# Determine the expected length where the rope tears, if possible (probability task)

Let $p(l)$ be the probability that a rope with an integer length $l$ doesn't tear for a given load. Assume we have that $p(l_1+l_2)=p(l_1)p(l_2)$ for all integer $l_1,l_2 > 0$ and $p(2) = \frac{1}{2}$. Can you now determine the expected length where the rope tears?

Remark: The rope gradually increases by one unit.

This is task from exam but I have no solution. For practice I try solve it but I think you don't can determine expected length for the problem.

We have $p(2)$, we can express this different: $p(1+1)$.

By this we know that $p(1) = \frac{1}{\sqrt{2}}$. Then for $p(3)$ we can do same and get $p(3) = p(1+2) = p(1)p(2)=\frac{1}{\sqrt{2}} \cdot \frac{1}{2}$

we continue like that for other values.. In general we have probability function $$p(i) = \left(\frac{1}{\sqrt{2}}\right)^i$$

Now problem is sum of all $p(i)$ should provide $1$ but

$$\sum^{\infty}_{k=1}\left(\frac{1}{\sqrt{2}}\right)^i = \frac{1}{1-\left(\frac{1}{\sqrt{2}}\right)}-1= 2.414213562$$

But not sure about that I find this very complicated to read and understand?

$p(i)$ gives the probability that the rope doesn't tear at stage $i$, given that it has not torn yet, so we should expect all $p(i)$ to sum to $1$. Instead, we can calculate $P(i)$, the probability that the rope does tear at stage $i$, by $P(i) = p(1)\cdot p(2)\cdots p(i-1)\cdot(1-p(i))$. We can expect all $P(i)$ to sum to $1$.