An exponential limit shape of random $q$-proportion Bulgarian solitaire

TL;DR

This paper introduces a generalized random Bulgarian solitaire model with variable parameters and proves that, under certain conditions, it exhibits an exponential limit shape as the number of cards grows large.

Contribution

It extends Popov's model by allowing both the proportion and probability parameters to vary with the number of cards, establishing conditions for an exponential limit shape.

Findings

Under specified conditions, the model converges to an exponential limit shape.

The results generalize previous fixed-parameter models to variable-parameter settings.

The paper provides rigorous proof of the limit shape behavior.

Abstract

We introduce \emph{$p_n$-random $q_n$-proportion Bulgarian solitaire} ($0<p_n,q_n\le 1$), played on $n$ cards distributed in piles. In each pile, a number of cards equal to the proportion $q_n$ of the pile size rounded upward to the nearest integer are candidates to be picked. Each candidate card is picked with probability $p_n$, independently of other candidate cards. This generalizes Popov's random Bulgarian solitaire, in which there is a single candidate card in each pile. Popov showed that a triangular limit shape is obtained for a fixed $p$ as $n$ tends to infinity. Here we let both $p_n$ and $q_n$ vary with $n$. We show that under the conditions $q_n^2 p_n n/{\log n}\rightarrow \infty$ and $p_n q_n \rightarrow 0$ as $n\to\infty$, the $p_n$-random $q_n$-proportion Bulgarian solitaire has an exponential limit shape.

Figures

Equations

$$\partial \lambda (x)={\lambda }_{\lfloor x\rfloor +1}.$$

$$\partial \lambda (x)=\frac{1}{{\lambda }_1}\partial \lambda (xn/{\lambda }_1)=\frac{1}{{\lambda }_1}{\lambda }_{\lfloor xn/{\lambda }_1\rfloor +1}.$$

$$\underset{n\to \infty }{\lim }{\pi }_n\left\{\lambda \in \mathcal{P}(n):|{\partial }^{a_n}\lambda (x)-\varphi (x)|<\epsilon \right\}=1$$

$$\underset{n\to \infty }{\lim }{\pi }_n\left\{\lambda \in \mathcal{P}(n):\underset{x\in [a,b]}{\sup }|{\partial }^{a_n}\lambda (x)-\varphi (x)|<\epsilon \right\}=1$$

$$\mathcal{L}(\{x:f(x)>t\})=\mathcal{L}(\{x:f^{\ast }(x)>t\})$$

$$\Vert \partial \mathrm{ord}\alpha -f{\Vert }_{\infty }\le \Vert \partial \alpha -f{\Vert }_{\infty }.$$

$$\Vert \partial \mathrm{ord}\alpha -f{\Vert }_{\infty }=\Vert (\partial \alpha {)}^{\ast }-f^{\ast }{\Vert }_{\infty }\le \Vert \partial \alpha -f{\Vert }_{\infty }.$$

$${\tilde{\rho }}^{(n)}(\lambda )=\sum _{\begin{array}{c}\alpha \in \mathcal{W}(n)\\ \mathrm{ord}\alpha =\lambda \end{array}}{\pi }_n(\alpha ).$$

$$\underset{n\to \infty }{\lim }{\pi }_n\left\{\alpha \in \mathcal{W}(n):|\partial \alpha (x)-\varphi (x)|<\epsilon \right\}=1.$$

$$\underset{n\to \infty }{\lim }{\pi }_n\left\{\alpha \in \mathcal{W}(n):|(\partial \mathrm{ord}\alpha )(x)-\varphi (x)|<\epsilon \right\}=1.$$

$$A=\bigcup _{\lambda \in L}\{\alpha \in \mathcal{W}(n):\mathrm{ord}\alpha =\lambda \}$$

$${\pi }_n(A)$$

$$=\sum _{\lambda \in L}{\pi }_n\{\alpha \in \mathcal{W}(n):\mathrm{ord}\alpha =\lambda \}$$

$$=\sum _{\lambda \in L}\sum _{\begin{array}{c}\alpha \in \mathcal{W}(n)\\ \mathrm{ord}\alpha =\lambda \end{array}}{\pi }_n(\alpha )$$

$$=\sum _{\lambda \in L}{\tilde{\rho }}^{(n)}(\lambda )$$

$$={\tilde{\rho }}^{(n)}(L).$$

$$\underset{n\to \infty }{\lim }{\tilde{\rho }}^{(n)}\left\{\lambda \in \mathcal{P}(n):|\partial \lambda (x)-\varphi (x)|<\epsilon \text{ for all }x>0\right\}=1.$$

$$\frac{p_n{q}_n^{2}n}{\log n}\to \infty \text{ as }n\to \infty .$$

$$p_n{q}_n\to 0\text{ as }n\to \infty .$$

$$p_n{q}_n\to 0\text{and}\frac{p_n{q}_n^{2}n}{\log n}\to \infty \text{as }n\to \infty .$$

$$P(\left|X-\mu \right|\ge \gamma )\le 2\exp \left(-\frac{{\gamma }^2}{3\mu }\right).$$

$$D=\lceil 14\frac{\log n}{pq}\rceil \text{and}M=\lceil \frac{n^2}{p}\rceil .$$

$$P(X<n)$$

$$\le 2\exp \left(-\frac{n^2}{3}(1+o(1))\right),$$

$$1-2n\exp \left(-\frac{n^2}{3}(1+o(1))\right)\to 1$$

$$\frac{1}{q_{n}n}\max \left\{N({\alpha }^{(D+1)}),\dots ,N({\alpha }^{(D+M)})\right\}\to 0\text{ in probability},$$

$$P_1$$

$$\displaystyle\leq 2\exp\left(-\frac{\Bigl{(}Dp-\frac{\log n}{q}\Bigr{)}^{2}}{3Dp}\right)=2\exp\left(-\frac{(\frac{14\log n}{q}-\frac{\log n}{q})^{2}}{3\cdot 14\frac{\log n}{q}}\right)=2n^{-\frac{13^{2}}{42}\frac{1}{q}}<2n^{-4}.$$

$$P_2:=n{P}_1=2n^{-3}.$$

$$N({\alpha }^{(D)})\le \frac{qn}{\log n}+D=\frac{qn}{\log n}+14\frac{\log n}{pq}=qn\left(\frac{1}{\log n}+14\frac{\log n}{p{q}^{2}n}\right)=o(qn),$$

$$\frac{1}{qn}\max \left\{N({\alpha }^{(D+1)}),\dots ,N({\alpha }^{(D+M)})\right\}<\epsilon$$

$$\underset{k\in [D+1,D+M]}{\max }\frac{|{\alpha }_1^{(k)}-p_n{q}_{n}n|}{p_n{q}_{n}n}\to 0\text{ in probability}$$

