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.
6 Z: U5 J/ E, U( `1 owww2.tvboxnow.com公仔箱論壇, y$ {) x& _/ l- K. w6 N) C% j6 @
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. tvb now,tvbnow,bttvb2 H+ E$ O' V: U7 V, O( c7 q
2 O# r6 X, e* r, O. s! x. Rtvb now,tvbnow,bttvbNow #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.
% g% f5 x5 g0 {. STVBNOW 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。
! L* e7 V' S, c#3 did the obvious choice 40 divided by 2 =20, so he picked 20; w8 b9 c" n! ]' A1 s( H/ ^
3 z6 G) d. z4 Q2 S4 E2 s! t- k公仔箱論壇#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
9 e( A6 M6 u& a2 C4 XTVBNOW 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。www2.tvboxnow.com9 `5 i1 g6 m2 O& ^$ f6 t* p# p' [
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well./ z6 |6 e' W' r) X8 x% g" S4 U
& K0 h. I+ H! g! K$ cwww2.tvboxnow.comEnded all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
