I've been stuck on this expected value question for some time now, for some reason:

Suppose we have two bins. Bin A has $k$ balls and Bin B has $n-k$ balls. For each step, we randomly pick a ball from one of the bins (so each ball has a $1/n$ chance of being selected each time) and move it to the other bin. What is the expected number of steps required until there are $k$ balls in Bin A again?

As for my work, I let $E_i$ be the expected number of steps required until there are $k$ balls in Bin A, if we started out with $i$ balls in Bin A. Then I found that $$E_i=\frac inE_{i-1}+\frac{n-i}nE_{i+1}+1$$ for $i$ for $1\le i\le n-1$. For $0$ and $n$, I found that \begin{align*}E_0&=E_1+1\\E_n&=E_{n-1}+1.\end{align*} When I tried to solve this system of equations, I got some really ugly constants that I couldn't figure out how to simplify. In particular, I found that \begin{align*}E_1&=E_2+\frac{n+1}{n-1}\\E_2&=E_3+\frac{n^2+n+2}{(n-1)(n-2)}\\E_3&=E_4+\frac{n^3+5n+6}{(n-1)(n-2)(n-3)}.\end{align*} I feel like this might not be the right approach, so feel free to suggest something completely different. Thanks!

Edit: I've edited my initial conditions, but I still can't get it to work out.

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    $\begingroup$ Your initial conditions are wrong. $E_0=1+E_1$, for instance. The ball must be chosen from the full bin. $\endgroup$ – saulspatz Jul 22 at 16:18
  • $\begingroup$ @saulspatz Apparently, I copied in the incorrect initial conditions, but I was actually using the correct ones. Still not sure how to do this :( $\endgroup$ – boink Jul 22 at 16:59
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    $\begingroup$ You need to work $k$ into the picture. For $i=k\pm1$, one of the terms $E_{i-1}$ or $E_{i+1}$ in your recurrence should be replaced by $1$. $\endgroup$ – Steve Kass Jul 22 at 17:11
  • $\begingroup$ @SteveKass Okay, that makes sense to me. I'm sorry, but I don't quite see how that solves the problem, since all that happens is I have split this problem into two (smaller) problems, where the endpoints are $0$ and $k-1$ (or $k+1$ and $n$), rather than $0$ and $n$. If someone could explain this, that'd be brilliant! Thanks :) $\endgroup$ – boink Jul 22 at 17:41
  • $\begingroup$ I posted what I think are the right recurrence relations as an answer, as it's too long for a comment. $\endgroup$ – Steve Kass Jul 22 at 19:18

Your recursion is only right is we define $E_i$ such that $E_k=0$ (hence, not the expected time to get $k$ balls again, but to simply get $k$ balls). Then the recursion is indeed

$$ E_i=\frac inE_{i-1}+\frac{n-i}nE_{i+1}+1 \hskip{1cm} i\ne k, \; 0\le i\le n \tag 1$$ with $E_i=0$ if $i=k$ or $i<0$ or $i>n$.

Once we have $E_i$, the required expected time is $F_k=\frac knE_{k-1}+\frac{n-k}nE_{k+1}+1$.

The linear system $(1)$ has $n$ variables and $n$ equations, it can be solved numerically (for example, writing the system in matricial form). But I'm not sure about a symbolic solution.

Alternatively, and perhaps a little more elegant, this can modelled as a Markov chain with $n+1$ states, and the expected time between visits can be obtained from its limiting distribution. Again, this is easy... numerically.

Update: this is an Ehrenfest model. The following is adapted/copied from here.

Let $C_{i, j}$ be the expected time to get from $i$ balls to $j$ balls in the first bin, with $C_{i,i}=0$

Then, for $0<i<n$

$$ C_{i,n}=\frac{i}{n} C_{i-1,n}+\frac{n-i}{n} C_{i+1,n}+1 \tag 2$$

(like eq. $(1)$ , a little more general). Or

$$ n(C_{i,n}-C_{i+1,n} - 1 ) =i( C_{i-1,n}-C_{i+1,n} )\tag 3$$

But we also know, by linearity of expectation

$$ C_{i,n}=C_{i,i+1} +C_{i+1,i+2} + \cdots + C_{n-1,n} \tag4$$

Putting both eqs together:

$$ n (C_{i,i+1} -1)=i(C_{i-1,i} +C_{i,i+1}) \implies A_i(n-i)=i A_{i-1} +n\tag{5}$$ where $A_i=C_{i,i+1}$. With the initial condition $A_0=1$ this is solved by

$$A_i=\frac{\sum_{t=0}^{i}\binom{n}{t}}{\binom{n-1}{i}} \tag6$$

By simmetry of the bins, $$B_i \triangleq C_{i,i-1} = C_{n-i,n-i+1}=A_{n-i}\tag 7$$

Finally our desired expected time is

$$F_k = 1 + \frac{k}{n}A_{k-1} + \frac{n-k}{n}B_{k+1}=1 + \frac{k}{n}A_{k-1} + \frac{n-k}{n}A_{n-k-1}\tag8$$

...which, operating, gives $$F_k=\frac{2^n}{\binom{n}{k}} \tag9$$

This result is surprisingly (embarrasingly) simple, so perhaps I messed up something or either there should be a much simpler derivation... (Granted, this derivation solves a more general problem: it gives the expected time from any values of the start and end ocuppancy values)

Update 2: I wrote the result above and I saw this: if we regard this as a Markov Chain, but with the $2^n$ states corresponding to which (labeled) balls are in bin $A$, then the chain is reversible (the matrix is doubly stochastic) and hence the states are equiprobable in the equilibrium distribution, with $p=1/2^{n}$ each. Then the $n+1$ macro states composed of those states having $k$ balls have total probability $\binom{n}{k}/2^n$, and the mean recurrence time is $2^n/\binom{n}{k}$


[Not a full answer, but too long for a comment.]

The correct recursion for this problem is a little different than yours, which doesn’t account for actually reaching the goal of $k$ balls in the bin.

Let $E(i)$ be the expected number of steps needed to reach $k$ balls in bin A for the first time in the future, if there are currently $i$ balls in bin A and $n-i$ balls in bin B.

For $i$ equal to $0$ or $n$:

\begin{equation} E(0) = \begin{cases} 1+E(1) & \mbox{ if } k\neq1 \\ 1 & \mbox{ if } k=1 \end{cases} \end{equation}

\begin{equation} E(n) = \begin{cases} 1+E(n-1) & \mbox{ if } k\neq n-1 \\ 1 & \mbox{ if } k=n-1 \end{cases} \end{equation}

For $0< i < n$,

\begin{equation} E(i) = \begin{cases} \frac in E(i-1)+\frac{n-i}nE(i+1)+1 & \mbox{ if } k\notin\{i-1,i+1\} \\ \frac in +\frac{n-i}n(1+E(i+1)) & \mbox{ if } k=i-1 \\ \frac{n-i}n +\frac in(1+E(i-1)) & \mbox{ if } k=i+1 \end{cases} \end{equation}


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