$$P_3:=P(|{\alpha }_1^{(k)}-pqn|>\epsilon pqn)$$

Full text

**AN EXPONENTIAL LIMIT SHAPE OF RANDOM $q$-PROPORTION BULGARIAN SOLITAIRE

Kimmo Eriksson

**Mälardalen University, School of Education, Culture and Communication,

Box 883, SE-72123 Västerås, Sweden

kimmo.eriksson@mdh.se

**Markus Jonsson111Corresponding author

**Mälardalen University, School of Education, Culture and Communication,

Box 883, SE-72123 Västerås, Sweden

markus.jonsson@mdh.se

**Jonas Sjöstrand

**Royal Institute of Technology, Department of Mathematics,

SE-10044 Stockholm, Sweden

jonass@kth.se

Abstract

We introduce $p_{n}$-random $q_{n}$-proportion Bulgarian solitaire ($0<p_{n},q_{n}\leq 1$), played on $n$ cards distributed in piles. In each pile, a number of cards equal to the proportion $q_{n}$ of the pile size rounded upward to the nearest integer are candidates to be picked. Each candidate card is picked with probability $p_{n}$, independently of other candidate cards. This generalizes Popov’s random Bulgarian solitaire, in which there is a single candidate card in each pile. Popov showed that a triangular limit shape is obtained for a fixed $p$ as $n$ tends to infinity. Here we let both $p_{n}$ and $q_{n}$ vary with $n$. We show that under the conditions $q_{n}^{2}p_{n}n/{\log n}\rightarrow\infty$ and $p_{n}q_{n}\rightarrow 0$ as $n\to\infty$, the $p_{n}$-random $q_{n}$-proportion Bulgarian solitaire has an exponential limit shape.

1 Introduction

The game of Bulgarian solitaire has received a great deal of attention, see reviews by Hopkins [10] and Drensky [2]. The Bulgarian solitaire is played with a deck of $n$ identical cards divided arbitrarily into a number of piles. A move consists of picking a card from each pile and letting these cards form a new pile. If piles are sorted in order of decreasing size, every position in the solitaire is equivalent to a Young diagram of an integer partition of $n$.

Popov [15] considered a random version of Bulgarian solitaire defined by a probability $p\in(0,1]$, such that one card from each pile is picked with probability $p$, independently of the other piles. We will refer to this stochastic process on configurations as $p$-random Bulgarian solitaire. The probabilities of configurations converge to a stationary distribution. Popov showed that as $n$ grows to infinity and configuration diagrams are downscaled by $\sqrt{n}$ in both dimensions, the stationary probability of the set of configurations that deviate from a triangle with slope $p$ by more than $\varepsilon>0$ tends to zero. In this sense, random configurations has a limit shape.

The objective of the present paper is to study such limit shapes in a generalization of random Bulgarian solitaire.

1.1 $q_{n}$-proportion Bulgarian solitaire

Olson [13] introduced a generalization of Bulgarian solitaire in which the number of cards that are picked from a pile of size $h$ is given by some non-negative valued function $\sigma(h)$. Eriksson, Jonsson and Sjöstrand [3] recently studied the special case when $\sigma$ is well-behaved in the sense that $\sigma(1)=1$ and both $\sigma(h)$ and $h-\sigma(h)$ are non-decreasing functions of $h$. In particular, they studied a special case that they called $q_{n}$-proportion Bulgarian solitaire, defined by the rule $\sigma(h)=\lceil q_{n}h\rceil$. This means that from each pile we pick a number of cards given by the proportion $q_{n}$ of the pile size rounded upward to the nearest integer. To illustrate the effect of the parameter $q_{n}$, set it to $0.3$ and consider the configuration $(6,2,2,1)$. From the first pile we pick $\lceil 0.3\times 6\rceil=2$ cards; similar calculations give that 1 card is picked from each of the other three piles. Note that for $q_{n}\leq 1/n$ exactly one card is always picked from each pile, retrieving the ordinary Bulgarian solitaire.

As $n$ tends to infinity, Eriksson, Jonsson and Sjöstrand [3] determined limit shapes of stable configurations of $q_{n}$-proportion Bulgarian solitaire: In case ${q_{n}^{2}n\rightarrow 0}$, the limit shape is triangular, which generalizes the limit shape result for the ordinary Bulgarian solitaire. For other asymptotic behavior of $q_{n}$, other limit shapes were obtained. Specifically, in case $q_{n}^{2}n\rightarrow\infty$, the limit shape is exponential. The intermediate case $q_{n}^{2}n\rightarrow C>0$ produces a family of limit shapes interpolating between the triangular and the exponential shape.

1.1.1 $p_{n}$-random $q_{n}$-proportion Bulgarian solitaire

We shall examine a $p_{n}$-random version of $q_{n}$-proportion Bulgarian solitaire, in which the proportion $q_{n}$ (rounded upward) of cards in a pile are only candidates to be picked, each of which is picked only with probability $p_{n}$, independently of all other candidate cards. This process will be denoted by $\mathscr{B}(n,p_{n},q_{n})$. Note that in the special case of a fixed $p$ and for $q_{n}\leq 1/n$, this process is equivalent to Popov’s $p$-random Bulgarian solitaire.

Our focus will be on establishing a regime in which $p_{n}$-random $q_{n}$-proportion Bulgarian solitaire has an exponential limit shape.

2 The concept of limit shapes

In this section we give the precise definitions of the limit shapes we consider. Let $\mathcal{P}(n)$ be the set of integer partitions of $n$. For any partition $\lambda\in\mathcal{P}(n)$ with $N=N(\lambda)$ positive parts $\lambda_{1}\geq\lambda_{2}\geq\dotsc\geq\lambda_{N}>0$, define $\lambda_{i}=0$ for $i>N(\lambda)$, and the diagram of $\lambda$ as the Young diagram oriented such that the parts of $\lambda$ are represented by left and bottom aligned columns, weakly decreasing in height from left to right. For example, ​​​​ \ytableausetupboxsize=5pt, aligntableaux=bottom {ytableau}

&

is the diagram of the partition $(2,1,1)$. We define the diagram-boundary function of $\lambda$ as the nonnegative, weakly decreasing and piecewise constant function $\partial\lambda:\mathbb{R}_{\geq 0}\rightarrow\mathbb{R}_{\geq 0}$ describing the boundary of $\lambda$, given by

[TABLE]

Following [4] and [16], the diagram is downscaled using some scaling factor ${a_{n}>0}$ such that all row lengths are multiplied by $1/a_{n}$ and all column heights are multiplied by $a_{n}/n$, yielding a constant area of 1. Following [3], we shall consistently make the choice $a_{n}=n/\lambda_{1}$, such that the height of the diagram is scaled to 1.

