How can Distinguishable objects behave as if they were Indistinguishable? I am mentally disabled: I cannot imagine indistinguishable marbles. However, I can imagine that there exist sequences of (random) manipulations on real (distinguishable) marbles causing them to end up in real (distinguishable) boxes according to the Bose-Einstein distribution: For example, in case of $2$ marbles and $3$ boxes the following six arrangements are supposed to be equally likely:
$$\begin{gather} &\_\_\_ &\_\_\_ &\_\_\_\\ &.. & & \end{gather}$$
$$\begin{gather} &\_\_\_ &\_\_\_ &\_\_\_\\ & &.. & \end{gather}$$
$$\begin{gather} &\_\_\_ &\_\_\_ &\_\_\_\\ & & &.. \end{gather}$$
$$\begin{gather} &\_\_\_ &\_\_\_ &\_\_\_\\ &. &. & \end{gather}$$
$$\begin{gather} &\_\_\_ &\_\_\_ &\_\_\_\\ &. & &. \end{gather}$$
$$\begin{gather} &\_\_\_ &\_\_\_ &\_\_\_\\ & &. &. \end{gather}$$
(Arrangements of this type can be listed in the case of $n$ marbles and $m$ boxes. So the Bose-Einstein distribution can be defined in general.)
Can one describe manipulations that result in the Bose-Einstein distribution? Restriction on the manipulations: one cannot choose a whole arrangement of marbles. Only individual boxes and individual marbles can be picked.
 A: (This answers the general question for $m$ particles in $n$ boxes.)
Bose-Einstein statistics (BES) can be generated by allocating particles one-at-a-time to $n$ boxes numbered $1,...,n,$ by repeatedly applying the following rule, starting with no particles yet allocated:

*

*If $r$ particles have been allocated, producing particle counts $(r_k)_{k=1,..,n}$ in the $n$ boxes, then allocate the next particle to box-number $X_r$, where $X_r$ is chosen from $\{1,...,n\}$ according to the probability distribution specified by $$P[X_r=k]={r_k+1\over r+n}\,\pmb 1_{k\in\{1,...,n\}}.\tag{1}$$
Sketch of proof:
Let $R_r=(r_1,...,r_n)$ be the particle-counts when $r$ particles have been allocated according to the above procedure. An allocation $R_r$ obeys BES if  $$P[R_r=(r_1,...,r_n)]={r!\,(n-1)!\over (n+r-1)!}$$
for all $(r_1,...,r_n)\in \{0,1,...,r\}^n$ such that $\sum_{k=1}^nr_k=r$ (i.e., if the distribution is uniform on the set of all $\binom{n+r-1}{r}$ such $n$-tuples).
The proof proceeds by noting that $R_0$ trivially obeys BES, and then showing that if $R_{r-1}$ obeys BES then so does $R_{r}.$ Thus, we have
$$\begin{align}&P[R_r=(r_1,...,r_n)]\\[1ex]
&=\sum_{k=1}^nP[R_r=(r_1,...,r_n)\mid R_{r-1}=(r_1,..,r_k-1,..,r_n)]\,P[R_{r-1}=(r_1,..,r_k-1,..,r_n)]\\[1ex]
&=\sum_{k=1}^nP[X_{r-1}=k]\,P[R_{r-1}=(r_1,..,r_k-1,..,r_n)]\\[1ex]
&=\sum_{k=1}^n{(r_k-1)+1\over(r-1)+n}\,{(r-1)!(n-1)!\over(n+(r-1)-1)!}\\[1ex]
&=\sum_{k=1}^nr_k{(r-1)!(n-1)!\over(n+r-1)!}\\[1ex]
&={r!(n-1)!\over(n+r-1)!}
\end{align}$$
QED
I adapted the above from a derivation in the following article:
Ijiri Y, Simon H A. Some distributions associated with Bose-Einstein statistics. Proc Natl Acad Sci U S A. 1975;72(5):1654-1657.

Aside:  Letting $\ell_r$ be the location (box-number) of the $r$th particle allocated, a calculation similar to the one above shows that $P[\ell_r=k]={1\over n}$ for all $k\in\{1,...,n\};$ i.e., $\ell_1,...,\ell_m$ are identically (though not independently) $\text{Uniform}\{1,...,n\}:$
$$\begin{align}&P[\ell_r=k]\\[1ex]
&=\sum\limits_{(r_1,...,r_n)\in\{0,...,r\}^n:\\ r_1+...+r_n=r}P[R_r=(r_1,...,r_n) \ \text{ and }\ R_{r-1}=(r_1,..,r_k-1,..,r_n)]\\[1ex]
&=\sum\limits_{(r_1,...,r_n)\in\{0,...,r\}^n:\\ r_1+...+r_n=r}\bigg(P[R_r=(r_1,...,r_n)\mid R_{r-1}=(r_1,..,r_k-1,..,r_n)] \cdot P[R_{r-1}=(r_1,..,r_k-1,..,r_n)]\bigg)\\[1ex]
&=\sum\limits_{(r_1,...,r_n)\in\{0,...,r\}^n:\\ r_1+...+r_n=r}{r_k\over n+r-1}\cdot {(r-1)!(n-1)!\over(n+r-2)!}\\[2ex]
&\overset{(*)}{=}{(r-1)!(n-1)!\over(n+r-1)!}\sum\limits_{(r_1,...,r_n)\in\{0,...,r\}^n:\\ r_1+...+r_n=r}r_k\\[1ex]
&={(r-1)!(n-1)!\over(n+r-1)!}\binom{n+r-1}{n}\\[1ex]
&={1\over n}
\end{align}$$
QED
(I found the sum $\sum_{...}r_k$ in (*) by imagining a matrix whose rows are all the $n$-tuples being summed over. By symmetry, every column has the same sum, which is the very sum $\sum_{...}r_k$ required. Thus the grand total of the column sums is $n\cdot\sum_{...}r_k,$ and this must equal the grand total of the row sums, which is $r\cdot\binom{n+r-1}{r}.$ Hence, the required sum is $\sum_{...}r_k={r\over n}\binom{n+r-1}{r}=\binom{n+r-1}{n}.$)
A: At least for the case with 2 marbles into 3 boxes, you need an 'attractive force' on the marbles so that after the first marble is tossed in some random box, the second marble is more likely to go into the same box as the first marble than either of the two empty boxes.
This can be worked out with conditional probability:
$$
\begin{aligned}
\frac{1}{6}=P(\text{both in first box})&=P(X_2=1 \text{ and } X_1=1)\\
&=P(X_2=1|X_1=1)\cdot P(X_1=1)\\
&=P(X_2=1|X_1=1)\cdot \frac{1}{3}.
\end{aligned}
$$
Thus $P(X_2=1|X_1=1)=1/2$. Then let it fall into one of the two remaining boxes with equal probability as well.
This gives
$$
P(x_2=i \text{ and } X_1=i) =\frac{1}{6}
$$
for $i=1,2,3$, and
$$
P(x_2=j \text{ and } X_1=i) =\frac{1}{6}
$$
for any $i\neq j$.
I imagine this pattern can be extended to there being more marbles and boxes, but don't know for certain. I'll think about it.

