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.
8 h+ x7 h R, ^! R4 j
. I0 `0 M1 m" N: g: g; a. \5.39.217.76Starting 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.
8 _: ~" J! {, N& K
8 y! X/ v% B! K0 Ptvb now,tvbnow,bttvbNow #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.
0 @3 [. Z- h" O- h
9 l' D+ _& S, x. j( P$ Z# {6 k# _/ M9 ~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。( l/ b, f0 R, y5 \
$ m, q2 T/ T' A2 K4 m7 Q0 utvb now,tvbnow,bttvb#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
9 b# K& K0 H C0 v# w0 r
) [9 C! t# J9 s- m% C$ U T) {#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
6 Y+ @- I8 o/ mtvb now,tvbnow,bttvb
1 X" `; @; [; B( W* E5.39.217.76Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
1 C1 |( T* `2 ~8 e# E& b8 Z {" ?
