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.tvb now,tvbnow,bttvb* F4 T e5 v( a: g2 [
* m% U$ {9 b* A, u8 L+ l8 h4 W B, iStarting 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.
+ m' }& K$ W. FTVBNOW 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。' F5 b: U+ R6 Q) h. ~# L
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.
2 R. Y) g; X" P% htvb now,tvbnow,bttvb公仔箱論壇4 G4 f+ j: H9 M4 ^: W3 e& O! V
#3 did the obvious choice 40 divided by 2 =20, so he picked 205.39.217.772 [5 u6 U) x6 K$ D5 m% J
* D6 T$ q4 d# M% w公仔箱論壇#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
" o$ {* x* j8 ~ @ S3 K5 c4 n
( d5 p2 \% T8 G! J% T! G, h5.39.217.77#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
% z3 ?' Y+ L# f8 @7 g公仔箱論壇. L, w. E# u- m, R( l" t6 F' k$ j
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
