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條問題重覆左...

第一位仁兄的計法錯了！正解如在重覆的問題的答案...

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a. 想像將n點分別隨機放入 365 日，然後要不重覆~
第一點成功的機率是 365 / 365 = 1
第二點成功的機率是 364 / 365 (第二個人不可和第一個人重覆生日嘛)
第三點成功的機率是 363 / 365 (第三個人不可和前面兩人重覆生日)
...
第 n 點成功的機率是 (365 - n + 1) / 365。
Part a 的答案：將以上機率乘起來就是了~

b. 如果你將 a. 的答案當成 (1-0.5) = 0.5 ，你會發現 n 的值會影響方程的 Degree，所以不可以用一般方法計... 唯有試不同的 n 了。
當 n = 15，Part a 的概率就是約 0.747，
當 n = 25，Part a 的概率就是約 0.431，
所以答案就是 15 和 25 之間了~
那麼 (例如) 如 n = 22 則概率 > 0.5，n = 23 則概率 < 0.5 ，答案是 22 還是 23？
題目其實希望有同生日的概率「最少」有 0.5，但 Part a 計的是沒有同生日的概率，所以我們希望 Part a 的概率小於，而不是大於 0.5。所以 n = 23。

真正的 n 就由你自己試了~

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貓朋，我支持你！
解任何概率題的正確態度應是重視其邏輯結構而不是重視其答案。

a)
The pobability of my birthday is 29thFeb = Pa = 1/ (365x4 +1) = 1/1461
The pobability of my birthday is not 29thFeb = Pb = 365x4/ (365x4 +1) = 1460/1461
The probability that no one (of N people) same as mine
= Pb x {(1460 -4)/1461 }^N + Pa x (Pb)^N
=1460 x{ (1456)^N + (1460)^N } /(1461)^(N+1)

b)
Let Pc =1460 x{ (1456)^N + (1460)^N } /(1461)^(N+1) , then
Pc < 2 x( 1460/1461)^(N+1) and
Pc > 2 x (1460/1461) x (1456/1461)^N
Take 0.5 = 1 -Pc and hence Pc=0.5
Solving, we have
N < 2024
N > 404
Try N=(2024+404)/2 =1214 , we have Pc = 0.45 appx.
and continue by bisection method to find the maxium Pc such that Pc <0.5, we have N =1085

2006-12-11 20:19:00 補充：
Although the answer is 23, I was very disappointed , not because my answer was not selected, but the answer has neglected the existence of 29th Feb. and the people born at this date. Computation error is no big deal but neglecting the existence of other people is unbelievable.

a) P(no one share the same b-day) = 1 - P(someone share the same b-day)
= 1 - 1 / 365 * N
= ( 365 - N ) / 365
b) P(someone share the same b-day)
= N / 365 = 0.5
N = 182.5
Hence... at least 183 people.