nightrider

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.tvb now,tvbnow,bttvb: S' G5 o0 a' @' M
公仔箱論壇; L$ v1 R/ p* T& r$ H
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.  / b1 D) Z+ f' P9 a  V( M9 M

+ b% F. W, F6 u7 F- a3 pwww.tvboxnow.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.
- {7 c* e8 _1 F8 a9 Ztvb now,tvbnow,bttvb/ P2 v1 L, m: [+ M, _) ]1 h
#3 did the obvious choice 40 divided by 2 =20, so he picked 20/ H+ k7 S  w# v7 i0 E
) ?+ ]' o% l# r& _2 [, {: A
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.( c' |) z5 q* X1 r

2 T: d$ S# w* O4 t+ ]#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.www.tvboxnow.com' H0 ~. Q# I$ U9 e' w9 X
公仔箱論壇1 f2 a9 X6 K4 ^9 c
Ended all have the same number and all died.