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.
" p8 `+ ]$ R; W& ~: x$ F9 Wtvb now,tvbnow,bttvb+ T9 A, k7 e" I7 B
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.
: l$ Y; T2 G' X* d5.39.217.77:8898
1 U, }# Z6 l! \tvb now,tvbnow,bttvbNow #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.5.39.217.77:8898* Q! m3 h$ ]( r S
) g/ `" J& C6 S B$ L, r, |$ r#3 did the obvious choice 40 divided by 2 =20, so he picked 20
1 F2 ~; j9 z& C; F% x0 A* @- }公仔箱論壇
2 x2 [1 Z& Z8 A/ ]& WTVBNOW 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。#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.5 i0 L) @1 C0 K1 ^2 I g* k! k
公仔箱論壇7 y0 g2 A5 A; G2 Y
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.5.39.217.77:8898! s8 P1 N* z' M' g4 {. S2 p
! I4 U6 G) [0 V- d0 G) J2 ]Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
