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., ~, Q D) V$ t5 P& L
) J! g# o6 Y; P; O8 ATVBNOW 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。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. tvb now,tvbnow,bttvb6 m1 L1 C- j) P3 j
# Y1 R+ w' V4 A: ]7 G! z
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.
/ B. I5 Q. r" t8 otvb now,tvbnow,bttvb5.39.217.763 V9 O4 L" k! X8 Y- a
#3 did the obvious choice 40 divided by 2 =20, so he picked 20
+ Q3 H6 S- {& Jtvb now,tvbnow,bttvb5.39.217.76# p; U3 n3 G$ e7 v+ W6 N
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.公仔箱論壇- }1 n: r6 f2 g. s! M
: k1 L! U! D7 ?8 e: s- i8 O5.39.217.76#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
, e9 s+ l5 h& D; L2 R8 K/ s* l# z
% W+ A/ E- T6 \" @- X& |, k1 C/ xEnded all have the same number and all died. |