Thus, given a partition $\lambda$, define the rescaled diagram-boundary function of $\lambda$ as the nonnegative, real-valued, weakly decreasing and piecewise constant function $\partial{\lambda}:\mathbb{R}_{\geq 0}\rightarrow\mathbb{R}_{\geq 0}$ given by

[TABLE]

The $p_{n}$-random $q_{n}$-proportion Bulgarian solitaire $\mathscr{B}(n,p_{n},q_{n})$ (with $p_{n},q_{n}\in(0,1]$) can be regarded as a Markov chain on the finite state-space $\mathcal{P}(n)$. Let us denote the sequence of visited states by $(\lambda^{(0)},\lambda^{(1)},\dotsc)$. In the truly random case of $p_{n}<1$, it is straightforward to verify that this Markov chain is aperiodic and irreducible. It is well-known that an aperiodic and irreducible Markov chain on a finite state-space has a unique stationary distribution $\pi$ and that starting from any initial state the distribution of the $i$th state $\lambda^{(i)}$ converges to $\pi$ as $i$ tends to infinity. We denote by $\pi_{n,p_{n},q_{n}}$ the stationary measure of the Markov chain $(\lambda^{(0)},\lambda^{(1)},\dotsc)$ on $\mathcal{P}(n)$ given by $\mathscr{B}(n,p_{n},q_{n})$ for $p_{n}<1$.222Readers acquainted with the limit shape literature may wonder whether the stationary measure has the property of being multiplicative, in the sense of interpretable as the product measure on the space of integer sequences restricted to a certain affine subspace [7]. The multiplicative property is useful in limit shape problems and related problems [1, 5, 8, 14, 16]. However, such techniques will not be used here as $\pi_{n,p_{n},q_{n}}$ is unlikely to be multiplicative in general. When we refer to a limit shape of the process $\mathscr{B}(n,p_{n},q_{n})$ for $p_{n}<1$ as $n$ grows to infinity, we shall mean the limit shape of the stationary measure $\pi_{n,p_{n},q_{n}}$. The intuitive sense of this concept is that when the solitaire is played on a sufficiently large number of cards for sufficiently long the configuration will almost surely be very close to the limit shape after suitable downscaling. Following Vershik [16], a sequence $\{\pi_{n}\}$ of probability distributions on $\mathcal{P}(n)$ is said to have a limit shape $\phi$ if the downscaled diagrams approach $\phi$ in probability as $n$ grows to infinity. The exact condition for convergence can vary. Consistent with Yakubovich [17] and Eriksson and Sjöstrand [4], we shall use the definition that

[TABLE]

for all $x>0$ and all $\varepsilon>0$.333Vershik [16] and Erlihson and Granovsky [6] used a stronger condition for convergence toward a limit shape, namely that

$\lim_{n\rightarrow\infty}\pi_{n}\left\{\lambda\in\mathcal{P}(n):\sup_{x\in[a,b]}|\partial^{a_{n}}{\lambda}(x)-\phi(x)|<\varepsilon\right\}=1$

should hold for any compact interval $[a,b]$, and any $\varepsilon>0$.

3 The approach of ordering piles by time of creation

It will sometimes be useful to explicitly order piles by time of creation rather than by size. Here we develop this approach.

When parts are not sorted by size, a configuration is not represented by an integer partition but by a weak integer composition: an infinite sequence $\alpha=(\alpha_{1},\alpha_{2},\dotsc)$, not necessarily decreasing, of nonnegative integers adding up to $n$. Let $\mathcal{W}(n)$ denote the set of weak compositions of $n$. We define the diagram, the diagram-boundary function $\partial\alpha$, and the rescaled diagram-boundary function $\partial{\alpha}$ of a weak composition $\alpha$ in exact analogy to the way we defined them for partitions in Section 2. For example, the diagram of $\alpha=(3,0,2,4,1,0,0,\dotsc)$ and the corresponding function graph $y=\partial\alpha(x)$ are shown in Figure 1. Also, for a weak composition $\alpha=(\alpha_{1},\alpha_{2},\dotsc,\alpha_{N},0,0,\dotsc)$ we define the number of parts $N=N(\alpha)$ disregarding the trailing zeros.

3.1 Connecting the limit shapes of compositions and partitions

We shall now connect compositions with partitions. For any $\alpha\in\mathcal{W}(n)$, define the operator $\operatorname{\textup{{ord}}}$ as the ordering operator that arranges the parts of $\alpha$ in descending order, thus yielding a partition. We shall now prove that such sorting of the piles by size respects the convergence to a limit shape. The proof uses some basic theory of symmetric-decreasing rearrangements, see for example [9, Ch. 10] or [11, Ch. 3]. For any measurable function $f\colon\mathbb{R}\rightarrow\mathbb{R}_{\geq 0}$ such that $\lim_{x\rightarrow\pm\infty}f(x)=0$, there is a unique function $f^{\ast}\colon\mathbb{R}\rightarrow\mathbb{R}_{\geq 0}$, called the symmetric-decreasing rearrangement of $f$, with the following properties:

• •

$f^{\ast}$ is symmetric, that is, $f^{\ast}(-x)=f^{\ast}(x)$ for all $x$,

• •

$f^{\ast}$ is weakly decreasing on the interval $[0,\infty)$,

• •

$f^{\ast}$ and $f$ are equimeasurable, that is,

[TABLE]

for all $t>0$, where $\mathcal{L}$ denotes the Lebesgue measure,

• •

$f^{\ast}$ is lower semi-continuous.

In particular, if $f$ is a symmetric function that is weakly decreasing and right-continuous on $[0,\infty)$ and tends to [math] at infinity, then $f^{\ast}=f$.

Lemma 1**.**

Let $\alpha\in\mathcal{W}(n)$ be a weak composition of $n$ and let $f:\mathbb{R}_{\geq 0}\rightarrow\mathbb{R}_{\geq 0}$ be a right-continuous and weakly decreasing function such that $f(x)\rightarrow 0$ as $x\rightarrow\infty$. The downscaled diagram-boundary functions before and after sorting of the weak composition satisfy the inequality

[TABLE]

where $\|\cdot\|_{\infty}$ denotes the max-norm \|f\|_{\infty}=\sup\big{\{}|f(x)|:x\geq 0\big{\}}.

Proof.

The intuition of the lemma should be obvious from Figure 2. To be able to use the standard machinery of symmetric rearrangements, we consider the functions $f$, $\partial{\alpha}$, and $\partial{\operatorname{\textup{{ord}}}\alpha}$ as being defined on the entire real axis by letting $f(x)=f(|x|)$ and analogously for $\partial{\alpha}$, and $\partial{\operatorname{\textup{{ord}}}\alpha}$.

Since $f(x)\rightarrow 0$ as $x\rightarrow\infty$, its symmetric-decreasing rearrangement $f^{\ast}$ is defined and, since $f$ is weakly decreasing and lower semi-continuous, we have $f^{\ast}=f$. Similarly, $\partial{\operatorname{\textup{{ord}}}\alpha}(x)\rightarrow 0$ as $x\rightarrow\infty$ and is weakly decreasing, so $(\partial{\operatorname{\textup{{ord}}}\alpha})^{\ast}=\partial{\operatorname{\textup{{ord}}}\alpha}$. Moreover, $(\partial{\alpha})^{\ast}=\partial{\operatorname{\textup{{ord}}}\alpha}$ must hold because the operator $\operatorname{\textup{{ord}}}$ arranges the composition parts in descending order.

Now, since symmetric rearrangements decrease $L^{p}$-distances for any $1\leq p\leq\infty$ (see for example [11], Section 3.4), we obtain

[TABLE]

∎

Clearly, Lemma 1 holds true also when the max-norm is replaced by the weaker convergence condition used in our limit shape definition (2).

