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.5.39.217.778 ]9 l: e  H, l( y) J1 E" h5 g5 T+ V
公仔箱論壇6 t5 L* j* F2 m
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.  
7 l/ [, w' s6 x9 R3 D# K! h5.39.217.77公仔箱論壇( a( a& f1 Y% j& |0 p: f) ?
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.
* O# A" m% j* K& P9 e  y) y7 B) p  |% _
#3 did the obvious choice 40 divided by 2 =20, so he picked 20tvb now,tvbnow,bttvb/ c5 V) W1 X+ X% g
公仔箱論壇. L6 }  {! Y/ K/ {/ `  c
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.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。3 Q, _( o/ N$ \# Y

& d2 L) @. f9 P! u/ M, ~5.39.217.77#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.) x' C+ L, p) A2 Z( }/ ~: C
tvb now,tvbnow,bttvb2 Z8 L  u" N2 E+ G5 O: b9 ]
Ended all have the same number and all died.