tag:blogger.com,1999:blog-2237906900456925233.post5416175477636731200..comments2023-10-18T23:36:58.396+08:00Comments on 數學資料庫手記: 投票選冠軍的概率問題--錯處在哪?(上)MathDBhttp://www.blogger.com/profile/18353093974880979999noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-2237906900456925233.post-18586103210639298162008-09-01T23:57:00.000+08:002008-09-01T23:57:00.000+08:00Right, it is not 0.5!>其中一位觀眾想用概率的知識計算他獲獎的機會。 &g...Right, it is not 0.5!<BR/><BR/>>其中一位觀眾想用概率的知識計算他獲獎的機會。 <BR/>>整個投票過程裏,每位觀眾的角色都是對稱的。<BR/>It looks the 觀眾(V) knows each voters should have the same prob. (p) in his calculation.<BR/><BR/>so, by the basic prob. law,<BR/>sum of all voters p = 1<BR/><BR/>p = 1/1001<BR/>----<BR/><BR/>Refer to Andy reply, the 觀眾 V join the game or not, the winning prob. of (A, B) won't change.<BR/>(if we accept both A and B are winners if they have equal votes)Pophttps://www.blogger.com/profile/17406316084081487132noreply@blogger.comtag:blogger.com,1999:blog-2237906900456925233.post-49882138047789198822008-08-31T23:48:00.000+08:002008-08-31T23:48:00.000+08:00 對,你看出了原因。可是你給出的概率不對。假設他投 A 而其他人投票都隨機投票,A 只需多得 50... 對,你看出了原因。可是你給出的概率不對。假設他投 A 而其他人投票都隨機投票,A 只需多得 500 票便獲勝。由二項式定理可知,A 勝出的概率為 [C(1000,500) + C(1000,501) + ... + C(1000,1000)]/[2^(1000)],而非 0.5 + 1/1001。Andy Chanhttps://www.blogger.com/profile/00595673356023147822noreply@blogger.comtag:blogger.com,1999:blog-2237906900456925233.post-36277280752520289192008-08-31T16:12:00.000+08:002008-08-31T16:12:00.000+08:001)>概率不多於 1/1002This is not the exact prob. of e...1)<BR/>>概率不多於 1/1002<BR/>This is not the exact prob. of each 觀眾 winning the prize, so 1/1002 * 1001 <> 1<BR/><BR/>2)<BR/>其中一位觀眾想用概率的知識計算他獲獎的機會。他假設所有人投票都是隨機的<BR/>So, the other 1000 votes are random.<BR/>Before he votes, the winning prob. of <BR/>參賽者(A, B) should be 1/2.<BR/>But his OWN vote is not random for him, so after he voted, the winning prob. of A (say he votes A) is 0.5 + 1/1001 [and Prob(B) = 0.5-1/1001Pophttps://www.blogger.com/profile/17406316084081487132noreply@blogger.com