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.
% T) C* Z' T& @* J' _9 a
4 a$ o0 X: F* W6 {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。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. 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。' u* l6 S4 q# H
; _4 a, x( [8 W: R# m( F0 r公仔箱論壇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.7 x$ a7 a9 c8 i+ P
公仔箱論壇. U% U7 I, {4 O6 u+ G& {
#3 did the obvious choice 40 divided by 2 =20, so he picked 20os.tvboxnow.com* R1 K5 }% z$ |# W
; _0 O& k! |; d/ v: |! i. t#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
, @" b! t( \1 v2 C) K: S公仔箱論壇公仔箱論壇% B( v* o; o- d6 J5 c: ^
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
3 g* y+ W o! l& z: b4 _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。& H& H' T4 E) y7 K3 d
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
