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.5.39.217.768 M; A! U- R% r" F4 `
公仔箱論壇; T+ J, M( ]- S1 ?! ^
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,bttvb+ t7 O( j: ]6 N- F3 R6 H- h1 Y
) S$ M: X% V. @. ?0 N. I) I- hNow #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.
: w, F5 _9 r, C公仔箱論壇; C4 H- s. D# {; M1 v
#3 did the obvious choice 40 divided by 2 =20, so he picked 20
- p Q. |# G4 W3 A m* \公仔箱論壇, }. _5 ^" o' [
#4 base on knowing the median rule take 60 divided by 3 =20, so he picked 20 as well.
* Q) s1 N3 Z/ I- ~8 v5.39.217.76tvb now,tvbnow,bttvb l3 y. U, g/ O1 M6 N# _& }
#5 same as above, he takes 80 divided by 4 =20, picked 20 as well.1 X( z. h. h2 Y
5.39.217.76( k. U. u6 B6 W; |/ S- A
Ended all have the same number and all died. |
發表於 2006-12-9 12:41 AM
| 
