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.
5 Z E7 B x: i& y5 C. }5 C, tTVBNOW 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。
6 }; q* h4 G! n6 GStarting 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.
& C. p/ I4 k7 v3 {8 ~( A% ^tvb now,tvbnow,bttvb0 }" l% N9 |8 s% ~
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.
9 M( \* F8 c( _1 |7 y/ STVBNOW 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。
; A' w1 u# z" z- btvb now,tvbnow,bttvb#3 did the obvious choice 40 divided by 2 =20, so he picked 20
; Y& W5 p l8 z! S+ J( X& a( Dtvb now,tvbnow,bttvb/ d. O5 I6 @5 `' I
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.1 ^# c+ g+ }- y8 D6 C: ^: f
公仔箱論壇2 s& T! d/ B' v% I" a
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
/ R/ \7 J8 |- p6 i! z2 a
! J* e4 D) C8 Y, r' \/ P% r( REnded all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
tvb now,tvbnow,bttvb) Y" U) W3 `* {
