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.& t9 G% j" G6 w( \' n
j8 Q; y/ i6 ~) w! O& N9 O5 J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.
& ~$ c: ?. @; I! u- T. \' otvb now,tvbnow,bttvb2 x9 ?/ v7 s/ M }( @8 S- O5 }
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。( w# Z2 r$ T& Y3 O8 F6 K
7 M0 g2 S2 d* r1 m- E5 f
#3 did the obvious choice 40 divided by 2 =20, so he picked 20 p) b* k- {! e8 i$ j5 p
4 g2 o9 O* y* W( U6 |5.39.217.76#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
& B) H0 R1 r# z9 J+ H% p. Q4 ]0 e1 G
% x/ X0 _- c4 r; r#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.公仔箱論壇5 Y1 c" d+ Z$ P8 C7 m
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。2 m# i. `- L7 a6 f/ W8 w5 Y
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
