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+ I' w. C$ g, W; |1 d, `! t
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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.
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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.tvb now,tvbnow,bttvb5 T6 E' H& e% [. j/ R
4 y% r( _: q$ o# x; Q& @公仔箱論壇#3 did the obvious choice 40 divided by 2 =20, so he picked 20tvb now,tvbnow,bttvb6 U" t/ N) \0 h- u2 H
" ?& j3 t7 }8 G1 @, n) Q! `5.39.217.76#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
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#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.5.39.217.768 O1 b' w7 @$ ^
. @ y( R0 {# \3 @' Y# b/ r7 Ptvb now,tvbnow,bttvbEnded all have the same number and all died. |
發表於 2006-12-9 12:41 AM
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