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.3 P* n# _6 \% w. l8 f1 ?& t

9 y: Q4 w( Q% o$ d& z5.39.217.77Starting 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.  
* \  Q, }0 x4 _0 s  h& ]# n# u公仔箱論壇公仔箱論壇- q% u7 u9 F4 V4 J9 I5 }) \
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.
/ c6 P5 v$ N7 Ytvb now,tvbnow,bttvb( [! U' `% o" k# Q7 [" c" z
#3 did the obvious choice 40 divided by 2 =20, so he picked 20公仔箱論壇! r( n! o$ I) G$ L! s# V

! f% I3 H3 ?, J' qtvb now,tvbnow,bttvb#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.tvb now,tvbnow,bttvb+ ~; d. |: M* Q+ a8 Y2 a% L
) E" w9 F) M3 Y! K. M
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
2 E3 V4 W0 h" _, s. F; Qtvb now,tvbnow,bttvb
8 z( N, I/ f/ j5 s  g5 b公仔箱論壇Ended all have the same number and all died.