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.+ e+ k4 V E$ T+ l+ h" a
4 f, e" N8 U4 }$ P! l& A6 NTVBNOW 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。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. / N; q4 H- k$ x0 }6 _- j, N! M- _
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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.www2.tvboxnow.com" F5 |& X4 r2 R" k$ C/ o: f
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。9 i$ e: O! V1 A% \
#3 did the obvious choice 40 divided by 2 =20, so he picked 20
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#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.公仔箱論壇& I) s3 n, Q' B* ?1 b
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#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.www2.tvboxnow.com& V! Y f1 f" W2 e$ g' ?6 V0 z {
( P1 i( l' S0 F6 TEnded all have the same number and all died. |
發表於 2006-12-9 12:41 AM
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