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.
/ U, a, J& A+ U5.39.217.77:8898* |# b0 l( w' ]( @7 M4 m+ G9 ^
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. U) Y. O9 f5 I6 S# G
! p$ g: k9 L* C5.39.217.77:8898Now #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.
; A9 j$ m: Y' d) O2 i3 ^4 \- k公仔箱論壇
5 }, p# q4 M. O' A1 s公仔箱論壇#3 did the obvious choice 40 divided by 2 =20, so he picked 20
+ Z7 C! E) a9 K, \& Wtvb now,tvbnow,bttvb
4 ?, g/ ` w: q- E3 J#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.tvb now,tvbnow,bttvb; R3 C1 O& ^2 T! ?# C
5 S N) a& ~" k& A5 T! R4 ?
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.& T, g) j: z3 w2 P, Q
$ I$ W$ P. D/ v/ X- c0 F kEnded all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
公仔箱論壇& J ]4 u! H [/ {# x H
