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.www3.tvboxnow.com7 C# i7 A& Y7 ~' s2 a
! F' D# Q/ L. x" }! xStarting 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.
2 o1 ~% i1 [3 I3 ~tvb now,tvbnow,bttvb8 @, U. A- M$ ^$ Z, S1 ?
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.
# F3 w- u) }$ E0 g `6 n% ~' o公仔箱論壇www3.tvboxnow.com6 b. g5 j/ L& J; h; s6 R
#3 did the obvious choice 40 divided by 2 =20, so he picked 20
/ s# U2 C7 v; @- S3 y公仔箱論壇
+ W9 J) n8 u6 }tvb now,tvbnow,bttvb#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.' n3 d8 q5 z% l6 f! j' x
公仔箱論壇: ?5 i. b. U8 k! r
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
( K. T/ N) o1 F+ h& `7 y' P( {tvb now,tvbnow,bttvb
2 h" W3 j- d) D" w w4 j公仔箱論壇Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
* G& ?: V6 m2 t8 ^& M( I( H" U
