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.公仔箱論壇9 J" r; W: f; m9 x
www3.tvboxnow.com+ o) [ @# h( w I; A; D
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.
0 B( @7 a# |6 D/ e& k0 x6 Z! D7 u1 j* 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.
1 P9 w6 G3 E1 x- t6 Z6 R公仔箱論壇
4 d j- Z- t4 {2 ?1 L( Q. B8 Fwww3.tvboxnow.com#3 did the obvious choice 40 divided by 2 =20, so he picked 20
" r3 C' l7 g6 n" t. b# u. Ewww3.tvboxnow.com
3 Q3 x) ~4 g- I公仔箱論壇#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.5 |1 F' [( w4 \, b5 C. N. m& C! C
公仔箱論壇9 {. i& |6 x) O( J) _, X/ s. W
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.6 E7 O/ t; M8 d
7 I$ E' j4 P7 i$ V2 VEnded all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
www3.tvboxnow.com! }' w; t; u: d5 W' e
