# 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

Let $P(A > B) = 0.7$ and $P(B > C) = 0.6$, and suppose that these events are independent. Then

• $P(A > B > C) = P(A > B) \cdot P(B > C) = 0.42$ in which case $\color{blue}{A > C}$.
• $P(B > A,C) = P(B > A) \cdot P(B > C) = 0.18$ in which case $\color{blue}{A > C}$ or $\color{red}{C > A}$.
• $P(A,C > B) = P(A > B) \cdot P(C > B) = 0.28$ in which case $\color{blue}{A > C}$ or $\color{red}{C > A}$.
• $P(C > B > A) = P(B > A) \cdot P(C > B) = 0.12$ in which case $\color{red}{C > A}$.

If we further assume that in those two events where we do not know which of $A,C$ is larger, both events occur with the same probability, then $\color{blue}{P(A > C) = 0.42 + \frac{1}{2}(0.18 + 0.28) = 0.65}$ and $\color{red}{P(C > A) = 0.35}$.

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I think $P(B>A,C)=P(B>A)P(B>C)$ may be wrong, because the two events $B>A$ and $B>C$ are not independent. Can you explain @chtlp's simulation? –  lqhl May 5 '12 at 14:44
I assumed the two events are independent, so by definition they are independent. This could be if e.g. $B$ is "fixed" and the distributions of $A,C$ depend on the value of $B$. (Example: If $A,C = B + U(-1,1)$, with $U(a,b)$ random variables uniformly distributed on $[a,b]$ the statement would be true.) –  TMM May 5 '12 at 15:30
This seems plausible, but I do not believe that A>C and C>A are equally likely in the middle cases, because of the conditional information we are given - the probabilities in the middle cases are not symmetrical. –  Ronald May 5 '12 at 15:42
@Ronald: From the description of the question, I assumed that the observations are fixed, and with probability $0.7$ the first observation is correct, and with probability $0.6$ the second ovservation is correct. With only these conditionals I do not see why $A > C$ or $C > A$ would be more likely in the two middle cases. (Of course "both events occur with the same probability" above is just an assumption, but without assumptions we cannot get an answer.) –  TMM May 5 '12 at 17:16
If we observe that P(A>B) = 0.7 then we somehow get the impression that A is larger than B. If we observe that P(B>C) = 0.6 then we similarly observe that B is larger than C, but not so strongly. I believe this makes a difference. If the observations are assumed to be fixed, as you say, then I think the simulation results seem to be more correct. I will check it independently. –  Ronald May 5 '12 at 23:11

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