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.
. i2 Q% N6 t: D; R% [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。
/ R) n5 K5 Q7 V- g" I6 p$ `: A9 O# H公仔箱論壇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. 7 N) O a( A/ B. o. S V7 s" S
# d$ {" C& x- t. [2 ]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.
; G. N0 B. U/ E0 {& n) E公仔箱論壇: y( H+ I2 ~ A) z
#3 did the obvious choice 40 divided by 2 =20, so he picked 20
8 P" o; Z7 _9 J, E2 `% G公仔箱論壇
3 W5 ^3 h7 M6 p" H7 Q+ q0 F公仔箱論壇#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well./ y! t u9 v3 g7 O2 ?0 a7 x
+ K- @! U* r4 k: p
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
* s* H+ x) F1 A- T4 E% J, M公仔箱論壇
" _) L3 R8 G7 M/ ?) B: q4 M( }Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
5.39.217.76' U" Y. Y! A B B3 j/ e
