# Probability of an element being the maximum element in an array?

I've come up with a problem that seems so simple but I could not find where the problem is.

Assume we have three elements $A$, $B$, and $C$. Each element is a random number from $(0,1)$. What is the probability that $A$ is the maximum element in this array?

I think it can be expressed as $P( \{A>B\} \cap \{A>C\} )$, where $\{A>B\}$ and $\{A>C\}$ are independent events. Then seems the probability is 1/4. But surely we know the probability should be 1/3.

Where's the problem and what is the correct approach to calculate this probability?

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In your calculation, you presumably used symmetry to conclude (correctly) that $P(A>B)=1/2$. If twofold symmetry is OK to exploit, what's wrong with exploiting threefold symmetry? –  André Nicolas Jan 10 '12 at 1:19

The answer is clearly $1/3$. What is written below may make that fact more obscure rather than less.

The joint density function is $1$ on the unit cube, and $0$ elsewhere. We can replace $>$ by $\ge$ without changing the probability.

The probability can be expressed as the iterated integral $$\int_{x=0}^1 \left(\int_{y=0}^x \left(\int_{z=0}^x 1\,dz\right) \,dy\right)\,dx.$$

Or else we can note that unit cube has volume $1$. Our probability is the volume of the part $K$ of the unit cube such that $x \ge y$ and $x \ge z$. That volume is the iterated integral above. But we can find the volume of $K$ more geometrically by noting that $3$ copies of it fit together to make up the unit cube.

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$\{A > B\}$ and $\{A > C\}$ are not independent events.

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Then how can we calculate it then? –  derekhh Jan 10 '12 at 0:41
@derekhh: You did, it is $1/3$. –  André Nicolas Jan 10 '12 at 0:48
@AndréNicolas: Er, I'm trying to see if it is possible to calculate the probability in the basic approach...But failed... –  derekhh Jan 10 '12 at 0:49

I'm trying to see if it is possible to calculate the probability in the basic approach

If $A = \alpha$, then the conditional probability that $\{B < \alpha, C < \alpha\}$ is $\alpha^2$ by independence. So, the unconditional probability can be found by multiplying by the pdf of $A$ and integrating. ($\int_0^1 \alpha^2 d\alpha = 1/3$). If you don't like conditional probabilities,

$$P(A > B, C) = \int_0^1 \int_0^\alpha \int_0^\alpha f(\alpha, \beta, \gamma) \mathrm d\gamma \mathrm d\beta \mathrm d\alpha = \int_0^1 \int_0^\alpha \int_0^\alpha 1 \mathrm d\gamma \mathrm d\beta \mathrm d\alpha = \frac{1}{3}$$

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I assume that the random variables $A$, $B$ and $C$ are independent. Since these are continuous random variables, with probability $1$ they are all different. There are $6$ disjoint events, the possible orderings: $A < B < C$, $A < C < B$, $B < A < C$, $B < C < A$, $C < A < B$, $C < B < A$, each of which (by symmetry) has probability $1/6$. All the other events can be written as unions of these. So e.g. $P((B<A) \cap (C<A)) = P(B < C < A) + P(C < B < A) = 1/3$, while $P(B<A) = P(B < C < A) + P(B < A < C) + P(C < B < A) = 1/2$.

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There are six equally probable orders for independent identically distributed continuous random variables:

• {A > B > C}
• {A > C > B}
• {B > A > C}
• {B > C > A}
• {C > A > B}
• {C > B > A}

plus various orders with some equality, each of probability $0$.

So clearly the probability that $A$ is the maximum element is $\frac{2}{6} = \frac{1}{3}$.

As opt says, $\{A > B \}$ and $\{A > C \}$ are not independent events. For example $\Pr(\{A > B\} |\{A > C\}) = \frac{2}{3}$ rather than $\frac{1}{2}$ and so $\Pr( \{A>B\} \cap \{A>C\} ) = \frac{2}{3}\times \frac{1}{2}=\frac{1}{3}.$

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