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 w3 u' `( Q1 ]' ]* p5.39.217.77:8898
+ ?3 b9 H7 H% |, z% V( ^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. 3 n3 j- n1 k# @3 r# `& F
) @$ b- [5 n2 N3 \
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.
# t+ t% F. s* T+ ]# \2 v( U+ w& x
# i3 _# B! i! J- N" \/ I+ _2 J# r3 A7 a7 hTVBNOW 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。#3 did the obvious choice 40 divided by 2 =20, so he picked 20
6 f- j4 s* X; r* p, L6 }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。; I" d3 {6 E: _4 h
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
v" R9 l( F3 _公仔箱論壇
9 @. l; I( m4 R#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
+ B, |6 d7 _5 O4 w8 T( {tvb now,tvbnow,bttvbTVBNOW 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。# K8 v. s$ H9 b0 a! B
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
% Q: a( R V( _2 G0 x
