I'm interested in N-Player Gambler's Ruin game with p = 0.5 and for each iteration :
- randomly pick two players
- make them exchange one chip
The result I'm interested in is the ranking probability of the players.
After running some simulations, it seems to me that for 3-player case considering chips stacks of [5, 3, 2] or [50, 30, 20], the probability for each player to be ranked first, second or third is the same (the computation time is not).
Is it true that for N players the ranking probabilities will remain constant when multiplying the stacks by any strictly positive integer number ?
If so, can you point out a known result from which it can be deduced ?
For $n=2$, scaling the initial stack list by a positive integer factor doesn't affect the distribution of winning probabilities.
For $n > 2$, the winning probabilities are still unaffected by a scaling of the initial stack list.
To explain the invariance, choose a given player and regard the other $n-1$ players as a team. Then the invariance for $n=2$ implies the invariance for $n > 2$.
However for $n > 2$, we can't claim such invariance for ranks greater than $1$.
To demonstrate this, take $n=3$.
For a given initial stack list $X=[x_1,x_2,x_3]$, let $P_X$ be the $3{\times}3$ matrix whose $ij$-th entry $P_X[i,j]$ is the probability that player $i$ finishes with rank $j$.
Then for $X=[1,2,3]$ we get $$ {\large{P_X}} = \pmatrix{ \dfrac{1}{6} &\dfrac{259}{1182} &\dfrac{121}{197} \\ \dfrac{1}{3} &\dfrac{241}{591} &\dfrac{51}{197} \\ \dfrac{1}{2} &\dfrac{147}{394} &\dfrac{25}{197} \\ } $$ whereas for $Y=[2,4,6]$ we get $$ {\large{P_Y}} = \pmatrix{ \dfrac{1}{6} &\dfrac{537964919}{2460583662} &\dfrac{252086911}{410097277} \\ \dfrac{1}{3} &\dfrac{502077053}{1230291831} &\dfrac{106039167}{410097277} \\ \dfrac{1}{2} &\dfrac{306154879}{820194554} &\dfrac{51971199}{410097277} \\ } $$ Comparing $P_X,P_Y$, while the first column entries (the winning probabilities) are the same, the entries for the other columns (ranks $2$ and $3$), though numerically very close, are not the same.
As regards how the above matrices were computed, here's an example to illustrate the process . . .
Assume an initial stack list of $[a_0,b_0,c_0]$, where $a_0,b_0,c_0$ are positive integers with $b_0\le c_0$.
Let $s=a+b+c$.
For positive integers $a,b,c$ with $b\le c$ and $a+b+c=s$, let $p[a,b,c]$ be the probability that player $1$ finishes with rank $3$ (i.e., player $1$ is the first to tap out), given a current stack list of $[a,b,c]$.
For each such triple $a,b,c$ we have the equation \begin{align*} p[a,b,c] =&\left({\small{\frac{1}{6}}}\right)\Bigl(\\[0pt] &{\phantom{\left({\small{\frac{1}{6}}}\right)\Bigl(}}{\phantom{+}}\;\,p[a-1,b+1,c]+p[a+1,b-1,c]\\[0pt] &{\phantom{\left({\small{\frac{1}{6}}}\right)\Bigl(}}+p[a-1,b,c+1]+p[a+1,b,c-1]\\[0pt] &{\phantom{\left({\small{\frac{1}{6}}}\right)\Bigl(}}+p[a,b-1,c+1]+p[a,b+1,c-1]\\[0pt] &{\phantom{\left({\small{\frac{1}{6}}}\right)}}\;\,\Bigr)\\[4pt] \end{align*} where on the $\text{RHS}$ we apply the substitutions $$ \left\lbrace \begin{align*} p[0,v,w]&=\,1\\[4pt] p[u,0,w]&=\,0\\[4pt] p[u,v,0]&=\,0\\[4pt] p[u,v,w]&=\,p[u,w,v]\;\,\text{if $v > w > 0$}\\[4pt] \end{align*} \right. $$ Thus we have a system of linear equations, one equation for each qualjfying triple $a,b,c$.
Then for an initial stack list of $[1,2,3]$ we get a system of $6$ linear equations in $6$ unknowns which yields $$ p[1,2,3]=\frac{121}{197}\approx .6142131980 $$ and for an initial stack list of $[2,4,6]$ we get a system of $30$ linear equations in $30$ unknowns which yields $$ p[2,4,6]=\frac{252086911}{410097277}\approx .6147002800 $$