number 4 has the best chance to live& k! f' f: ~6 [% f1 `! K/ Y
assume number 1, 2, 3 and 4, pick number a, b, c and d.1 D) R- R$ F% i5 [* }9 Q6 n
number 4 knows how many left in the bag which is 100-a-b-c
5 d. T5 h- h$ n0 x* ^/ ktake the average of (100-a-b-c)
+ H7 M# b8 X& N+ {5.39.217.76then number 4 knows a or b or c must be greater or less than average of (100-a-b-c)tvb now,tvbnow,bttvb' M8 c: L) @2 W7 y: j( x: T6 W
and number 4 can squeeze himself into the average and what number 5 picks.
) N& h. B. g: P& c: P5.39.217.76so number 4 will have the best chance.
4 {/ [1 s2 s. U' q: e# P* Rtvb now,tvbnow,bttvb5 b, c9 D3 I% s
i hope this is a good logic hehe |
發表於 2006-12-12 09:29 AM
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