If you want the generic formula for the number of combinations that result in $b_0$ empty boxes, $b_1$ boxes with 1 marble, $b_2$ boxes with 2 marbles, etc. and the associated probabilities, I have solved for the general formula in this stackexchange post.

Another example:
Consider tossing 3 marbles into 2 boxes. If we think of it as tossing them sequentially, then the first is equally likely to go into either box (probability 1/2), the second marble is twice as likely to go into the box with the first marble (probability 2/3), and the third marble is equally likely to go in either box if they are both occupied, and three times as likely to fall into the box containing both marbles one and two if they fell into the same box (probability 3/4). This gives all arrangements equal probability
A: Let $q \in (0,1)$ be some arbitrary parameter.
Let $X_1,\dots,X_m$ be independent geometrically distributed random variables with parameter $q$. For instance, each $X_i$ can be obtained by tossing a (biased) coin until you get heads and counting the number of tails before it happened.
For all $i \in \{1,\dots,m\}$, put $X_i$ marbles in the $i$-th box.
The resulting arrangement of marbles follows the Bose-Einstein distribution of $N = X_1 + \cdots X_m$ marbles in $m$ boxes.
If you aim for a deterministic number $n$ of marbles, just repeat this process until $N=n$. The choice of $q$ can be optimized in order to maximize the probability of this event.
A: Having received two nice answers, I am scetching here the one I've made up in the mean time:
Choose one ordered pair of boxes with equal probability. Then take one ball from the first box and put it into the second [of the ordered pair]. If the first is empty do nothing. (This one can be generalized like the camaredes' solutions.)
A: I know it might be a little bit too late but still want to discuss this with people if possible:
Different results might due to different assumptions, one example is Bertrand paradox. (Trick here is how we define "equally probable") And I guess this could also be explained in that way.
In Bose-Einstein distribution, if we labeled the balls to distinct the balls:
$$|1|\:\:\:|2|\:\:\:|\:\:|  \: and \: |2|\:\:\:|1|\:\:\:|\:\:|$$
are same one situations, which means which ball goes into which box does not matter.
In the above comment: "Here is a "manipulation" which does not lead to the desired distribution. Choose a marble (distinguishable marbles!) such that every marbles are equally likely to be chosen. Then choose a box such that every boxes are equally likely to be chosen. Find the marble chosen in the boxes and put in the box chosen. In the case of 3 boxes and 2 marbles the described manipulation will lead to the following distribution on the arrangements listed above: 1/9, 1/9, 1/9, 2/9, 2/9, 2/9. But I want ,manipulations that lead to 1/6, ... 1/6."
In the comment situation, if we labeled the balls to distinct them:
$$|1|\:\:\:|2|\:\:\:|\:\:|  \: and \: |2|\:\:\:|1|\:\:\:|\:\:|$$
are different situations. But
$$|1 2|\:\:\:|\:\:|\:\:\:|\:\:|  \: and \: |2 1|\:\:\:|\:\:|\:\:\:|\:\:|$$
are viewed as the same.
This means that in our assumption, which ball is in which box matters but the order that they entered the box does not matter. That is why we have 9 as the denominator.
However, if we change the assumption a little bit -- the box can record the order in which things went into it. Then we will get three more situations, so, 12 in total:
$$|1 2|\:\:\:|\:\:|\:\:\:|\:\:|$$
$$|2 1|\:\:\:|\:\:|\:\:\:|\:\:|$$
$$|1|\:\:\:|2|\:\:\:|\:\:|$$
$$|1|\:\:\:|\:\:|\:\:\:|2|$$
$$|2|\:\:\:|1|\:\:\:|\:\:|$$
$$|\:\:|\:\:\:|1 2|\:\:\:|\:\:|$$
$$|\:\:|\:\:\:|2 1|\:\:\:|\:\:|$$
$$|\:\:|\:\:\:|1|\:\:\:|2|$$
$$|2|\:\:\:|\:\:|\:\:\:|1|$$
$$|\:\:|\:\:\:|2|\:\:\:|1|$$
$$|\:\:|\:\:\:|\:\:|\:\:\:|1 2|$$
$$|\:\:|\:\:\:|\:\:|\:\:\:|2 1|$$
Then the probability will be back to $\frac{1}{6}$ again.
Hope this make sense.
