Well, depends on if the probabilities overlap. So:
P(A)=.5 P(A&B)=.1 P(&~B)=.2
is Dutch-Bookable
It seems closer to the solvability of a system of linear equations. Depends on what kind of probabilities you get? Like if you have
P(A), p(B), p(C), p(A&B)=P(B&C)=P(A&C)=0, it’s trivial.
But if you have
P()=1/4
then you’ve got trouble.
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Well, depends on if the probabilities overlap. So:
P(A)=.5 P(A&B)=.1 P(&~B)=.2
is Dutch-Bookable
It seems closer to the solvability of a system of linear equations. Depends on what kind of probabilities you get? Like if you have
P(A), p(B), p(C), p(A&B)=P(B&C)=P(A&C)=0, it’s trivial.
But if you have
P()=1/4
then you’ve got trouble.