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.5.39.217.77:8898! ?, \4 i3 \& K9 E0 K7 B
$ g9 t! P7 r( v2 B4 [* j; xTVBNOW 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。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. 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。) `" m+ ]9 U/ H
) o# C1 ?! n2 w! x1 M. SNow #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.
h3 A/ b5 p/ A/ r9 W3 ~" c
8 b5 C$ P' M0 O4 g* y: c; \#3 did the obvious choice 40 divided by 2 =20, so he picked 20& h; q, r6 k" R* A
' d! C& j- Q4 p! B公仔箱論壇#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.0 e! k1 {: l, ]* a7 p) f3 I: k8 k4 N, ]
; O m* q1 Y1 o: E
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.公仔箱論壇7 ` O1 w8 R c W. w
5.39.217.77:8898% S! M) i6 o/ b( T& S3 }9 g7 G
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
公仔箱論壇( C8 L1 B6 k1 n: `0 R y
