I know that the answer is $\dfrac14$ and also a beautiful way to calculate it.

However, where am I going wrong with the following argument:

Assume that the first cut is at length $x$ and the second one is at length $y$ and that the line segment is of unit length.

  1. For a triangle, $0 < x < \dfrac12$. Therefore, $x$ can be independently selected with probability $\dfrac12$.
  2. For each $x$ in Step 1, $\dfrac12 < y < x+\dfrac12$ for a triangle. That is, $y$ will work when it is in the aforesaid range of length $x$. On the other hand, the total possible values for $y$, for any given $x$, are $x < y < 1$. The length of this interval is $1-x$. Therefor for each $x$ in Step 1, $y$ can be selected with probability $\dfrac{x}{1-x}$.

Now, $\dfrac {\int_0^{1/2} \dfrac{x}{1-x} dx}{\dfrac12} = 0.386$

$\therefore$ For $0 < x < \dfrac12$, the average value of probability of selecting $y$ to form a triangle is $0.386$.

Since the events 1 and 2 are independent, the final probability for both the events combined would be $\dfrac12 \times 0.386 = 0.193$.


You seem to have assumed $0 \le x \le y \le 1$.

If so, you would then be wrong to assume that the marginal density for $X$ (the lower of the two random numbers) is constant.

It is in fact $f_X(x)=2-2x$ and so your calculation should be $$\dfrac{\int\limits_{x=0}^{1/2} (2-2x) \dfrac{x}{1-x} \, dx}{\int\limits_{x=0}^{1} (2-2x) \, dx} = \frac14$$

  • $\begingroup$ Thank you! Could you please also guide on the logical arguments that could help me arrive at that density function? $\endgroup$ Feb 18 '20 at 8:04
  • 1
    $\begingroup$ @RiteshSingh The probability both values are above $x$ is $(1-x)^2$ so $F_X(x)=1-(1-x)^2 = 2x-x^2$ and thus $f_X(x)=2-2x$ $\endgroup$
    – Henry
    Feb 18 '20 at 8:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.