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.1 K1 M- ~2 w; I# p; y1 L* h
7 {3 U; _7 A8 z8 q+ t% d0 Vtvboxnow.comStarting 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 z2 D, e0 u$ }9 S1 F& O- O5 N" \- X ~tvboxnow.com公仔箱論壇; M v. g8 {6 i1 B
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.0 S& E- U+ c& Z# G2 }2 K# X
- {/ l8 a& j. \( `, v& M9 c( X0 U#3 did the obvious choice 40 divided by 2 =20, so he picked 20
5 e0 v! k; x/ P# f! q3 U; rTVBNOW 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。! b. L/ D$ E% t- ~
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
7 D. d) X. g+ Q- _% ?tvb now,tvbnow,bttvb
# Q$ }5 Z3 j8 V/ ~公仔箱論壇#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
$ m/ w1 K% Y' H" |
1 ]2 ?; z! D* ?* W, {3 {0 GEnded all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
