Probability that $n$ random walks (1D) intersect a single point 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 &quot;cosmological&quot; [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 &quot;equally likely outcomes&quot;, 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)
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 $n&lt;3$ have not been graphed while Riemann zeta $\zeta(&lt;1)$ 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.
I now realise that your rebuttal of my analysis in your comment above at Probability that $n$ random walks (1D) intersect a single point was in fact valid, as I had made a mistake in my analysis. I assumed incorrectly that I could simply add together the probabilities of collision after one step, two steps etc. to get the total probability of collision. However, these events are not mutually exclusive, since random walks can collide at multiple steps, so if the probability of collision at step 1 was 0.6 and the probability of collision at step 2 was 0.5, I would already have a probability exceeding 1. Hence, the correct process is to work out the probability of non-collision before step m multiplied by the probability of collision at step m, and only then add together the probabilities. Alternatively, I could just calculate the probability of non-collision at or before step m by multiplying together probabilities, then subtract it from one, and take the limit as m tends to infinity.

Edit:
This method describes collision random walk with probability per step is: [-1,1] 50%.
s: step number in random walk, n: number of intersecting random walks.
Probability intersecting at single step position:
$$p(s)= \sum_{k=0}^{s} \left(\frac{1}{2^{s}}  \binom{s}{k}\right)^{n}$$
Total probability over all steps:
$$p(n)=\sum_{s=1}^{\infty}\sum_{k=0}^{s} \left(\frac{1}{2^{s}}   \binom{s}{k} \right)^{n}$$

I have now created a spreadsheet to calculate the answer to your question, which is:




n
p(n)




1
1


2
1


3
0.907472937


4
0.382922701


5
0.147129979


6
0.062758395


7
0.028516044


8
0.013419184


9
0.006446336


10
0.003136593


11
0.001538874


12
0.000759198




I am not sure how accurate my results are, but if you need better accuracy, you could extend the spreadsheet tables as required.
My spreadsheet uses data tables, and it appears that Excel cannot handle nested data tables, so the table above was produced by manually typing in the first twelve input values for n and copying the result. For a larger data set, a quick script could be written to automate the process.
The spreadsheet is available at https://groups.io/g/UnsolvedProblems/message/12251.
I can now see that your question is related to Pólya's Random Walk constants, but because you are essentially asking a different question, I don't think the answer to your question should directly equate to Pólya's Random Walk constants.
In terms of the normal approximation you mentioned in your analysis, the conditions of the Central Limit Theorem state under which circumstances a normal approximation applies, which can be found at https://en.wikipedia.org/wiki/Central_limit_theorem.