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.
% R+ S! O" f- z: H9 ?5.39.217.77:8898
. O, m4 U6 n; G5 n8 q2 A; U0 L5.39.217.77:8898Starting 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. tvb now,tvbnow,bttvb: B% L' [6 Z& r0 }: N1 M. B
5.39.217.77:8898* F& n8 o/ ?6 o9 p' _4 c0 N
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.2 h1 O0 e! y" |: G8 ]
" g, G: q, E4 G' M9 U公仔箱論壇#3 did the obvious choice 40 divided by 2 =20, so he picked 20TVBNOW 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。: k" h8 y& @ S$ N0 b
7 ? P7 T1 }5 J#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.9 t4 I2 j/ N' x. ~3 y; H# B c
/ d; `) s3 k0 V4 e4 n# p' N公仔箱論壇#5 same as above, he takes 80 divided by 4 =20, picked 20 as well./ g# |5 L3 r* \
tvb now,tvbnow,bttvb' x% p5 l; V8 n1 o+ }
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
公仔箱論壇0 D, C% T$ x1 W. r7 V: u