Lemma 2**.**

For any distribution $\pi_{n}$ on $\mathcal{W}(n)$, define a corresponding distribution $\tilde{\rho}^{(n)}$ on $\mathcal{P}(n)$ by

[TABLE]

If $\phi$ is a limit shape of $\pi_{n}$ on $\mathcal{W}(n)$ then $\phi$ is also a limit shape of $\tilde{\rho}^{(n)}$ on $\mathcal{P}(n)$.

Proof.

The assumption that $\phi$ is a limit shape of the distribution $\pi_{n}$ on $\mathcal{W}(n)$ means that

[TABLE]

for all $x>0$. By virtue of Lemma 1 we can replace $\alpha$ with $\operatorname{\textup{{ord}}}\alpha$ in this formula:

[TABLE]

The set $A:=\left\{\alpha\in\mathcal{W}(n):|(\partial{\operatorname{\textup{{ord}}}\alpha})(x)-\phi(x)|<\varepsilon\text{ for all }x>0\right\}$ can be written as a disjoint union of equivalence classes with respect to sorting:

[TABLE]

where $L=\{\lambda\in\mathcal{P}(n):\,|\partial{\lambda}(x)-\phi(x)|<\varepsilon\text{ for all }x>0\}$. The $\pi_{n}$-probability measure of $A$ is

[TABLE]

From (4) we have that $\lim_{n\rightarrow\infty}\pi_{n}(A)=1$. Because $\pi_{n}(A)=\tilde{\rho}^{(n)}(L)$, we can conclude that also ${\lim_{n\rightarrow\infty}\tilde{\rho}^{(n)}(L)=1}$, that is,

[TABLE]

This means that $\phi$ is a limit shape of the distribution $\pi_{n}$ on $\mathcal{P}(n)$. ∎

4 Three regimes

Recall from Section 1.1 the $q_{n}$-proportion Bulgarian solitaire developed in [3], where the limit shape is triangular when $q_{n}^{2}n\rightarrow 0$, exponential when $q_{n}^{2}n\rightarrow\infty$ and an interpolation between the two when $q_{n}^{2}n\rightarrow C>0$.

The $p_{n}$-random $q_{n}$-proportion Bulgarian solitaire seems to share this property of three regimes of limit shapes. Specifically, in Section 7 we conjecture the limit shape to be triangular when $p_{n}q_{n}^{2}n\to 0$, exponential when $p_{n}q_{n}^{2}n\to\infty$ and an interpolation between the two (a piecewise linear function graph that depends on $C$) when $p_{n}q_{n}^{2}n\to C>0$.

The focus in this paper is the exponential regime of the $p_{n}$-random $q_{n}$-candidate Bulgarian solitaire, i.e. the case $p_{n}q_{n}^{2}n\to\infty$ as $n\to\infty$. However, with the proof technique we employ we will prove the stronger statement that the limit shape holds even when the configurations are considered elements of $\mathcal{W}(n)$, i.e. even without sorting the piles of a configuration according to size to create a partition in $\mathcal{P}(n)$. We will instead require the stronger condition $p_{n}q_{n}^{2}n/{\log n}\to\infty$ as $n\to\infty$. By virtue of Lemma 2, the limit shape will also hold for partitions.

5 The exponential limit shape

Here we investigate the limit shape of configurations in the $p_{n}$-random $q_{n}$-candidate Bulgarian solitaire $\mathscr{B}(n,p_{n},q_{n})$ in the regime

[TABLE]

Our main result, Theorem 1, says that, under the additional asymptotic property $p_{n}q_{n}\to 0$ as $n\rightarrow\infty$, the boundary function of the diagram, downscaled, will resemble the exponential shape $e^{-x}$ asymptotically almost surely (a.a.s.), i.e. with a probability that tends to 1 as $n\to\infty$. See Figure 3. Throughout this section, “a.a.s.” can be read as “with a probability that tends to 1 as $n\to\infty$”. Also, the asymptotic notations $o$ and $O$ will always be with respect to $n\to\infty$.

We shall see that the condition $p_{n}q_{n}^{2}n/{\log n}\to\infty$ implies that the rounding effect in computing the number of candidate cards is negligible. Thus, the number of candidate cards will tend to $q_{n}n$ as $n\to\infty$. This in turn means that the expected number of picked cards will eventually be close to $p_{n}q_{n}n$, thus $\lambda_{1}\approx p_{n}q_{n}n$ is the size of the pile created in a move of the solitaire. Recall from Section 2 that the scaling factor we employ is $a_{n}=n/\lambda_{1}=\frac{1}{p_{n}q_{n}}$. Thus, if $p_{n}q_{n}$ is bounded away from zero, then the scaling $\frac{1}{p_{n}q_{n}}$ is bounded and hence cannot transform the jumpy boundary diagrams into a smooth limit shape. Therefore, we also require

[TABLE]

On the other hand, if $p_{n}q_{n}$ tends to zero too fast, the pile sizes will be small and their random fluctuations will be large. For instance, the new pile after each move has a size drawn from the binomial distribution $\textup{Bin}(K,p_{n})$, where $K\approx q_{n}n$ is the number of candidate cards, with relative standard deviation $\sim 1/\sqrt{p_{n}q_{n}n}$. The requirement (5) guarantees that $p_{n}q_{n}$ does not tend to zero too fast.

Theorem 1**.**

For each positive integer $n$, pick $q_{n}$ and $p_{n}$ with $0<p_{n},q_{n}\leq 1$ and a (possibly random) initial configuration $\lambda^{(0)}\in\mathcal{P}(n)$. Let $(\lambda^{(0)},\lambda^{(1)},\dotsc)$ be the Markov chain on $\mathcal{P}(n)$ defined by $\mathscr{B}(n,p_{n},q_{n})$, and denote its stationary measure by $\pi_{n,p_{n},q_{n}}$. Suppose

[TABLE]

Then $\pi_{n,p_{n},q_{n}}$ has the limit shape $e^{-x}$ under the scaling $a_{n}=(p_{n}q_{n})^{-1}$.

The proof of Theorem 1 heavily relies on the following version of Chernoff bounds. For a proof, see for example [12].

Chernoff Bound**.**

For $n\geq 1$ and $0<p\leq 1$, let $X\sim\textup{Bin}(n,p)$ and set $\mu=E(X)=np$. Then, for any $0<\gamma<\mu$,

[TABLE]

The idea of the proof of Theorem 1 is the following.

We will use the approach developed in Section 3, i.e. card configurations in the solitaire will be represented by weak integer compositions and the piles are ordered with respect to creation time, i.e. if $\alpha\in\mathcal{W}(n)$ is the current configuration in the solitaire, then $\alpha_{1}$ was the last formed pile, $\alpha_{2}$ the pile that was formed two moves ago, etc. With this representation, some piles may be empty, so one may imagine each pile being placed in a bowl and the bowls are lined up in a row on the table. In each move of the solitaire, the new (possibly empty) pile is put in a new bowl to the left of all old bowls. As mentioned in Section 4, we shall prove Theorem 1 as a limit shape result for diagram-boundary functions of compositions. Thus, throughout this section, each configuration of $n$ cards will be represented by an element of $\mathcal{W}(n)$. Also, in the following we may abbreviate $p=p_{n}$ and $q=q_{n}$ unless the dependency on $n$ is crucial.

