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.
4 ?' p4 z* [. v% @) Z' ]( VTVBNOW 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。公仔箱論壇9 X! c2 n. ]: ^- r1 ^
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. 公仔箱論壇/ L3 l% s9 y8 d$ U4 g. A: Q# y
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。* ~4 H/ c6 n1 ?- ^9 y
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.
2 N" N2 D3 N& F( A4 b9 X
: p& e! s m+ t( I( B& b/ x#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。( O+ o5 t( K2 r! ^! U
2 F9 s+ I* [1 P5 e; K" h6 z
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.$ [( x( j" U6 h+ r" o
. |7 Z/ [6 n X% R4 [& M6 n0 ^
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
' w# d3 G/ r# ~+ q. J* l) A5.39.217.77% R6 ?. U7 I4 c0 G) E( E
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
u3 ?& ^ U( P; D# c) {# c
