎老師票數的期望值 = 18 × 40 = 5，當計算獲得票數多於期 望值的百分比時，由於期望值較小，獲得票數不用太多，已經令百 分比很大，例如，獲得票數是 9 票，這樣，獲得票數多於期望值的 百分比 9−55 × 100% = 80%，足以成為最受學生歡迎老師的得主。
若你是學生會的幹事，你會怎樣處理最受學生歡迎的老師選舉 呢？
參考：
1. 維基百科網站
http://zh.wikipedia.org/w/index.php?title=%E6%9C%9F%E6%9C
%9B%E5%80%BC&variant=zh-hk
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Game of Chance
學校名稱：佛教善德英文中學 2008 至 2009 年度借調老師：程國基
Introduction
Have you ever played a game called “Da Feng Chui”(the big wind blows 大風吹) during school Christmas party? This game is very exciting and everybody will enjoy it. Is there any strategy to win the game?
How to play the game called “Da Feng Chui”(the big wind blows 大風吹)? The rule is very simple. At the beginning, all participants sit in a circle except one person comes out and stands at the middle of the circle.
Exciting moment comes when that person says something that someone have. Those participants need to leave their own seats immediately and find another seat as quickly as possible. At the same time, that person can find a seat to sit. Clearly, there will be somebody who cannot find his/her own seat; he/she will then receive punishments. Since the game requires participants with quick responses and fast movements, there will be many interesting things happening and they may feel a lot of fun. Apart from the amusement of the game, some mathematics can be found from this activity.
Let's think about a simple example: There are only Mary(M), Susan (S), John(J), Dick(D) and Peter(P) joining the game. John is selected to come out first. The following figure (a) is used to show the situation:
J D
M
S
P Figure (a)
J D
M
S
P Figure (b)
If John says "Those wear shoes, they must leave", all participants will
leave their own seats. In order to make John find no seats, it may be: Mary
⇒Susan; Susan ⇒ Dick; Dick ⇒ Peter; Peter ⇒ Mary, where ⇒ means moves to the seat of somebody as shown in figure (b).
Do you know totally how many ways are there for Mary, Susan, Dick and Peter to find the new seats in order to prevent John to get a seat?
The following is one of the ways presented in tabular form:
Mary Susan Dick Peter Mary 8 X
Susan 8 X
Dick 8 X
Peter X 8
8 – means that each player cannot take his/her original seat again.
X – means that the player on left hand side takes the seat of another player.
For example, refer to the table, Mary moves to the seat of Susan.
With the aid of the above table, you only need to consider the following cases.
Case 1: If Mary has chosen Susan's seat, Susan will have 3 choices to move.
If Mary and Susan have chosen their new seats, Dick will have only 1 choice left. If Mary, Susan and Dick have chosen their new seats, it is clear that only 1 way is provided for Peter to move.
Case 2: If Mary has chosen Dick's seat or Peter's seat and Susan has chosen Mary's seat, Dick and Peter both will have only 1 choice to select.
Case 3: If Mary has chosen Dick's seat and Susan has chosen Peter's seat, Dick will have 2 choices to move. If Dick has chosen his new seat, Peter will have only 1 choice left.
Case 4: If Mary has chosen Peter's seat and Susan has chosen Dick's seat, Dick will have 2 choices to move. If Dick has chosen his new seat, Peter will have only 1 choice left.
Hence, the total number of ways for these players to select other seats
= 1× 3 × 1 × 1 + 2 × 1 × 1 × 1 + 1 × 1 × 2 × 1 + 1 × 1 × 2 × 1
= 9
Although the above calculation is not so complicated, it will be more difficult for us to count and we need to consider more cases if there are more players joining this game. That is why we are going to use“probability”
in order to solve this problem in a more systematic way. You may have learnt that for P (A∪ B) = P (A) + P (B) − P (A ∩ B) any events A, B.
How about P (A∪ B ∪ C ∪ D) for any events A, B, C, D? Do we have a similar result? Yes and it can be obtained by the “inclusion – exclusion principle”for probability:
P (A∪ B ∪ C ∪ D)
= P (A) + P (B) + P (C) + P (D)− P (A ∩ B) − P (A ∩ C)
−P (A ∩ D) − P (B ∩ C) − P (B ∩ D) − P (C ∩ D) +P (A∩ B ∩ C) + P (A ∩ B ∩ D) + P (A ∩ C ∩ D) +P (B∩ C ∩ D) − P (A ∩ B ∩ C ∩ D)
Let A = the event that Mary takes her original seat.
B = the event that Susan takes her original seat.
C = the event that Dick takes his original seat.
D = the event that Peter takes his original seat.
From the above formula and with the fact that P (A) = P (B) =
Hence, P(none of 4 players Mary, Susan, Dick and Peter take their original seats)
= 1− P (A ∪ B ∪ C ∪ D)
= 1−(
1−2!1 + 3!1 −4!1)
= 38
Since the number of ways that these participants select their seats with-out restrictions = 4!.
Hence the number of ways that none of the participants take their orig-inal seats = 4!×38 = 9.
Up to now, the problem seems to be solved and you may feel every-thing OK. However, I can tell you that Mathematics does not stop here.
You may learn more if you investigate further…
Let's look some interesting results below.
From the above, for four players
Mary Susan Dick Peter
P (none of 4 players take their original seats) = 1−(
1−2!1 +3!1 −4!1) . With some calculations for only 2 players or 3 players, we have the following expressions:
P (none of 2 players take their original seats) = 1−( 1− 2!1) P (none of 3 players take their original seats) = 1−(
1− 2!1 +3!1)
Observe all these expressions. Do you agree that they all seem to fol-low a pattern?
Yes, that is P (none of n players take their original seats)
= 1−( where n is any positive integer.
Although the proof of (*) is omitted here, the proof will be similar as in the case when n = 4. Try by yourself if you like!
We can now generalize a formula for the number of ways that none of n players move to their original seats.
By using original definition of probability, P (none of n players take their original seats)
= Number of ways that none ofnplayers move to their original seats Number of ways thatnplayers find their seats without any restrictions
Combined it with (*), 1−1!1 +2!1 −3!1 + . . . +(−1)n!n
= Number of ways that none ofnplayers move to their original seats
n
Hence, the number of ways that none of n players move to their original seats
This value of (**) has a name called subfactorial and denoted by !n but you don't confuse with n factorial n!.
From (**), when n starts from 0, we have !0 = 1, !1 = 0, !2 =1,
!3 = 2, !4 = 9,…You may find the value increasing rapidly.
For example, !10 = 1334961.
The above "Seat" problem has its history. It is originated from a clas-sical old problem first proposed by the French mathematician Pierre Re-mond de Montmort in 1708. This old problem was recorded in his treatise
The above "Seat" problem has its history. It is originated from a clas-sical old problem first proposed by the French mathematician Pierre Re-mond de Montmort in 1708. This old problem was recorded in his treatise