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.公仔箱論壇7 S% O- y O/ ~7 J/ }! e6 k) k
& a! u6 }5 g/ R- P$ m9 [) rStarting 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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, z1 N; o0 [) ] H' e- k7 I x/ hNow #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.
7 q! m* h' Z u+ ?4 \公仔箱論壇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。 G- L$ J6 M9 _% V) J
#3 did the obvious choice 40 divided by 2 =20, so he picked 20% q2 S7 y* P5 W* w2 B, q
9 N6 C8 V _3 [/ Q- JTVBNOW 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 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.tvb now,tvbnow,bttvb5 D& M+ l5 S- I* h3 F- a
! R2 E3 e4 K- C M% J6 e* ~8 Y+ wos.tvboxnow.com#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.os.tvboxnow.com4 u/ f! o ~4 v
0 e4 w0 [2 u! M, `3 O5 ztvb now,tvbnow,bttvbEnded all have the same number and all died. |
發表於 2006-12-9 12:41 AM
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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 a4 r7 j$ x8 w/ i- O, \, X
