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.6 ?1 Q1 k7 F0 K6 |; l; B/ @
+ n6 h( Y; t) G9 E8 w公仔箱論壇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. . d- U3 G' Z- _! v8 ]% ^* t- Z
; T4 @! j& k& e: {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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#3 did the obvious choice 40 divided by 2 =20, so he picked 205.39.217.76- K4 N. C) N' U) M
, |" X. Z) }1 f; I5.39.217.76#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
. p8 g' m: g& w5 }+ ?$ ?TVBNOW 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。
7 ~$ r, v" j, x& o! q) m! Z5.39.217.76#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.# J! i. j7 T8 K* k0 i5 v
TVBNOW 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。1 m x. s1 }% |# R
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
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