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 g( V; B$ G$ j! h7 ~$ a6 M+ VTVBNOW 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 P3 A' G, f( T2 F0 K
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 j; \3 j+ |, \9 r0 V: E
9 F: L$ u" k. k1 m3 T1 }7 pTVBNOW 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。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) K7 ~4 `$ x- [4 F: S, h
x( F; ?9 l- F! Q' h2 H#3 did the obvious choice 40 divided by 2 =20, so he picked 20
& L) l3 U* m0 u+ R2 U4 R公仔箱論壇TVBNOW 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。, d) k) i/ k1 t# d' j; Z
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
6 D' {, y+ k' D- [公仔箱論壇' E8 `+ \, C9 D1 G
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
/ }* `; R7 }4 W) b# m: R! e5.39.217.76 s* u/ c; v3 h& D
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
