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86#
發表於 2003-7-19 04:04 AM
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actually for the diamond question, i do have a totally different approach.
if the answer is either A97, B0, C1, D2, E0
OR A97, B0, C1, D0, E2. D and E knows that they only can get 2 diamonds and D gets nothing at all! So if D and E are smart, i say they will work together! since if they vote no all the way to the end, they will split the diamonds 50 50 each. so my approach is as follows
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D,E (50,50)
C,D,E (49,0,51) or (49,51,0) 1 diamond should be enough to break the team.
//
Then if B is in charge, there is no way he can top C's offer so B dies for sure if he is in charge. Therefor B has to vote whatever A's plan is.
//A can split the diamond this way, 48,0,0,52,0 or 48,0,0,0,52. A will vote yes, B will vote yes because he wants to stay alive, and whoever gets 52 diamonds (D or E) will vote yes since he is getting more from this deal and it is enough to betray the other person. |
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