Assume a configuration $\alpha=(\alpha_{1},\alpha_{2},\dotsc,\alpha_{N},0,0,\dotsc)$ of $n$ cards with $N=N(\alpha)$ piles (so that $\sum_{i=1}^{N}\alpha_{i}=n$) in the solitaire $\mathscr{B}(n,p,q)$. The number of candidate cards in the next move is $\kappa:=\sum_{i=1}^{N}\lceil q\alpha_{i}\rceil$. We denote the rounding effect in pile $1\leq i\leq N$ by $R_{i}:=\lceil q\alpha_{i}\rceil-q\alpha_{i}$ and the total rounding effect by $R:=\kappa-qn=\sum_{i=1}^{N}R_{i}$.

Clearly, $R<N$ (since $R_{i}<1$ for any $i$), i.e. the total rounding effect is bounded above by the number of piles. The first thing we will do is to make sure that after a sufficient number $D$ of moves from the initial configuration $\alpha^{(0)}$, the number of piles $N(\alpha^{(D)})$ in the resulting configuration $\alpha^{(D)}$ is much smaller than $qn$ a.a.s. (so that the number of candidate cards $\kappa$ is approximately $qn$ and thus the total rounding effect $R$ is negligible). In Lemma 4 we show that it is possible to choose such a $D$, namely $D=c\frac{\log n}{pq}$ for any $c\geq 14$.

We also need to make sure that the number of piles stays $o(qn)$ for sufficiently many additional moves $M$, long enough to establish the convergence of the overall shape. Lemma 4 will also guarantee that $M=\lceil n^{2}/p\rceil$ suffices for this purpose.

Thus, in the following we shall use

[TABLE]

If the number of piles stays $o(qn)$ during $M$ moves so that the number of candidate cards stays approximately $qn$, the newly formed pile in each of these moves will have expected size $pqn$. Our proof technique involves studying the evolution of such a pile (which will follow an exponential decay in size). Therefore we need to additionally make sure that no old piles (which could potentially be much larger than $pqn$) remain after these $M$ moves. Lemma 3 shows that, in fact, after $M$ moves all piles in the starting configuration have disappeared a.a.s.

Lemma 3**.**

*Let $M$ be given by (8). From any initial configuration $\alpha\in\mathcal{W}(n)$, after $M$ moves in the solitaire $\mathscr{B}(n,p_{n},q_{n})$, all piles in $\alpha$ have been consumed a.a.s. *

Proof.

Consider a pile of size $n$. The size of this pile after $M$ moves is statistically dominated by $\max(n-X,0)$ where $X\sim\textup{Bin}(M,p_{n})$ whose expected value is $E(X)=Mp_{n}=\lceil n^{2}/p_{n}\rceil p_{n}>n$. Therefore, the probability that the pile remains after $M$ moves is ${P(X<n)}$ with the bound

[TABLE]

where we used the Chernoff bound (7). Thus, since any given pile in $\alpha$ has size $\leq n$, and the number of piles (of any size in any configuration) is $\leq n$, the probability that all piles in $\alpha$ have been consumed after $M$ moves is at least

[TABLE]

which concludes the proof. ∎

Lemma 4**.**

Let $n,p_{n},q_{n}$ and an initial configuration $\alpha^{(0)}$ be given in the solitaire $\mathscr{B}(n,p_{n},q_{n})$. Then

[TABLE]

*where $D$ and $M$ are given by (8). *

Proof.

Let us abbreviate $p=p_{n}$ and $q=q_{n}$. We will first prove that all piles of size at most $q^{-1}\log n$ disappear with high probability after $D$ moves, making sure that there are not many small piles in $\alpha^{(D)}$. Consider a pile of size at most $q^{-1}\log n$ in $\alpha^{(0)}$. Note that every nonempty pile decreases by at least $1$ with probability at least $p$ in each move. Therefore, after $D$ moves the number of picked cards from this pile statistically dominates $X\sim\textup{Bin}(D,p)$ with expected value $Dp=14q^{-1}\log n$. Using the Chernoff bound (7), the probability that this pile remains after $D$ moves is at most

[TABLE]

Since there can be at most $n$ piles of size at most $q^{-1}\log n$, the probability that not all piles of size at most $q^{-1}\log n$ have disappeared after $D$ moves is bounded by

[TABLE]

Let us now turn our attention to the number of piles after these $D$ moves. By the above, all piles smaller than $q^{-1}\log n$ have disappeared with high probability. Clearly, the number of piles larger than $q^{-1}\log n$ can never be more than $\frac{n}{q^{-1}\log n}=\frac{qn}{\log n}$. Also, during the process of these $D$ moves, at most $D$ new piles have been formed. (Exactly $D$ piles have been formed but some may have disappeared in the process.) Thus, for the total number of piles $N(\alpha^{(D)})$ in the configuration $\alpha^{(D)}$ after $D$ moves, with probability at least $1-2n^{-3}$, we have

[TABLE]

where we used the assumption (5) in the last step. It follows that, for any $\varepsilon>0$,

[TABLE]

with probability at least $1-2n^{-3}M\geq 1-\frac{2}{pn}\to 1$ since $pn\to\infty$. (That $pn\to 0$ is also a consequence of the assumption (5).) ∎

Lemma 4 asserts that the number of piles remains to be $o(qn)$ during the $M$ moves from $\alpha^{(D)}$ to $\alpha^{(D+M)}$, hence the number of candidate cards remains to be $qn$ (a.a.s.) during the same moves. Therefore the number of picked cards (which equals the size of the newly formed pile), remains of expected size $pqn$. In Lemma 5 we prove that the actual number of picked cards in each of these $M$ moves does not deviate (relatively) from $pqn$.

Lemma 5**.**

Let $n,p_{n},q_{n}$ and an initial configuration $\alpha^{(0)}$ be given in the solitaire $\mathscr{B}(n,p_{n},q_{n})$. Let $D$ and $M$ be given by (8). Then

[TABLE]

*as $n\to\infty$. *

Proof.

Let us abbreviate $p=p_{n}$ and $q=q_{n}$. Let $\varepsilon>0$ and let $\kappa$ be the number of candidate cards in $\alpha^{(k-1)}$ for some $k=D+1,\dotsc,D+M$. Recall that the total rounding effect in computing the number of candidate cards is bounded above by the number of piles. It therefore follows from Lemma 4 that $\kappa=nq(1+o(1))$. The new pile size is $\alpha_{1}^{(k)}\sim\textup{Bin}(\kappa,p)$. Then, using the triangle inequality and the Chernoff bound (7) we have

[TABLE]

where the last equality is derived as follows. By (5), $\log n=o(pqn)$ and hence $\log n^{a}=o(pqn)$ for any $a\geq 1$. Since $pqn\to\infty$, this means that $\exp(-pqn)$ tends to zero faster than $\exp(-\log n^{a})$, i.e., $\exp(-pqn)=o(1/n^{a})$. Since $np\to\infty$, we therefore also have $\exp(-pqn)=o(p/n^{a})=o(1/M)$. The next to the last equality follows from the fact that $\varepsilon pqn$ dominates over $|pqn-\kappa p|$ (since $|pqn-\kappa p|=|pqn-nq(1+o(1))p|=pqn\cdot o(1)$).

