(Apology for the title, I struggled to find a better way to describe the question in a single sentence.)

Let there be $m$ identical black balls $\Huge\bullet$, $n$ identical white balls; consider randomly ordering them. For example, with $m = 6, n = 4$: $$ \overbrace{\Huge\bullet\Huge\bullet\Huge\bullet}\;\underbrace{\Huge\circ\;\,\Huge\circ}\;\Huge\bullet\Huge\circ\Huge\circ\Huge\bullet\Huge\bullet $$

And let $r$ be the number of groups of continuous black balls (over braced), and $s$ be the number of groups of continuous white balls (underbraced). So in the example, $r = 3, s = 2$.

What is the probability of $\mathrm{P}(r, s)$ with some given $r, s$? what about $\mathrm{P}(r)$ given some $r$?

My thoughts:

So the probability $\mathrm{P}(r, s)$ should be

  • If $r = s$, the ordering may start with either black or white, and end with another, so $2 \times f(m, n, r, s)$

  • If $| r - s | = 1$, it must start and end with the color of more groups, so $f(m, n, r, s)$.

  • If $| r - s| > 1$, it is impossible, so its $0$.

Where $$ f(m, n, r, s) = \frac{\small\text{ways to divide } m \text{ balls into } r \text{ distinct groups of one or more balls} \;\times\; \text{ways to divide } n \text{ balls into } s \text{ groups of one or more balls}}{\text{all ways to order two types of balls}} \\ = \displaystyle\frac{\displaystyle\binom{m - 1}{r - 1}\binom{n - 1}{s - 1}}{\displaystyle\binom{m + n}{m}} $$

And for $\mathrm{P}(r)$, since for any $r$, $s$ can only be one of $r + 1, r, r - 1$ to have probability, I simply substituted $s$ with them, summed it up and got the result: $$ \frac{(n - 1)!\,(m - 1)!\,m!\,n!}{(m + n)!\,(m - r)!\,(r - 1)!\,} \left[ {2 \over (n - r)!\,(r - 1)!} + {1 \over (n - r + 1)!\,(r - 2)!} + {1 \over (n - r - 1)!\,r!}\right] \\[1ex] = \frac{(n - 1)!\,(m - 1)!\,m!\,n!}{(m + n)!\,(m - r)!\,(r - 1)!} \left[ n^2 + n \over (n - r + 1)!\,r! \right] $$

But these are both extremely bulky, and very complicated, so I wonder is there a better approach to solving this problem, better with a result not requiring conditions? Or did I do something terribly wrong?

  • $\begingroup$ You can't casually invoke Stars and Bars like this, as the various outcomes are not generally equi-probable. $\endgroup$
    – lulu
    Aug 3, 2022 at 11:14
  • $\begingroup$ I would start by working it out recursively (might help to distinguish those strings that begin with $b$ from those that begin with $w$). If nothing else, that will make it easy to reliably compute a whole lot of examples. $\endgroup$
    – lulu
    Aug 3, 2022 at 11:15
  • $\begingroup$ As a quicker way to see that your formula can not be correct, let $m$ be large and $n=1$. Then of course $s=1$ but $r$ can only be $1$ or $2$, depending on where the single white ball is placed. Your formula gives a non-zero value for any $r$ from $1$ to $m$. $\endgroup$
    – lulu
    Aug 3, 2022 at 11:55


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