Let us suppose that we have a sample of 2 random distinct numbers $I=\{z_1,z_2\}$ that are generated from a uniform distribution with support in $[0,1]$. Let's call $d = \max{I}$ the maximum of the randomly generated sample.

I want to compute the probability that $z_1\leq r$ for some $0\leq r\leq1$ ($r$ is not a random variable) given that $z_1 \neq d$, i.e.

$$ p(z_1 \leq r | z_1 \neq d) $$

To easily compute this probability, we can notice that if $z_1\neq d$, then $z_1$ is the minimum, and therefore

$$ p(z_1 \leq r | z_1 \neq d) = p(\min I\leq r) = 1-p(z\geq r)^2 = 1 - (1-r)^2 $$

I have confirmed this result numerically on Mathematicathat you can check with the following code

checkDistribution[r_] := 
 Module[{win = 0, loss = 0, list, max, i, n = 2}, 
  For[i = 1, i <= 10000, i++,
   list = RandomSample[Range[100 n], n]/(100 n) // N;
   max = Max[list];
   If[list[[1]] != max, 
    If[list[[1]]^(n - 1) <= r, win = win + 1, loss = loss + 1];];
  Return[{r, win/(win + loss)} // N]]
points = Table[checkDistribution[r][[{1, 2}]], {r, 0, 1, 0.01}];
Show[points // ListPlot, Plot[2 r - r^2, {r, 0, 1}, PlotStyle -> Red]]

that returns

enter image description here

I want however to compute this probability without using the fact that $z_1$ is the minimum, but only using our knowledge that $d$ is the maximum. We should then consider the distribution of the maximum of two random variables. Since $d$ is the maximum, we have

$$ p(d\leq r) = r^2\\ p(d>r) = 1-r^2 $$

Now, we have two cases

  1. $d>r$, in which case $p(z_1 \leq r | z_1 \neq d) = p(d>r) p(z_1<r) = (1-r)\times r$
  2. $d\leq r$, in which case $p(z_1 \leq r | z_1 \neq d) = p(d<r) p(z_1<d) = r^2 \times 1$

but the sum of these two terms does not give the answer. Where is the mistake?

I want to solve this problem in the other way, because I want to generalize it to a set of 3 numbers $I=\{z_1,z_2,z_3\}$ and compute

$$ p(z_1 \leq r | z_1 \neq d)\,. $$ In this generalization, I don't know if $z_1$ is the minimum of the distribution.


4 Answers 4


Let's name the events: $A \equiv z_1 \le r$, $B \equiv z_1 \ne d$ , $C \equiv d > r$ and its complement $\bar C \equiv d \le r$ .

Then $$P(A|B)= P(C) P(A | B C) + P(\bar C) P(A | B\bar C) \tag 1$$

Now, you've already computed $P(\bar C)=r^2$, and also $ P(A | B \bar C) = 1$

Hence it's true that

$$P(A|B)= (1-r^2) P(A | B C) + r^2 \tag 2$$

Now, you are implicitly assuming $P(A | B C) = P(A)=r$. However that is wrong.

What is true is that $P(B | A C)=1$ which implies

$$P(A | B C) = \frac{P(A B C)}{P(B C)}=\frac{P(A C)}{P(B C)}=\frac{P(C|A)P(A)}{P(B|C)P(C)}=\frac{2r}{r+1} \tag 3$$

because $P(C|A)=1-r$, $P(A)=r$, $P(C)=1-r^2$ , $P(B|C)=P(B)=\frac12$

Hence finally $P(A|B) = 1 - (1-r)^2$ as expected.

  • $\begingroup$ Thank you so much $\endgroup$
    – apt45
    Mar 17 at 22:31
  • $\begingroup$ Can you explain me why $P(C|A) = 1-r$ and not $P(C|A) = 1-r^2$? And why does $P(B|AC)=1$ imply the last equation? $\endgroup$
    – apt45
    Mar 18 at 13:18

The applicable relations are: $$\begin{align}&P(z_1\le r\mid z_1\ne d)\\ &= P(d>r)P(z_1\le r\mid z_1\ne d\text{ and }d>r)+P(d\le r)P(z_1\le r\mid z_1\ne d\text{ and }d\le r)\\ &= P(d>r)P(z_1\le r\mid z_1\ne d\text{ and }d>r)+P(d\le r)P(z_1\lt d\mid z_1\ne d\text{ and }d\le r)\\ &\ne P(d>r)P(z_1\le r)\phantom{xxxxxxxxxxxxxx}+P(d\le r)P(z_1\le d)\\ \end{align}$$

You've attempted to use the inequality as an equality.


Both 1. and 2. are wrong: for 1. you get.

$p(z_1\leq r, d>r | z_1 \neq d)= \frac{p(z_1 \leq r, d>r , d\neq z_1 )}{p(z_1 \neq d)}=\frac{p(z_1 \leq r, z_2 \geq r )}{2}= \frac{r(1-r)}{2}$

  1. Is a bit tricker but

$p(z_1\leq r, d<r | z_1 \neq d) = \frac{p(z_1 \leq r, d<r , d\neq z_1 )}{p(z_1 \neq d)} = \frac{p(z_1 \leq z_2 \leq r )}{2}$

With $p(z_1 \leq z_2 \leq r )= \int^r_0 (\int^{z_2}_0 1 dz_1) dz_2 = \frac{1}{2}r^2$

thus : $ p(z_1\leq r, d>r | z_1 \neq d) + p(z_1\leq r, d<r | z_1 \neq d) = \frac{r(1-r)}{2} + \frac{\frac{1}{2}r^2}{2} = 1-(1-r)^2$

Which is your initial result. It might to think about the problem geometricly when lookin at several points.


If your intuitive argument is unconvincing, you can check by focusing not on the distribution of $d$, but rather on the already known distributions of $z_1$ and $z_2$.

$$\def\P{\operatorname{\sf P}}\begin{align}\P(z_1\leqslant r\mid z_1\neq d)~&=~\P(z_1\leqslant r\mid z_1<z_2)\\[1ex]&=~2\P(z_1\leqslant r, z_2>z_1)\\[1ex]&=~\textstyle 2\int\limits_0^{\min\{r,1\}}\int\limits_u^1\,\mathrm dv\,\mathrm d u\\[1ex]&=~\textstyle 2\int\limits_0^{\min\{r,1\}} (1-u)\,\mathrm du\\[1ex]&=~(2r-r^2)\,\mathbf 1_{0\leqslant r\leqslant 1}+\mathbf 1_{1<r}\\[1ex]&=~\bigl(1-(1-r)^2\bigr)\,\mathbf 1_{0\leqslant r\leqslant 1}+\mathbf 1_{1<r}\end{align}$$

Which confirms the result of your intuition.


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