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. Z$ t1 u! q0 c0 q/ j" N# Y( ]2 I& A! K
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。! w u6 c- x) c
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! j8 G/ h- p6 }" g4 u/ W' A, M
tvb now,tvbnow,bttvb, W( B3 g( s' U/ u5 d5 d! Q- B
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.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。1 ]) I2 Y* y9 y0 e8 T
: ]$ a* D8 u( ?( g( w0 d: a$ N3 C公仔箱論壇#3 did the obvious choice 40 divided by 2 =20, so he picked 20tvb now,tvbnow,bttvb1 F% E0 q: f, |8 j# b; P F% f; o
tvb now,tvbnow,bttvb! Q* q# \( L# P, Y% \ ?6 Q
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.3 S8 G9 J2 u( R
: ^! ?0 q: j% ?1 s0 D$ J' p& s公仔箱論壇#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
, ?1 E$ r; U0 Z9 ], o+ ]公仔箱論壇6 t* X: T% A/ K! G/ }
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
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。5 c3 u3 X5 H6 f4 A$ r
