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,bttvb3 N7 j' @/ b" Y: t7 S
3 q# n2 ?8 L5 Wtvb 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.
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6 S2 l3 q0 w: q6 Q* H公仔箱論壇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.
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! ?% Y* [5 j' @0 B k6 ?$ L#3 did the obvious choice 40 divided by 2 =20, so he picked 20
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#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
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1 x9 z8 J4 ?2 x+ m7 @7 j. ktvb now,tvbnow,bttvb#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.tvb now,tvbnow,bttvb# q) U# P+ k; _9 l2 ?- f/ q
$ c1 c8 R$ M3 lEnded all have the same number and all died. |
發表於 2006-12-9 12:41 AM
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