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.
! |* p' v, X/ B; t0 {4 F' o: STVBNOW 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。5.39.217.77. n m0 J8 A) }5 i8 P5 C
Starting 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.
! L# X: b+ |3 p5 F3 D公仔箱論壇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。+ V& E7 \$ S ~" c) `) H( S4 ~
Now #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.tvb now,tvbnow,bttvb7 ?9 Y0 Y( i% T8 W8 U# u
! p' }0 \9 D$ D" F5 j* l公仔箱論壇#3 did the obvious choice 40 divided by 2 =20, so he picked 20/ i1 k7 F' L, N
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#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.! E. }: F- E, h9 e( d3 t1 l' C
/ x1 X+ w' P/ c+ `$ Ctvb now,tvbnow,bttvb#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
! L0 I' [7 Z! YTVBNOW 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。
7 N1 K' \( v9 CTVBNOW 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。Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
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