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.
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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. . R0 }8 D3 W' e$ D$ E
2 M" B- R2 a7 u, l* ~. T3 m( rNow #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.公仔箱論壇6 Z5 \1 h3 X* u' a
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#3 did the obvious choice 40 divided by 2 =20, so he picked 205.39.217.77:8898; {7 \+ `+ H3 G, {; O* u: y
$ T& N/ M3 C, t j) Y }# ]" k0 }/ a5.39.217.77:8898#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.5.39.217.77:8898, ^ ^, e/ k" g- o8 h; ]& s4 ^
, T1 v% G" o6 R2 `6 v! f8 Z公仔箱論壇#5 same as above, he takes 80 divided by 4 =20, picked 20 as well., S) ^# j7 o! [* d: L8 R
! r9 \. v Q: L/ p! g) F% E5.39.217.77:8898Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
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