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.
, D/ @1 Q6 Q- T6 J/ V, }& O' f# } L' N, E
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.
1 c Y3 r5 c, Q: D4 {' \6 B# S7 \8 L- H! F, Y
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.公仔箱論壇' L' |0 `6 c! H6 m+ M
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。, f2 J7 ?8 y d. t) O- x( S, H
#3 did the obvious choice 40 divided by 2 =20, so he picked 20
0 H) k) o0 l; Y( ]os.tvboxnow.com
4 R! U$ O- e; U& h2 k& Z) r- L* Ktvb now,tvbnow,bttvb#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
; }$ @9 d( V( S& hos.tvboxnow.comos.tvboxnow.com; d0 C5 l, u" b, l- Q) _1 s" x$ |
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.
/ }: h8 Q2 R$ W# q4 ^& X公仔箱論壇公仔箱論壇6 T5 K7 C( C' w, G
Ended all have the same number and all died. |