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! L# F5 Q1 O( v V, w8 T
% T" \& E% ^/ i' u公仔箱論壇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. 5.39.217.77:8898* y1 a; V7 i6 \, H, z- S% K
/ g9 I: X. y. Y$ s- `tvb 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.
9 A+ D6 I5 \& J/ e- i
8 ^3 g9 E8 }/ K- Otvb now,tvbnow,bttvb#3 did the obvious choice 40 divided by 2 =20, so he picked 20
* T1 t7 `$ y4 [1 w" ?! F8 xtvb now,tvbnow,bttvb9 N5 F/ r/ g/ h1 r' F
#4 base on knowing the median rule take 60 divided by 3 =20, so he 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。. j3 \ z2 \5 s" G1 p6 O
! ~1 ~9 k- k( d' _( V公仔箱論壇#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
P- b+ n: {& s8 w8 d6 n- PTVBNOW 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。
1 U2 ~+ s+ a* z: ctvb now,tvbnow,bttvbEnded all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
