Probability that $n$ random walks (1D) intersect a single point

Question

I am struggling finding a relation to determine the probability $n$ random walks (1D) intersect in a single point at step $s$. In the method below my attempts. My method is somewhat intuitive based. I am looking for more rigorous proof.

note:
This question arises from someone who claims that a matching cumulative digit sum of: $\pi$, $e$ and $\varphi$ (golden ratio) is unique and "cosmological" [1]. I tend to disprove it. This digit sum can be seen as a random walk if the constants are normal (every digit occurs with same frequency).

Method:

For every step $s$ on the random walk we can determine the probability density function if we know the standard deviation on every step $s$.

The standard deviation of a single step can be calculated it's a discrete uniform distribution "equally likely outcomes", where $q$ is the number of outcomes e.g. the number of digits $[0,1,2,3,4,5,6,7,8,9]$, $q=10$:

$$\sigma=\sqrt{\frac{q^{2}-1}{12}} $$

All the (1D) random walks start in the origin for this example. The standard deviation will grow with every step $s$, the variance is proportional to the number of steps [2].

$$Var(s)=s \cdot \sigma^{2}$$

$$\sigma(s)=\sqrt{s} \cdot \sigma$$

While the bins grow rapid I assume a normal approximation of the Binomial distribution.

$$f(x)=\int_{-\infty}^{\infty} {\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x}{\sigma }}\right)^{2}}dx $$

The probability that $n$ random walks intercept in a single point is (not sure):

$$p(s)=\int_{-\infty}^{\infty} \left[ {\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x}{\sigma }}\right)^{2}} \right]^n dx $$

With help of Wolfram Alpha [3] the solution is found for $n=3$ meaning the probability of $3$ point intersecting random walks.

$$p(s)=\frac{1}{2 \sqrt{3} \ \pi \ \sigma^{2}} \cdot \frac{1}{s}$$

The total probability $p(n)$ is proportional to the reciprocal sum of $s$. So the total probability is proportional to the harmonic series:

$$p(n)=\frac{1}{2 \sqrt{3} \ \pi \ \sigma^{2}} \cdot \sum_{s=1}^{\infty}\frac{1}{s}$$

This series diverges, meaning there are infinate point intersections of $n=3$ random walks. So a matching cumulative digit sum of $\pi$, $e$ and $\varphi$ is not unique, probability $\sim 8 \%$ for the first 1200 digits (see graph).

Question

Does anyone know the general formula for the probability $p(n)$ that $n$ random walks (1D) intersect a single point?

import numpy as np

#Elements of digits [0,1,2,3,4,5,6,7,8,9] rescaled to fit random walk
array=[-9,-7,-5,-3,-1,1,3,5,7,9]

#steps, in single random walk, x walks to intercept, number of trial to find intercept
steps=2500
xwalks=3
trials=1500

#Set output array to zero
count=np.full([steps],0)

for n in range(trials):

    #Identify initial array, set total array to zero
    w0=np.random.choice(array,steps)
    w0=np.cumsum(w0)
    total=np.full([steps],0)

    #Select x random walk check for intercept
    for m in range(xwalks-1):
    
        #Next current random walk
        w=np.random.choice(array,steps)
        w=np.cumsum(w)

        #Compare previous and current random walk
        eq=np.equal(w0,w)
        eq=eq.astype(int)

        #Count intercepts
        total=total+eq
    
        #Set current walk to previous
        w0=w

    #Sum all interceptions for all trials
    count=count+np.where(total==(xwalks-1),1,0)

#Print output
print(count)
print(np.sum(count))
print(np.sum(count)/trials)

Answer 1

The probability $n$ random walks starting in origin (1D) intersect a single point can be calculated with:

$$p(s)=\int_{-\infty}^{\infty} \left[ {\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x}{\sigma }}\right)^{2}} \right]^n dx$$

Solutions have been found with Wolfram Alpha 1. This solution can be written as a function of the number of steps $s$. So the probability $p$ at $s$ steps can be calculated with a empirical found formula:

$$p(s)= \frac{1}{g(n) \cdot \sigma ^{(n-1)} } \cdot s^{-\frac{1}{2}(n-1)} $$

Where the function for $g(n)$ is found as:

$$g(n)=\sqrt { n} \cdot (2 \pi)^{\frac{1}{2}(n-1)}$$

The total probability over all steps $s$ is only dependent upon the $p$-series formed by $s$. This summation can be calculated with the Riemann zeta function:

$$\zeta(\small{\frac{1}{2}(n-1)} \normalsize) =\sum\limits_{s=1}^{\infty}s^{- \frac{1}{2}(n-1)}$$

$$p(n)=\frac{1}{g(n) \cdot \sigma^{(n-1)}} \cdot \sum\limits_{s=1}^{\infty}s^{- \frac{1}{2}(n-1)} $$

Graph information

Below a table and a plot for the probabilities of $p(n)$ point coinciding random walks (1D). Details:

• The standard deviation is chosen: $10$ equal likely outcomes per step: $$\sigma=\sqrt{\frac{99}{12}}$$
• For infinate random walks $n\rightarrow \infty$ the $p$-series converges to $1$. This probability is plotted as: $p_{\infty}(n)$.
• $p(n)$ can be graphed as an continuous function.
• Values $n<3$ have not been graphed while Riemann zeta $\zeta(<1)$ are resulting in negative probabilities.

Observations:

• $n=3$ random walks have infinate single point intersections. $p$-series: $\sum\limits_{s=1}^{\infty}s^{-1}=\infty$, Riemann zeta: $\zeta(1)=\infty$ both have infinity probability $p(n)\rightarrow \infty$.
• $n=2$ random walks have infinate single point intersections for $p$-series: $\sum\limits_{s=1}^{\infty} s^{-\frac{1}{2}}=\infty$ , interpretation Riemann zeta is unclear: $\zeta(\frac{1}{2})=-1.46035...$ resulting in a negative probability $p(2)$.
• It is possible that $n\geq4$ random walks never point intersect. For $n=4$ only $\sim 0.35 \%$ of them will point intersect (with 10 equal likely outcomes per step).
• The probability $p(n)$ plot of $n$ random walks (1D) intersecting a single point is directly related to the $p$-series. A relation with Riemann zeta is unclear while the number of walks $n=a+ib$ would be a complex number.
• I learned this has to do with: recurrence and transience 2.

Any more rigorous proofs or information is welcome please.