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While I know how to do this by considering a square of side $1$ and finding area of $|x-y|\le \frac13$, my doubt lies in why another method I tried does not give the right answer.

Considering the rod to be on a number line where the left most point is at $x=0$ and the right most point is $x=1$, let $s$ be the distance from $x=0$ where A cuts. Now B can cut anywhere in the $1$ unit long line, and the required probability is $\frac 23$, as B has to be either $\frac 13$ to the left of A or $\frac 13$ to the right of A.

Now had the rod been divided discretely I would have summed the probabilities for all values of $s$. So I assume infinitesimally small discrete lengths $ds$ and sum their probabilities. Let the probability at $s$ be $f(s)$. From $x=a$ to $x=b$,

$$f(a)+f(a+ds)+f(a+2ds)+ \dots+f(b)=P$$ Multiplying both sides by $ds$, $$f(a)ds+f(a+ds)ds+f(a+2ds)ds+ \dots+f(b)ds=Pds$$ This is $$\int_a^b f(x)dx = Pds$$

So $P$ ends up being much greater than $1$ which means I have some flaw in my earlier deductions, but I cannot figure out where. Any help regarding this will be highly appreciated.

AltercatingCurrent
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    Not exactly your problem, but this is how things should be done. Your approach is immature. I recommend learning probability theory and measure theory to get better understanding. – Quý Nhân Jan 18 '25 at 04:59
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    If two people cut a rod there will be three pieces and at least one piece will be at least $1/3$ as long as the original rod. I guess you are interested only in the piece between the two cuts. – David K Jan 18 '25 at 06:48
  • You have two “other” approaches to the problem. For which of these do you want someone to tell you why it doesn’t work? – David K Jan 18 '25 at 06:52
  • @DavidK im sorry, the approach i am specific about is the one where i get the integral. the previous one where i mentioned 2/3 is basically valid for only 1/3<x<2/3, where f(s) is a constant function 2/3 and now i agree that it was kind of unnecessary. and yes, i want to know about the piece between the two cuts – AltercatingCurrent Jan 18 '25 at 10:01
  • If the position of A's cut is $s$ and $\frac13 < s < \frac23$, you want the probability B's cut is between $s-\frac13$ and $s-\frac13$. But if $s < \frac13$, you want the probability B's cut is between $0$ and $s+\frac13$. While if $\frac23 < s$, you want the probability B's cut is between $s-\frac13$ and $1$. Then do your integral(s) over $s$. – Henry Jan 18 '25 at 11:43
  • To begin with, when you say, “the probability at $s$,” you need to say clearly probability of what, and also what you mean by “at $s$.” When we do discrete probability with a probability mass function, the answer to the two questions is that $p_X(s)$ is the probability that $X$ has the value $s.$ – David K Jan 18 '25 at 22:58
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    Think of the problem differently, it may help: A and B mark random points on a ring of perimeter 1. Then C reveals that there is already a secret (or equivalently: random) point on the ring where it was welded together from a unit interval. Between the three segments of the ring, everything is symmetric! – Hagen von Eitzen May 26 '25 at 21:46

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I think the problem is that you're using the function $f$ to represent two very different things at two different times.

When we do discrete probabilities, there is a function sometimes called a probability mass function. A typical symbol for this function is $p_X$ where $X$ is the name of a discrete random variable with the desired probability distribution. Then $p_X(x)$ is exactly the probability that $X = x.$

There can only be countably many values of $x$ where $p_X(x)$ is non-zero, and the sum of all of these non-zero values must be $1.$

For a continuous distribution, there is a function called the probability density. A typical symbol for this function is $f_Y$ where $Y$ is the name of a continuous random variable with the desired probability distribution. The value of $f_Y(y)$ is not a probability. In order to get a probability, you have to take a definite integral of $f_Y.$ The integral over the entire real number line must be $1.$

Your idea of discretizing the probabilities along the rod is a good one; it is just in the details that you had trouble. To begin with, let's stick to standard analysis, so the gaps between the discrete values are equal to the positive real number $\Delta s$ and not the infinitesimal $ds.$

To get the probability that $a \leq s \leq b$ we say that we have a discrete variable $X$ defined on $a,$ $a + \Delta s,$ $a + 2\Delta s,$ and so forth up to $b$ where it also is defined; $X$ may be defined at points outside the interval $[a,b]$ as well. We use $X$ as the value of $s.$ We suppose the existence of a probability mass function $p_X,$ and then we can set $P$ to the probability that $a \leq X \leq b),$

$$ P = \mathbb P(a \leq X \leq b) = p_X(a) + p_X(a + \Delta s) + p_X(a + 2\Delta s) + \cdots + p_X(b). $$

This corresponds pretty well with a Riemann sum of a continuous variable $Y$ with probability density function $f_Y$:

$$ f_Y(a)\Delta s + f_Y(a + \Delta s)\Delta s + f_Y(a + 2\Delta s)\Delta s + \cdots + f_Y(b)\Delta s. $$

If you consider this very carefully you may notice that there are $n+1$ terms in this sum where a Riemann sum usually has $n$ terms, but let's gloss over this for the sake of intuition; as $n$ goes to infinity, the extra term goes to zero anyway. (If we were actually trying to use a finite Riemann sum to estimate the probability we could list the terms more carefully.)

Ignoring the extra term, we have a Riemann sum for the integral $$ \int_a^b f_Y(s) \,ds = \mathbb P(a < Y < b). $$

The main thing to remember is, typically $f_Y(s) \neq p_X(s).$ If you want the second cut to be distributed uniformly between $0$ and $1,$ then $f_Y(s) = 1$ whenever $0 < s < 1$ and $f_Y(s) = 0$ when $s < 0$ or $s > 1.$ But if you populate this same interval with $1000$ uniformly spaced discrete points which are the only points where $p_X(s) > 0$ and you want a uniform distribution over these points, then $p_X(s) = 1/1000$ at each of these points and $0$ everywhere else.

If we define $p_X$ and $f_Y$ this way, we find that the non-zero values of $p_X(s)$ are

$$ p_X(s) \approx f_Y(s) \Delta s $$

(only approximately equal because of that extra term, but the approximation approaches equality as $\Delta s$ goes to zero).

The factor $\Delta s$ as it approaches zero corresponds to the $ds$ in the integral.

So here's the intuition to take away from this: the "$ds$" factor is already in the discrete probabilities that you listed. The multiplication by $ds$ that you did put two factors of $ds$ in every term where you only want one for the integral. Just don't do that.

Instead, let the integral correspond to a Riemann sum, each of whose terms is the probability at a particular point in your discrete distribution. Therefore,

$$ \int_a^b f_Y(s) \,ds \approx p_X(a) + p_X(a + \Delta s) + p_X(a + 2\Delta s) + \cdots + p_X(b) = P $$

(not $P\,ds.$)

David K
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