Let's say that I have three ranges of numbers:

$$ A: [0.15, 0.26]\\ B: [0.25, 0.34]\\ C: [0.20, 0.35]$$

If a number was randomly picked from the range of each set, what is the probability that the number from set $A$ will be the largest? From set $B$? etc?

  • $\begingroup$ As the question is stated, there's not enough information to answer. Do you intend the 'picking' from those ranges to be done (i) uniformly? (ii) independently? $\endgroup$
    – Glen_b
    Jun 3, 2013 at 23:10

1 Answer 1


When you pick randomly and independently from three random variables $A$, $B$, and $C$, having (cumulative) distributions $F_A$, $F_B$, and $F_C$ and corresponding distribution functions $f_A$, $f_B$, and $f_C$, then by definition of independence the chance that all three numbers are less than some value $x$ equals

$$\Pr(\max(A,B,C) \le x) = F_A(x)F_B(x)F_C(x).$$

Differentiation with respect to $x$ (via the product rule) gives the PDF of the max as

$$f_{\max(A,B,C)} = f_A(x)F_B(x)F_C(x) + F_A(x)f_B(x)F_C(x) + F_A(x)F_B(x)f_C(x).$$

That sure looks like a decomposition corresponding to $A$, $B$, and $C$. Indeed, the first term (by definition) tells us that the chance that $x \lt A \le x+dx$ and $B \le x$ and $C \le x$ is $f_A(x)F_B(x)F_C(x)dx$. Integrating that over all $x$ would then give the chance that $A$ exceeds both $B$ and $C$.

In the present situation, $\int_\mathbb{R} f_A(x)F_B(x)F_C(x)dx = 17/8910 \approx .00191,$ for instance.

The formula clearly generalizes to any finite number of independent random variables having any distributions you please.

  • $\begingroup$ This is great. Thanks. I want to make sure I understand it. My ranges are actually coming from beta distributions. Am I right that f_A(x) is my PDF and F_A(x) is my CDF for the beta distribution A? Also, if I do it correctly, shouldn't the probabilities add up to 1? What I just tried didn't, so I think I'm doing something wrong. $\endgroup$
    – jasonlfunk
    Jun 3, 2013 at 18:01
  • 2
    $\begingroup$ Yes, you are correct about $f_A$ and $F_A$. If you add up the integrals you will be integrating the derivative of $f_{\max(A,B,C)}$ which works out to $F_A(1)F_B(1)F_C(1) - F_A(0)F_B(0)F_C(0)$ = $1-0$ = $1$ as expected. In your example (with uniform distributions) the chance that $B$ is largest equals $5627/8910$ and the chance that $C$ is largest is $3266/8910$; all three chances add to $1$. $\endgroup$
    – whuber
    Jun 3, 2013 at 19:03
  • 1
    $\begingroup$ +1 It's answers like these that keep me coming back here. $\endgroup$
    – Glen_b
    Jun 3, 2013 at 23:13

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