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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0 O* g: L |1 Q |& d6 _# \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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! ?2 W4 l5 N6 ]9 N5 P. q' atvb 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.
$ l' I& ?( C' c/ qTVBNOW 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。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。2 S; O/ B" R7 Q+ Y0 K3 S* H; w) w
#3 did the obvious choice 40 divided by 2 =20, so he picked 205.39.217.77:8898: r5 E% o4 s3 h. D
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#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.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 z7 ^# E( X. w3 g& P* q
4 n8 I* @6 \' Z. x6 W( F5.39.217.77:8898#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
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Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
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