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.
3 S% w& t, \1 |+ t) m! O+ o c2 h2 _/ \
9 ^' q! U# A n4 L2 Z, BStarting 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.
! l6 O$ L a; j% @8 oTVBNOW 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。7 u' s! Q# j& C4 J0 e
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.) j2 M w( @/ z1 B3 Q. N7 B& M2 F
5 p1 u$ T' n8 [( _tvb now,tvbnow,bttvb#3 did the obvious choice 40 divided by 2 =20, so he picked 20
, P: H A/ e; H* R* G# p' F- j$ [tvb now,tvbnow,bttvb5.39.217.77:8898& u+ w: w! l, x) X
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.% F$ V0 y: P/ M' w. R
9 a& l% T" Q( s5 u- A* b C5.39.217.77:8898#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.6 S; S/ S0 W7 W+ T( F3 ]
: k* z$ K- K- ^2 p% X2 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。Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
