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,bttvb3 n( j7 o3 P% l+ }1 ~1 d

. m& r; x( I+ ^4 P: P5 P( eStarting 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.  5.39.217.77! `8 U: ]( r; F5 G! Q) k
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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.. F" }0 m+ P* B, B. E2 g2 L

7 H* v. A) }& T1 K1 ]2 `# G* b#3 did the obvious choice 40 divided by 2 =20, so he picked 207 d3 w$ I' f6 T9 e7 a3 u( n

1 L5 |; A6 H9 ^: O5.39.217.77#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.6 b, F) Y& v0 U& Q3 t
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。$ v- Y- A" r. M2 k  y# M
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
0 b2 o3 c" t: k2 y4 _! fTVBNOW 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。& P3 K8 Z$ M9 t8 A5 _
Ended all have the same number and all died.