c*w
发帖数: 4736 | 1 两个将军,在三个战场上分配各自的士兵. 在一个战场上,
谁的士兵多谁赢. 赢得两个战场胜利的赢得整个战役的胜利.
两位将军有同样多的士兵 (很多, 可以认为连续). 那末他们
应怎样分配士兵才最有可能赢得战役?
这是一个很经典的game theory问题,可如果过去没见过, 做
着也不容易. | h*l
发帖数: 19 | 2 There is clearly a problem. Say, suppose (x1,x2,x3)
(x1<=x2<=x3)
is a good strategy, A will always use it since he asumes it
always
leads highest winning probability(故伎重演). Then B can just
choose
(x1+d,x2+d,x3-2d) for 100% win. So, A actually don't have a
optimzed
winning point even in propability sense. People's thinking
choosing
a point cover maximum area in the sense someone points out
in his
post before (outside of "three-trangle") is wrong! It
depends on
rival's strategy.
However, it' | z***e
发帖数: 5600 | 3 No, the optimal strategy does not have to be a deterministic
strategy ( (x1,x2,x3) as you said). On the other hand, if
there is a strategy such that can beat every deterministic
strategy by at least 50% chance, then this strategy will
beat
any kind of strategy by at least 50% (and actually should be
=50% a.e). From the original purposer
of the problem, it seems that this optimal strategy does
exist, which would be the most surprising result. I am
still
trying to figure out what probability m
【在 h*l 的大作中提到】
: There is clearly a problem. Say, suppose (x1,x2,x3)
: (x1<=x2<=x3)
: is a good strategy, A will always use it since he asumes it
: always
: leads highest winning probability(故伎重演). Then B can just
: choose
: (x1+d,x2+d,x3-2d) for 100% win. So, A actually don't have a
: optimzed
: winning point even in propability sense. People's thinking
: choosing
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