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. Z+ y$ t" r' {$ }0 e. ^$ i1 Q
4 ?; F; H% H# Y$ U3 W" V; Utvb now,tvbnow,bttvbStarting 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.
# D& a) z* V8 G% k* i! Ktvb now,tvbnow,bttvb
8 C6 _0 h% {0 e1 w# XTVBNOW 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。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.1 n, m8 {; M* r
, q% K8 @9 X1 a' c+ a% F0 v#3 did the obvious choice 40 divided by 2 =20, so he picked 20tvb now,tvbnow,bttvb6 k/ |" K7 i8 {: f+ `1 y
# B$ A3 b0 G( q3 y4 ^. vtvb now,tvbnow,bttvb#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.tvb now,tvbnow,bttvb7 I$ i$ j: o- [; |' v3 E3 ~& \
1 ?5 C7 f) {- q! z. b& s#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.www3.tvboxnow.com; Z: }) w9 `/ y1 v
+ ]. ~- t$ ^+ ~4 n( l8 j1 c2 k公仔箱論壇Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
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