Therefore, the probability that $|\alpha_{1}^{(k)}-pqn|>\varepsilon pqn$ for any $k$ during the entire process of $M$ moves is bounded by $MP_{3}=M\cdot o(1/M)=o(1)$. ∎

While playing the solitaire, there is a possibility that at some point there will be too many piles, and thereby the number of candidate cards will be bigger than $qn$ (and thus the size of the newly formed pile will be bigger than $pqn$). Lemmas 3 and 4 ensures that this never happens a.a.s. during the entire process of $M$ moves from $\alpha^{(D)}$ to $\alpha^{(D+M)}$.

There is also a risk that, even if there are suitably many ($qn$) candidate cards, the number of picked cards among them will deviate from $pqn$ due to random fluctuations (and thereby the size of the newly formed pile will deviate from $pqn$). Lemma 5 ensures that this never happens a.a.s. during the same period of $M$ moves.

Therefore, after $m:=D+M$ moves we have the following a.a.s.

• •

all current piles have been formed during the last $M$ moves, and

• •

all current piles had size $pqn$ when they were formed.

At this point, i.e. in the configuration $\Gamma:=\alpha^{(m)}$, the leftmost pile (of size $\Gamma_{1}$) was formed one move ago, the second pile from the left (of size $\Gamma_{2}$) was formed two moves ago, and so on. We shall prove that the size $\Gamma_{k}$ of the pile that was formed $k$ moves ago for any $k=1,2,\dotsc,m$ is $\Gamma_{k}=\Gamma_{1}(1-pq)^{k}=pqn(1-pq)^{k}$ a.a.s., i.e. the size decreases exponentially with $k$ with decay factor $1-pq$.

We will now consider the evolution of a given pile of size $A_{1}$ during $r\geq 1$ steps in the $p$-random $q$-proportion Bulgarian solitaire in the following way. We will need to keep track of each individual card in this pile. To this end, we label the cards $1,2,\dotsc,A_{1}$ starting from the top, and each card will keep their label throughout the process. Let $X_{i,k}\in\{0,1\}$ where $i=1,\dotsc,A_{1}$ and $k=1,\dotsc,r$ be independent Bernoulli random variables with $P(X_{i,k}=1)=p$.

Consider the following process. Let $A_{k+1}$ be the number of cards after $k$ moves. In each move $k=1,2,\dots,r$, we remove the card with label $i$ if $X_{i,k}=1$ and this card belongs to the candidate cards, i.e., the $\lceil qA_{k}\rceil$ top-most remaining cards. We will call this process a $q$-process. This process describes the evolution of a pile of size $A_{1}$ in the $p$-random $q$-proportion Bulgarian solitaire.

Using the same Bernoulli variables, for any real number $0\leq s\leq 1$, we define an $s$-threshold process in the following way. In each move $k=1,2\dots,r$, we remove the card with label $i$ if $X_{i,k}=1$ and $i\leq sA_{1}$. In this process, we let $A^{[s]}_{k+1}$ denote the number of remaining cards after $k$ moves. When it is relevant to indicate the initial pile size, an $s$-threshold process is called an $(s,A_{1})$-threshold process and the number of remaining cards after $k$ moves is denoted by $A^{[s,A_{1}]}_{k+1}$.

In the proof of Theorem 1, we will use two different $s$-threshold processes (for two different values of $s$) to over- and underestimate the sizes of $r+1$ consecutive piles in $\Gamma$ (corresponding to the $r$ steps in an $s$-threshold process). Both these processes will have the same desired limit shape and thus the limit shape of our solitaire will follow by the squeeze theorem. We first need a combinatorial lemma giving sufficient conditions for overestimation and for underestimation.

Lemma 6**.**

(i) If $\lceil sA_{1}\rceil\leq\lceil qA_{1}\rceil$, then $A^{[s]}_{k}\geq A_{k}$ for $k=1,\dotsc,r+1$.

(ii) If $(1-q)A^{[s]}_{r+1}\geq A_{1}-\lceil sA_{1}\rceil$, then $A^{[s]}_{k}\leq A_{k}$ for $k=1,\dotsc,r+1$.

Proof.

(i) A card that is removed at some step $\ell$ during the $s$-threshold process must have label $i\leq\lceil sA_{1}\rceil$, so in the $q$-process it belongs to the $\lceil qA_{1}\rceil$ candidate cards in the initial pile and hence it belongs to the candidate cards also at step $\ell$ and will be removed. Thus, every card removed in the $s$-threshold process is removed in the $q$-process too, and it follows that $A_{k}^{[s]}\geq A_{k}$ for $k=1,\dotsc,r+1$.

(ii) We show by induction over $r$ that, after $r$ steps, the remaining cards in the $s$-threshold process is a subset of the remaining cards in the $q$-process. Suppose $(1-q)A_{r+1}^{[s]}\geq A_{1}-\lceil sA_{1}\rceil$. Since $A_{r}^{[s]}\geq A_{r+1}^{[s]}$ we have $(1-q)A_{r}^{[s]}\geq A_{1}-\lceil sA_{1}\rceil$ which by the induction hypothesis implies that $A_{k}^{[s]}\leq A_{k}$ for $1\leq k\leq r$. It follows that $(1-q)A_{r}\geq A_{1}-\lceil sA_{1}\rceil$ which in turn implies that $A_{r}-(A_{1}-\lceil sA_{1}\rceil)\geq\lceil qA_{r}\rceil$. This latter inequality means that the $\lceil qA_{r}\rceil$ topmost cards before step $r$ in the $q$-process all have labels larger than $A_{1}-\lceil sA_{1}\rceil$. Thus, if a card is removed in step $r$ in the $q$-process it is also removed in step $r$ or earlier in the $s$-threshold process. This concludes the induction step. The base step $r=0$ is trivial. ∎

Recall that we are considering the configuration $\Gamma=\alpha^{(m)}$ after $m=M+D$ moves in the solitaire from the initial configuration $\alpha^{(0)}$. We will compare the sizes of $r+1$ consecutive piles in $\Gamma$ to the $r+1$ pile sizes in an $s$-threshold process. In order to make the comparison for all piles in $\Gamma$, this will be done for $r+1$ consecutive piles (which we will call an $r$-chunk) at a time. In each $r$-chunk the initial pile size is the corresponding pile size in the solitaire. In other words, $\Gamma_{1},\Gamma_{2},\dotsc,\Gamma_{r+1}$ will be compared to the pile sizes in an $(s,\Gamma_{1})$-threshold process (with initial pile size $\Gamma_{1}$); and $\Gamma_{r+2},\Gamma_{r+3},\dotsc,\Gamma_{2(r+1)}$ will be compared to the pile sizes in an $(s,\Gamma_{r+2})$-threshold process (with initial pile size $\Gamma_{r+2}$), and so on. Let us call the resulting union of $s$-threshold processes an $(r,s)$-union process. Thus, if we denote the pile sizes in this $(r,s)$-union process by ${U}_{1},{U}_{2},\dotsc$, we have

[TABLE]

We intend to use the $(r,s)$-union process to estimate the pile sizes in $\Gamma$. In an $s$-threshold process, starting with a pile of size $A_{1}$, the number of remaining cards $B$ above the level $A_{1}(1-s)$ after $r$ moves is binomially distributed: $B\sim\textup{Bin}(A_{1}s,(1-p)^{r})$. See Figure 4.

