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) a& f1 }# r! i+ N# k- u6 q
7 y6 c) Y$ b+ OStarting 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. + v% q4 \, t5 X8 N, v/ q& O
' `2 ]* z/ I8 k9 W+ C1 v9 \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.
# K! S Q5 }* i5.39.217.77:8898
& F% v/ w" l3 ~! \3 X; c8 W' s& s#3 did the obvious choice 40 divided by 2 =20, so he picked 20
7 s3 w# x0 @' ~6 F/ ~) QTVBNOW 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。2 l' O$ q& A% Z' }' N- }2 }+ D, n
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
, v2 r+ R% Y& w) A5.39.217.77:8898
9 |4 n8 u5 ? xtvb now,tvbnow,bttvb#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
( ?1 @0 G' q: C* zTVBNOW 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 ~! {) f7 V! A# Q+ ?
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
