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.
_8 {7 W- [4 D# }+ jwww2.tvboxnow.com# k' P5 \7 k/ ]& c0 V. t8 I
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.
5 i6 _+ t7 C; Q- x" j5 i5 e: Ytvb now,tvbnow,bttvb
: R* x- g& `. [: W- [# G' K" r公仔箱論壇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.tvb now,tvbnow,bttvb9 Y0 k: T$ a' f& M4 F
t' o2 [* R& v& h. M# ^tvb now,tvbnow,bttvb#3 did the obvious choice 40 divided by 2 =20, so he picked 20www2.tvboxnow.com2 V8 H2 d3 w: }( W' T" k/ b
( W( K: ? J4 ]) O$ p8 ztvb now,tvbnow,bttvb#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
9 J& O3 ?! ^! }/ \% T- k# f# ^9 D1 o8 v5 R
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
# o P# W+ } v5 w
% X) V- k: q* r4 R7 Ctvb now,tvbnow,bttvbEnded all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