Therefore we need to choose $r=r_{n}$ and $s=s_{n}$ in such a way that we have the following in each $s$-threshold process:

• I

The pile size $A_{k+1}$ is close to $A_{1}(1-pq)^{k}$ a.a.s. for all $k=1,\dotsc,r$, which we need to establish the wanted limit shape.

• II

At the same time $s$ must be close enough to $q$ to make the over- and underestimations tight enough.

To accomplish (I), clearly $r=r_{n}$ can at least not be chosen bigger than $1/p_{n}$, in fact we shall require $p_{n}r_{n}\to 0$ as $n\to\infty$, in order for the variance in the size of the last pile (after $r_{n}$ moves) in an $s$-threshold process to be small with high probability. However, we shall see that $p_{n}r_{n}$ may not tend to zero too fast. We will require

[TABLE]

(Recall from (5) that $p_{n}q_{n}^{2}n/{\log{n}}\to\infty$.) However, since $r_{n}$ is a positive integer for any $n$, if $p_{n}\not\to 0$ we cannot have $p_{n}r_{n}\to 0$, but will see that $r_{n}=1$ suffices in the case $p_{n}\not\to 0$. In other words, we will require

[TABLE]

To accomplish (II) we shall see that $s=q$ will suffice for the overestimation and $s=q(1+2pr)=q(1+o(1))$ for the underestimation.

One way of choosing $r_{n}$ such that (9) and (10) are fulfilled is

[TABLE]

This choice fulfills (10) since $(r_{n}-1)p_{n}<r_{n}p_{n}\leq\rho_{n}^{-1/3}\rightarrow 0$. That (9) is fulfilled is easily verified:

[TABLE]

as $n\to\infty$ by (5).

Our next lemma, Lemma 7, will bound the probability $P^{\prime}$ that an initial pile of size $I_{n}:=O(pqn)$ will, after $r_{n}$ moves in an $s_{n}$-threshold process, deviate from the expected size assuming exponential decay, when $s_{n}=q_{n}(1+o(1))$.

Since the number of piles is $\approx(p_{n}q_{n})^{-1}$, the number of $r$-chunks is $\approx(p_{n}q_{n}r_{n})^{-1}$. When using Lemma 7 we need the bound $P^{\prime}$ to hold for each chunk during all $M$ moves (where $M$ is given by (8)), specifically $P^{\prime}M/(p_{n}q_{n}r_{n})\to 0$ as $n\to\infty$. The probability in Lemma 7 is therefore bounded by $o(p_{n}q_{n}r_{n}/M)=o(p_{n}^{2}q_{n}r_{n}/n^{2})$. This is also why the pile size deviation $\varepsilon np_{n}q_{n}$ is scaled with the number of chunks, resulting in the deviation $(\varepsilon p_{n}q_{n}n)(p_{n}q_{n}r_{n})=\varepsilon p_{n}^{2}q_{n}^{2}nr_{n}$.

Lemma 7**.**

Let $(p_{n})_{n}$ and $(q_{n})_{n}$ be real sequences such that $0<p_{n},q_{n}\leq 1$ and $p_{n}q_{n}\rightarrow 0\text{ as }n\rightarrow\infty$. For each $n$, let also $B_{n}\sim\textup{Bin}(F_{n}s_{n},(1-p_{n})^{r_{n}})$ where $(F_{n})_{n}$ and $(s_{n})_{n}$ are real sequences such that

[TABLE]

and $F_{n}s_{n}$ is an integer for any $n$. Let also $(r_{n})_{n}$ be the sequence of positive integers in (11). Then, for all $\varepsilon>0$ we have

[TABLE]

Proof.

Let us abbreviate $F=F_{n}$, $B=B_{n}$, $p=p_{n}$, $q=q_{n}$, $r=r_{n}$ and $s=s_{n}$. Thus, we want to prove that

[TABLE]

We first note that the expected value $E(B)=Fs(1-p)^{r}$. Using the triangle inequality $|B+F(1-s)-F(1-pq)^{r}|\leq|B-E(B)|+|E-F(1-pq)^{r}+F(1-s)|$ we obtain

[TABLE]

By the Chernoff bound (7) we get

[TABLE]

For the indices $n$ for which $r_{n}>1$ we have $rp\leq 2(r-1)p\rightarrow 0$ and hence

[TABLE]

For the indices $n$ for which $r_{n}=1$, the relations in (14) are trivially true.

This means

[TABLE]

Thus the numerator in (13) can be written $[(\varepsilon+o(1))p^{2}q^{2}nr]^{2}$. By the assumptions in (12), the denominator in (13) can be written

[TABLE]

Putting these together, the bound (13) on $P$ can be written

[TABLE]

where (9) was used in the last step. Since $pqnr\to\infty$ (also by (9)) and $pqr\to 0$ (by (6) and (10)), we have $\frac{1}{pqr}=o(n)$ and hence $\log\frac{1}{pqr}=o(\log n)$. Therefore

[TABLE]

From this follows

[TABLE]

thus $P=o(p^{2}qr/n^{2})$. ∎

Note that Lemma 7 concerns an $s$-threshold process, i.e. only $r$ steps. In other words, it asserts that

[TABLE]

where $A_{1}=O(p_{n}q_{n}n)$ is the first pile size in an $r$-chunk and $A_{r+1}=(1-s_{n})A_{1}+B_{n}$ the last (see Figure 4). However, the deviation and the probability were chosen in such a way that they can be added over all $r$-chunks. This is done in Lemma 8 which bounds the probability for deviation for the entire union process. Specifically, we will show that, for any $C>0$, the piles in $\Gamma$ formed at most $\frac{C}{pq}$ moves ago, i.e. $\Gamma_{k}$ for $k\leq\frac{C}{pq}$, will follow an exponential decay a.a.s. The sizes of the piles formed more than $\frac{C}{pq}$ moves ago ($k>\frac{C}{pq}$) will be shown to be sufficiently small to be close enough to the tail in the exponential limit shape.

Lemma 8**.**

Let $U_{1},U_{2},\dotsc$ be the pile sizes in an $(r_{n},s_{n})$-union process corresponding to $\mathscr{B}(n,p_{n},q_{n})$, where the initial pile size is $U_{1}=O(p_{n}q_{n}n)$, and $r_{n}$ is given by (11) and $s_{n}=q_{n}(1+o(1))$. Let $M=\lceil n^{2}/p_{n}\rceil$. Then

[TABLE]

Proof.

