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.
# K- l+ |8 G, ETVBNOW 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 H7 I- H2 y* L. p) A8 I
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. www2.tvboxnow.com% A: @$ @- t3 t
1 [# F3 g1 m) I3 I P: M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.
4 {& a$ I) I7 D: w
/ h5 }! Z0 z; K& [1 | [! M#3 did the obvious choice 40 divided by 2 =20, so he picked 20
2 J7 N7 i9 B! V5 bwww2.tvboxnow.com5 D& m* c: h- U# J" Q: c
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
- Q) E! {7 M, y6 _www2.tvboxnow.com2 R4 @7 [& j O- \9 m/ n4 M
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
8 r+ S6 b2 b6 J9 v5 ~$ K& C9 l& F) V' X- m- {$ T, X u
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
www2.tvboxnow.com" i! X7 @ |# U0 C8 X( N; o
