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.
+ U% b$ L% ^5 W% Z5 I# xtvb now,tvbnow,bttvb
8 ~- V# `( @4 YStarting 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. 公仔箱論壇- {; ~: f, ]( v% h
9 R4 F9 U9 n* O5 ^/ x, j& r公仔箱論壇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.5.39.217.76+ Q$ f1 o" C, M1 J
公仔箱論壇( M+ |) w8 u, S3 I" G5 k. M1 Z7 R
#3 did the obvious choice 40 divided by 2 =20, so he picked 20
- t/ @- Q$ A: E; q, \' B1 U5.39.217.76
2 G6 U4 h @, \#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
0 f _/ U9 D* e m# o! ZTVBNOW 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。9 E# @4 J4 o; u7 ]
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
3 N$ ~5 B, l! W( U1 G+ e3 [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。tvb now,tvbnow,bttvb5 c7 j# R% W% o H% I" A4 x
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
tvb now,tvbnow,bttvb& b7 Y* P( e3 o7 x
