Ranking probability problem $A, B, C$ are independently sampled from an uniform distribution in $[0, 1]$.
We know $P(A > B) = 0.7, P(B > C) = 0.6$, what is $P(A > C)$?
Is this a well defined problem? Does it have a sensible answer?
EDIT: Suppose we have two careless observers.
An observer observes $A > B$ and there are 70% probability that she is right.
Another observer observes $B > C$ and there are 60% probability that she is right. So what is the probability of $A > C$ in the underlying event?
 A: I wrote following MATLAB code. Simulation results show the probability is around 0.602. I hope someone could confirm this with an analytic answer.
N = 1000000;

A = rand(N, 1);
B = rand(N, 1);
C = rand(N, 1);

p1 = 0.7;
p2 = 0.6;

c1 = rand(N, 1);
c2 = rand(N, 1);

ob1 = ((A > B) & (c1 < p1)) | ((A < B) & (c1 > p1));
ob2 = ((B > C) & (c2 < p2)) | ((B < C) & (c2 > p2));

ob = ob1 & ob2;

pos = ob & (A > C);


sum(pos) / sum(ob)

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I enumerate all the 6 possibilities of relative order of $A, B, C$. They all appear with probability 1/6.
The following lists shows with how much probability each case passes the two observers


*

*$A>B>C$, $0.7\times 0.6$

*$A>C>B$, $0.7\times 0.4$

*$B>A>C$, $0.3\times 0.6$

*$B>C>A$, $0.3\times 0.6$

*$C>A>B$, $0.7\times 0.4$

*$C>B>A$, $0.3\times 0.4$
Among them, $A>B>C$, $A>C>B$, $B>A>C$ are the valid cases. So
$\frac {0.7\times 0.6+0.7\times 0.4+0.3\times 0.6} {0.7\times 0.6+0.7\times 0.4+0.3\times 0.6+0.3\times 0.6+0.7\times 0.4+0.3\times 0.4} = 0.6027$
A: Ah. It depends strongly on the method for making those probabilistic observations.
For example:
If we observe that A=0.7, then we should note P(A>B)=0.7.
If we observe that C=0.4, then we should note P(B>C)=0.6.
(This is perhaps the most obvious, natural way of accessing those probabilities. An observation of B would affect both probabilities)
And, if those were our observations, then it's absolutely guaranteed that A>C.
P(A>C) = 1.
