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 [! L; v Z. ~8 l7 u
1 [$ p- [6 c, L5 S" ^, u- etvb now,tvbnow,bttvbStarting 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.
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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.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。4 r% [" b2 T% z7 O' t% e
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。0 N$ ], w& V, G; D6 k: d
#3 did the obvious choice 40 divided by 2 =20, so he picked 20
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# |, p( F; ~0 Z: @8 Etvb now,tvbnow,bttvb#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.公仔箱論壇, D N8 e F3 E8 G0 ^1 p
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#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.tvb now,tvbnow,bttvb( N" R9 T+ q7 S! I
: p2 C& Q& o+ ~6 v/ tEnded all have the same number and all died. |