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.
# Q/ H4 A. v2 t7 \2 l* ?4 a% E
( ~) T7 R! {. x& gTVBNOW 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。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. H6 g2 o! C; [2 c, x
4 v9 M7 o: k8 S+ Q$ Z* QNow #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.+ A/ O* S1 f4 d% F6 T; {4 o
% M0 R1 X7 | `9 r2 I#3 did the obvious choice 40 divided by 2 =20, so he picked 20tvb now,tvbnow,bttvb: Q/ H& C8 `( h L" p+ F- M( h
tvb now,tvbnow,bttvb* M v/ q/ s- q, K" T
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.5.39.217.77) a8 `+ l% e5 g
9 i! K1 R& M1 x) VTVBNOW 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.
* k) Y; E! n1 j2 j6 ?' ^9 ^, A' {1 K. r9 K
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
