Banach Matchbox Problem
distribution在這裏的意思實際上是概率，我最開始上課的時候打盹了，唉，沒有聽懂，學長給我解釋的是爲什麼可以用他的那種方法計算。但我還是沒有明白這題讓我幹什麼，應該怎麼做？我竟然就不懂裝懂，真是可悲。以後絕對不能這樣了。
然後我就google了一下這個著名的Banach Matchbox Problem。
發現了一個神奇的網站。數學問答網站
從這裏我得到了認真仔細的解答，就是在求解一個從1-n不同排序隊列的概率，令我驚訝的是這個問答竟然寫於1998年！
Date: 10/31/98 at 10:43:14
From: Doctor Anthony
Subject: Re: Probability
This is a classic problem, sometimes called the Banach Matchbox
problem. Note that the least number of ‘trials’ is n+1 and the maximum
number is 2n+1.
Suppose p = the probability that he uses the lefthand pocket and
q = the probability that he uses the righthand pocket. Then:
Prob(k=n) = C(n+1,n+1)p^(n+1) q^0 + C(n+1,n+1)p^0 q^(n+1)
= p^(n+1) + q^(n+1)
= 2(1/2)^(n+1) if p = q = 1/2
= (1/2)^n
If he has used 1 match from the other box, then during the first n+1
occasions he must have chosen n matches from one box and 1 match from
the second box. Then on the n+2 nd occasion he returns to the empty
box.
The probability of this is:
Prob(k=n-1) = C(n+1,n)p^n q p + C(n+1,n)p q^n q
= 2 C(n+1,n)(1/2)^(n+2) if p = q = 1/2
= C(n+1,n)(1/2)^(n+1)
= (n+1)(1/2)^(n+1)
If he has used 2 mmatches from the other box, then during the first n+2
occasions he must have chosen n matches from one box and 2 matches from
the second box. Then on the n+3 rd occasion he returns to the empty
box.
The probability of this is:
Prob(k=n-2) = C(n+2,n)p^n q^2 p + C(n+2,n)p^2 q^n q
= 2 C(n+2,n)(1/2)^(n+3) if p = q = 1/2
= C(n+2,n)(1/2)^(n+2)
The pattern is now clear:
P(k=n-r) = C(n+r,n)(1/2)^(n+r)
So if we want the answer in terms of k we replace r by n-k in this
expression:
P(k matches in other box) = C(n+n-k,n)(1/2)^(n+n-k)
= C(2n-k,n)(1/2)^(2n-k)