# 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?

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Wait, if they are all sampled from the same uniform distribution on $[0,1]$, how can we have $P(A > B) \neq 0.5$? – TMM May 5 '12 at 13:37
@TMM I edited the question. Is it well defined now? – lqhl May 5 '12 at 14:05
There is a potentially interesting Bayesian problem here, struggling to get out. – André Nicolas May 5 '12 at 14:29

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)


=======================update==============================

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

$\frac {0.7\times 0.6+0.7\times 0.4+0.3\times 0.6} {0.7\times 0.6+0.7\times 0.4+0.3\times 0.6+0.3\times 0.6+0.7\times 0.4+0.3\times 0.4} = 0.6027$

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What does "$A > C > B, 0.7 \times 0.4$" mean? Certainly it cannot mean $P(A > C > B) = 0.7 \cdot 0.4$. – TMM May 5 '12 at 15:35
@TMM It means the probability that $A>C>B$ passes the two observers. Since $A>C$, it passes the first observer probability with $70\%$ (she makes a correct observation) probability. Since $C>B$, it passes the second observer with $40\%$ probability (she makes a mistake). And the two events are independent. Hope this solves your problem :) – chtlp May 5 '12 at 15:41
Nope, it doesn't. The two observations are fixed, while the values of $A,B,C$ are not. So what does "the probability that [it] passes the two observers" mean? – TMM May 5 '12 at 15:45
@TMM Imagine we repeat sampling $\langle A, B, C\rangle$ many times, some of them fit the description $P(A>B)=0.7$, $P(B>C)>0.6$ (pass the observers''). And we want to know in these events, how many of them have $A>C$. – chtlp May 5 '12 at 16:16
+1: Despite the downvoting, the simulated answer and the maths is absolutely correct, under the assumption that the values of A,B and C are independent of the observed probabilities. Nice job. – Ronald May 5 '12 at 23:27

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.

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You are assuming $A$ is fixed in $P(A > B) = 0.7$, but it could also be that $B$ is fixed, e.g. $B = 0$ and $A = B + U(-0.3, 0.7)$ and $C = B + U(-0.4, 0.6)$ with $U(a,b)$ a uniformly distributed random variable on $[a,b]$. In that case $P(A > C) > 0.5$ but $P(A > C) \neq 1$. – TMM May 5 '12 at 17:24
As I said, it depends on the method of making the probabilistic observations. What I said is consistent with the observations, and I would argue is the most natural way for those observations to occur, but there are other possible cases. – Ronald May 5 '12 at 23:02