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.
+ w( X1 o R/ Btvb now,tvbnow,bttvb
* p9 B9 |( Y$ y2 n& u; htvb now,tvbnow,bttvbStarting 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.
R1 x% ^) }* d5 {. iTVBNOW 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。
5 S6 R, D4 v, h' M/ u9 f" P6 QNow #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.tvb now,tvbnow,bttvb( j7 ]2 n5 y( p/ y& I5 f5 D b R
5.39.217.77:88983 k2 L; m0 b4 V( H, L% O* L
#3 did the obvious choice 40 divided by 2 =20, so he picked 20
\0 I0 j! j! N3 \4 d+ O2 [公仔箱論壇( N$ X. T" |, X5 o" E1 t6 l- j1 N y' t
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
: C* g: W! E1 h7 ~/ [* q公仔箱論壇
; r- v. p+ x+ ]. ?4 s' Ntvb now,tvbnow,bttvb#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
& d' _. t, t& U
4 K/ p+ K+ c. w3 j; k+ Dtvb now,tvbnow,bttvbEnded all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
& \: u* u9 Z& v) D
