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.tvb now,tvbnow,bttvb4 x6 z, \& m# {1 i. P& Z c. x
; W, [; `; n5 @6 M- [$ e! d9 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. tvb now,tvbnow,bttvb/ n& @1 ]; \6 W2 h& U9 f
公仔箱論壇) q1 l6 V9 ~( a
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.
' o/ ^1 c" d; j! o3 |tvb now,tvbnow,bttvb( s/ _0 w# d B0 J6 d; D6 b3 A
#3 did the obvious choice 40 divided by 2 =20, so he picked 20
5 [8 m. g" ]2 x
# H4 w6 N+ Z+ T#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.tvb now,tvbnow,bttvb9 P. i( ~/ l/ q7 ^- e Y: p4 W
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。0 c* a5 f( f# o& {# o* ^2 e
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
" a' a; t% q8 ^! A( D$ U- h
! w: s$ `4 O: B0 d9 r% [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。Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
