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.
7 `5 ~$ ?7 J/ l' L& y2 z/ Etvb now,tvbnow,bttvb
) w! A+ H$ I# s+ htvb now,tvbnow,bttvbStarting 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. 0 M/ [( y" y- n" t/ g7 R2 w
5.39.217.76% B8 }# B$ S& X
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.0 \" H$ Q$ ~7 L# F( [
公仔箱論壇( f) ?1 \4 \1 b% I+ k
#3 did the obvious choice 40 divided by 2 =20, so he picked 20公仔箱論壇8 ]$ g( C$ ?. a% x, X7 E' q) z
+ q% ]9 u1 j) z5 v! M#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
P3 ?% B1 t; r8 Y7 `- c2 j; q; [5.39.217.76
( [) f1 M/ i! X% g: b- v- T公仔箱論壇#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.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。( x9 t5 T% m* K3 M! W5 \# K
! c0 S0 w2 S X4 V) Ltvb now,tvbnow,bttvbEnded all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
