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.
$ F& @& M1 l( W. z9 C9 V7 Rtvb now,tvbnow,bttvb
# |; F, _0 O2 @7 b# uTVBNOW 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. 公仔箱論壇: W, M+ b* X; @: L( g: q
+ A$ I- Z4 J' j( V% D$ G7 {& `. {5.39.217.77:8898Now #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.
' d" {! F- Z/ T1 t. MTVBNOW 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。4 A* ~* q$ X, o
#3 did the obvious choice 40 divided by 2 =20, so he picked 20
+ i4 M8 F, D- C7 E5 g V公仔箱論壇5.39.217.77:8898/ b# J! z2 Q1 w- z3 ^
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
l0 _; {; Y' V5 r$ x# F公仔箱論壇
1 O! |% h1 K4 q公仔箱論壇#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
% t% A! U) I* H9 R; u2 l3 o$ a. VTVBNOW 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。
! [$ s- |$ I/ p8 b O [; L7 OEnded all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
