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,bttvb7 h! r: S" W7 M, U) p
7 t3 C" h1 O: h% ^; o7 k7 a5.39.217.77Starting 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. 5.39.217.775 p" H& o$ R' U& t3 B( k I
! ~8 f2 ?: O3 C3 b& g( |3 F4 ltvb now,tvbnow,bttvbNow #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,bttvb4 q# H* C s& h. z* E$ p4 l, R5 a
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#3 did the obvious choice 40 divided by 2 =20, so he picked 201 E* `3 h/ h/ a" A$ N& ~
; V3 j' s4 U* v* u#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.公仔箱論壇. L& U# l9 d2 e3 T
$ L- s. J3 N# Atvb now,tvbnow,bttvb#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
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9 B1 e& T9 l6 t# ~! PEnded all have the same number and all died. |
發表於 2006-12-9 12:41 AM
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