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should be no2?? choose number that is less than the first one while greater than "leftover beans"/3. Which control wht no.3 & 4 &5 chooses/ A! L7 V: s( |# e9 E) j8 Q$ {

# J2 X  M9 i! V3 Y) G6 EIf no.1 takes very little bean.. e.g. 3 beans.. then no.2 just take one more than him.. bcos when the forth and fifth person is choosing, they should not be able to guess easily wht no.2 has chosen....if no.1 takes 3 beans.. (the logically minimum... since nobody will choose 1... and therefore 2 is the minmum that a normal person will choose... but since equilvalent is dead as well.. minimum will choose is 3).. then no.2 will take 4 beans.. no.3 should be able to tell that one person chose 3 & 4 beans and is highly likely to choose 4 too and hope that 4 &5 choose higher since it is harder to guess the highest and lowest when the numbers are so close together. Therefore no.2 should be the best position???
no.2
NO.3 RIGHT?
I would guess No.4~~couse he quite a lots choose to make no. 5 die~~~like only leave 2 beans at the bag~~when the bag is on No.4's hand and it ONLY got 49 OR LESS beans~~he will sure win~~~
Sorry~~that's wrong answer~~misunderstanding the Question~~~sorry~~~
no. 4
( ?1 f" c6 P8 M$ etvb now,tvbnow,bttvb其實照睇機率no.3都好大~~因為佢有最大能力control no.4 n 5~~because when the bag is on No.3's hand and it ONLY got 57 OR LESS beans~~he will sure win~~~but As 他們都是很聰明的人~~~第一個就唔會咁順灘咁簡多beans~~~如果佢簡好少beans~~like 10 or something(唔想計exactly幾多~haha)~~~~因為2,3,4 都係靠低過佢1粒再逼no.5 點簡都死~~~而no1簡拎少既話~其他人有可能成為最低要死果個~~而佢地都唔會點簡相同數目~~~因為冇point咁做(想同人一齊死就會)~~咁第4個就可以簡前3個人既平均數~~咁就一定中間~~希望岩啦~~~^^
no4
no one dies
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first 1 took 22, second 21, third 20, fourth 19, third 18. so these they wont have the same number of beans, of course 2 of them 5 have to die coz they got biggest n smaller number. seriously idk what is the question want us to answer? is it want them 5 all alive or 2 have them have to sarcarfice?
睇下ans~先!
i wanna see the answer[-O< [-O<
so hard~~公仔箱論壇* f' W. u) X# h. M( f5 e
thxthx
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