The best chance is #3 because all he needs is take #1 + #2 divide by 2. e.g. #1+#2=28 then he take 14. This is the best chance to avoid being highest or lowest and worst case scenario is he equals #1 & #2. However because all 5 knows this simple theory so I think they all died because they all ended up picking the same number of beans.TVBNOW 含有熱門話題，最新最快電視，軟體，遊戲，電影，動漫及日常生活及興趣交流等資訊。3 l8 _. n" S" N3 e* @5 q; I( u
, W3 g! H0 d7 `1 c1 P5 e! @tvb now,tvbnow,bttvbStarting from #1, he knows he cannot pick anything bigger then 49 because if he did, then #2 only have to leave 3 beans for #3 #4 & #5 then he'll live. e.g. #1 picked 53 then #2 picks 100-53-3=44. then A,C,D,E all died. e.g. #1=53, #2=(100-53-3)=44, #3=1, #4=1, #5=1. The best chance for #1 is to pick anything less then or equal the median 20 (100 divided by 5). In fact anything between 3-20 won't change the result. Let's say #1 pick 20. 6 v( V4 G5 H" a! @& C- q3 L' R: x
9 q% j$ J% Z; l. C$ Btvboxnow.comNow #2 knows whatever he picks, #3 will take the median between him & #1 e.g. now #1 picked 20, if he pick 6 then #3 will pick 13 putting him either being the lowest or highest. He cannot allow that so the best chance is to match #1, so he picked 20 as well.公仔箱論壇8 l2 ~- _3 l1 O) y
0 {, O4 L. K C; @#3 did the obvious choice 40 divided by 2 =20, so he picked 20
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) z& ] w1 x( W ]公仔箱論壇#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.公仔箱論壇& x- |. T- i$ ~, y4 i
7 v: w7 Q# o7 e; M9 r9 TTVBNOW 含有熱門話題，最新最快電視，軟體，遊戲，電影，動漫及日常生活及興趣交流等資訊。#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.: y( z! V7 ]' K' E( e3 C0 g8 D# |0 x
`+ D9 S+ Z% p/ h, C# e公仔箱論壇Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
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TVBNOW 含有熱門話題，最新最快電視，軟體，遊戲，電影，動漫及日常生活及興趣交流等資訊。) W2 ` g3 `4 e: i% h
