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.tvb now,tvbnow,bttvb/ v- i. ^( {& E0 \$ `
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。0 f2 A& _7 h# s( F' d6 S
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. tvb now,tvbnow,bttvb8 Y2 Z+ g) |8 `/ Z& E6 _' |
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。+ ^1 U2 R% _4 A+ [8 y" d
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.* ?. n2 `) p% P. s
3 ]6 a) G2 t( N H8 y#3 did the obvious choice 40 divided by 2 =20, so he picked 20
' u" n8 |% p! y0 y* c公仔箱論壇
, N; W: c' ]+ c1 y# p#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
, Z9 D' p. D) u3 T5.39.217.76tvb now,tvbnow,bttvb3 w' x! L% z: |, O% H
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
4 N! u, L3 E) }, }+ G: v' Vtvb now,tvbnow,bttvbTVBNOW 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。" w9 Y& z# j) f6 E
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
