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.5.39.217.77) V- g! B D8 R
J$ l, ]" x5 g; S. {
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. 5.39.217.77& J3 T" @( A* M. `$ w( o* r
0 C" w* m- s) H( A2 w) G* T5.39.217.77Now #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.
, y5 b' a4 }8 L- j8 G) b/ u0 z8 A公仔箱論壇
& @, p Q3 r0 y* k" e1 P#3 did the obvious choice 40 divided by 2 =20, so he picked 20
L6 d0 }- r3 Y# ctvb now,tvbnow,bttvb! Y# P% Y* q8 S) `
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
& v ^! H- j/ A0 Gtvb now,tvbnow,bttvbtvb now,tvbnow,bttvb/ d$ N7 Z$ X0 _9 P6 x
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
; z3 W% O( _7 D$ r; g# e$ m5.39.217.77
3 q& \& C! _" o3 q, O1 I" o公仔箱論壇Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
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。5 }; n' \2 j4 g! z8 s* u
