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.公仔箱論壇0 O1 m; d0 m/ L! _2 W
www3.tvboxnow.com7 [+ x' i M0 N. `" @
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.
! ^ ^6 Z% S( _' ?公仔箱論壇8 y; o% }: L/ N" g8 J
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.
4 b5 N) l& { L$ z; C9 @) r+ m( x! g4 V
#3 did the obvious choice 40 divided by 2 =20, so he picked 206 L' b* ]4 t6 a: x4 C8 S3 A
3 n% f; ~- e9 [2 W7 \1 @1 s3 c#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
. ?+ M3 F9 e0 k% Q9 }4 g公仔箱論壇0 v; [: z6 y5 @) F# k
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
' G& H E3 `4 O& x* C1 L
7 ]2 s! q8 s1 |# n7 PEnded all have the same number and all died. |