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.
( ~1 l7 v, z2 O6 V+ E# {( Xtvb now,tvbnow,bttvb' \8 }! L3 g! H9 @& z9 U9 a! M! C. l
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.
6 ~9 [& f2 W9 I; U* y8 m' q+ c
: A/ \# |1 ]6 v1 G8 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.+ h& E7 O4 g5 M2 Q
' O" Y5 B: Q8 A, X3 Stvb now,tvbnow,bttvb#3 did the obvious choice 40 divided by 2 =20, so he picked 20
7 B; z2 l, G4 ?" G5 j7 S5.39.217.76tvb now,tvbnow,bttvb; ?0 X! R" J! q+ @% n& U( p
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
# |* n% _0 b4 t' K) @8 {% OTVBNOW 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。8 V: X& \3 I( \$ w3 L+ M* _$ _
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.) H2 x, x' B: H9 I
: Y! H) Z/ j! ~6 D5 h5.39.217.76Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
tvb now,tvbnow,bttvb, g5 O' \3 j6 Y5 C. v( \/ l; U
