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.
! s% t/ p6 E. Z# d# k* F! a/ w* N5.39.217.77:88986 s- x2 |0 L6 k" n6 C' E
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.
! s# ^' m. l8 ]! h9 ?% C3 T. E5 `8 P! \* O
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.* R6 ?( l, ^8 W) Z2 a
公仔箱論壇/ Y2 N6 M, ~7 q/ {! C# B+ H/ {9 ?
#3 did the obvious choice 40 divided by 2 =20, so he picked 20公仔箱論壇& n- B9 p; C9 G4 f
8 |. i7 W, y, ^5.39.217.77:8898#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.5.39.217.77:8898; N: W+ S' H9 g
2 Z, T: }- f, ITVBNOW 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 same as above, he takes 80 divided by 4 =20, picked 20 as well." Z% N3 q/ K, @% h2 c; e! J$ E
2 v- H3 x8 H4 N; E6 y: e6 ^tvb now,tvbnow,bttvbEnded all have the same number and all died. |
發表於 2006-12-14 03:45 PM
| 