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.5.39.217.770 @# M- g- w, m. b

* N, G& M" x  f: v: k8 V5.39.217.77Starting 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.  5.39.217.77, W/ x/ t% F' E( j

3 j  B" ~4 w9 j) a  j+ ^( s公仔箱論壇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.5.39.217.774 C3 r( F! m( ]) c5 ^) k7 R

6 R# E  L; Z  L( Z' A$ I5.39.217.77#3 did the obvious choice 40 divided by 2 =20, so he picked 20
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8 c3 j- n' Q) x* [0 h2 t5.39.217.77#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
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#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.8 n9 {0 y( m* t4 @$ S8 R( p, E8 L
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Ended all have the same number and all died.