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.
2 c1 P W e/ l# C$ }% hTVBNOW 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。
0 G7 u7 h; F; X) }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.
( P+ W% B# m& i- jTVBNOW 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。
. b2 B- Y$ E4 Z1 Etvb 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.
( }& k0 M5 v( @$ _5 i3 cwww1.tvboxnow.com! P0 ~+ H% {# m) U* C0 R' z
#3 did the obvious choice 40 divided by 2 =20, so he picked 20公仔箱論壇; \, G' M: I' L V6 ~* ^
3 e; p: D- K( m" o" l$ [#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
# P$ R& `4 J/ P4 g& F8 k" [
5 R: P \, V0 x$ J/ w0 J4 ?" ]7 T公仔箱論壇#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
; S2 e O; Z9 p# s8 Q+ M; j( ~- q公仔箱論壇! @! S7 {8 I. y1 G8 z! x/ h& W
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
