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: L j5 Z: F9 P) B: k公仔箱論壇4 ]3 s/ Q7 |1 v6 Z
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 ]# ^* g, w4 k5 Dtvb now,tvbnow,bttvb
. R9 l6 V0 c `6 N% t/ ^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.www3.tvboxnow.com) g2 w% d0 W( s/ V+ ^; \% R
1 r6 d/ ]3 F" T6 O#3 did the obvious choice 40 divided by 2 =20, so he picked 20TVBNOW 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。- y0 s' L/ U3 R3 T; V* M
( y2 p2 f) ]0 U* l) ^! B; q' G2 {9 | ~公仔箱論壇#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.tvb now,tvbnow,bttvb$ ?4 w. A: F: v/ F3 t
8 i# K4 M4 H" J9 a#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.& c b3 b) s4 E3 r7 S! q% f
6 p: O, t5 ?5 y$ O2 }9 X8 dTVBNOW 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。Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
