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.
$ l5 S9 D% l1 P2 t0 a+ k
: n+ E: X) Y) K5.39.217.77:8898Starting 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 j8 p: c2 M6 |: r+ Y' b
5.39.217.77:8898$ ?. y. g$ v8 w# w; H' e
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.' n+ Q0 B0 q* u) v1 \7 |8 r3 J
% w+ s# Z# @9 X( J( w, G) b) ]6 w+ ^#3 did the obvious choice 40 divided by 2 =20, so he picked 20
* X8 M( Q \/ v W6 J6 c: F. X' T. S
# R8 e3 j# ]- D& Z6 i* t#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
+ F- ^5 ~, v# g. k5 f6 [$ Q9 YTVBNOW 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。
' J& H" v; T- R! J5.39.217.77:8898#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.tvb now,tvbnow,bttvb H0 O( S* k: \* r
7 S% X/ n# Q' |7 [5 I2 E8 s+ {tvb now,tvbnow,bttvbEnded all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
$ {& `& H7 Q2 i! Q. F5 E+ K$ W