As in the proof of Lemma 7, for the simplicity of notation we do not indicate in $p$, $q$, $r$ and $s$ the dependence on $n$. Let $C,\varepsilon>0$ and $\varepsilon^{\prime}=\varepsilon/C$. By the triangle inequality,

[TABLE]

Lemma 7 is now applicable for the first pile in each $r$-chunk (since $U_{1}\geq U_{2}\geq\dotsb$ and $U_{1}=O(p_{n}q_{n}n)$), so by its formulation (15), $|U_{k+r+1}-U_{k+1}(1-pq)^{r}|<\varepsilon^{\prime}p^{2}q^{2}rn$ with probability ${1-o(p^{2}qr/n^{2})}$. Thus,

[TABLE]

with probability $1-o(p^{2}qr/n^{2})$. We now note that the first term in the right hand side has the same form as the left hand side, only shifted with $r$ piles. Thus, by induction it follows that, for any positive integer $d$, we have

[TABLE]

with probability $1-o(p^{2}qr/n^{2})$. Thus, adding the probabilities for deviation for $k=r,2r,\dotsc,\eta r$, where $\eta=\lfloor\tfrac{C}{pqr}\rfloor$ we get

[TABLE]

We have thereby proved the claim in the lemma for $k=r,2r,\dotsc,\eta r$. If $k$ is not a multiple of $r$, suppose $dr<k<(d+1)r$ for some positive integer $d$. Then, since $pqr\to 0$ as $n\to\infty$ (which follows from (5) and (11)), we have $(1-pq)^{r}=1-pqr+o(pqr)$ and hence

[TABLE]

The lemma then follows by (5) and the fact that ${U}_{dr}\leq{U}_{k}\leq{U}_{(d+1)r}$. ∎

6 Proof of Theorem 1

Below follows the proof of Theorem 1, stated in Section 5.

Proof.

First, as in the previous section, let us consider $\mathscr{B}(n,p_{n},q_{n})$ as a process on $\mathcal{W}(n)$ rather than on $\mathcal{P}(n)$, and let $\alpha^{(0)}\in\mathcal{W}(n)$ be the weak composition representing the initial configuration of cards in the solitaire. Let also $M$ and $D$ be given by (8).

Let $(r_{n})_{n}$ be the sequence of positive integers given by (11) and let $(s_{n})_{n}$ be the sequence $s_{n}=q_{n}(1+2p_{n}r_{n})$. By Lemma 3 applied on $\alpha^{(D)}$, all piles present in $\alpha^{(D)}$ have disappeared in $\Gamma:=\alpha^{(D+M)}$ a.a.s. Let $\Gamma_{k}=\alpha^{(D+M)}_{k}$ for $1\leq k\leq M$ be the number of cards in the pile that was formed $k$ moves ago. By Lemma 5, each of these piles had size $O(np_{n}q_{n})$ a.a.s. when they were formed. Let $F_{n}:=O(np_{n}q_{n})$ be a sequence such that $F_{n}s_{n}$ is an integer for each $n$. Let $0<\varepsilon<1$ and choose $C_{n}$ such that $C_{n}>\frac{p_{n}q_{n}\log\varepsilon}{\log(1-p_{n}q_{n})}$.

Let $\check{U}_{1},\check{U}_{2},\dotsc$ be the pile sizes in the $(r_{n},s_{n})$-union process with initial pile size $\Gamma_{1}$. Using the fact that $p_{n}q_{n}r_{n}\to\ 0$, it is a straightforward computation to show that $s_{n}=q_{n}(1+2p_{n}r_{n})$ implies $(1-q_{n})((1-p_{n}r_{n})^{r_{n}}-\varepsilon p_{n}q_{n}r_{n})>1-s_{n}$ and therefore also

[TABLE]

By Lemma 7, the probability that $\check{U}_{1}(1-p_{n}r_{n})^{r_{n}}-\varepsilon\check{U}_{1}p_{n}q_{n}r_{n}<\check{U}_{r_{n}+1}$ is $P_{1}:=1-o(p_{n}^{2}q_{n}r_{n}/n^{2})$. Thus, with probability $P_{1}$ we have $(1-q_{n})\check{U}_{r_{n}+1}>(1-s_{n})\check{U}_{1}\geq\check{U}_{1}-\lceil s\check{U}_{1}\rceil$ so by Lemma 6(ii), the pile sizes $\check{U}_{1},\check{U}_{2},\dotsc,\check{U}_{r_{n}+1}$ in the first $r$-chunk of the $(r_{n},s_{n})$-process underestimate the pile sizes $\Gamma_{1},\Gamma_{2},\dotsc,\Gamma_{r_{n}+1}$ with probability $P_{1}$. In the next chunk, we have a new absolute threshold $s\Gamma_{r_{n}+2}=s\check{U}_{r_{n}+2}$. Since $\Gamma_{r_{n}+2}\leq\Gamma_{1}$, we have $(1-q_{n})\check{U}_{2r_{n}+2}>(1-s_{n})\check{U}_{r_{n}+2}$ with probability at least $P_{1}$, making Lemma 6(ii) applicable also for the second chunk to conclude that $\check{U}_{r_{n}+2},\dotsc,\check{U}_{2r_{n}+2}$ underestimate $\Gamma_{r_{n}+2},\dotsc,\Gamma_{2r_{n}+2}$ with probability at least $P_{1}$. Continuing in the same manner for the first $C_{n}(p_{n}q_{n}r_{n})^{-1}$ chunks, we conclude that the $(r_{n},s_{n})$-union process underestimates the solitaire with high probability:

[TABLE]

Let $\widehat{U}_{1},\widehat{U}_{2},\dotsc$ be the pile sizes in the $(r_{n},q_{n})$-union process with initial pile size $\Gamma_{1}$. By Lemma 6(i) (with $s_{n}=q_{n}$), the $(r_{n},q_{n})$-union process surely overestimates the solitaire in each chunk.

Taking the results for the $(r_{n},s_{n})$-union process and the $(r_{n},q_{n})$-union process together we have

[TABLE]

Now, applying Lemma 8 to both the pile sizes $\check{U}_{k}$ and to the pile sizes $\widehat{U}_{k}$ and using the squeeze theorem, we obtain

[TABLE]

Thus, the probability that $|\Gamma_{k}-\Gamma_{1}(1-p_{n}q_{n})^{k}|<\varepsilon np_{n}q_{n}$ for the first $\tfrac{C_{n}}{p_{n}q_{n}}$ piles throughout all $M$ moves from $\alpha^{(D)}$ to $\alpha^{(D+M)}$ is $1-M\cdot o(1/M)=1-o(1)$.

For piles $k>\tfrac{C_{n}}{p_{n}q_{n}}$, the exponential decrease (with decay factor $1-pq$) in pile size will yield piles smaller than $np_{n}q_{n}(1-p_{n}q_{n})^{\frac{C_{n}}{p_{n}q_{n}}}<\varepsilon np_{n}q_{n}$ (by our choice of $C_{n}$). Thus, the pile sizes themselves are below $\varepsilon np_{n}q_{n}$.

In summary; playing sufficiently many moves of $\mathscr{B}(n,p_{n},q_{n})$, the resulting composition diagram will a.a.s. be arbitrarily close to the boundary diagram of the composition $\alpha$ where $\alpha_{k}=np_{n}q_{n}(1-p_{n}q_{n})^{k-1}$ for all $k=1,2,\dotsc$. The corresponding boundary function is $\partial{\alpha}(x)=np_{n}q_{n}(1-np_{n}q_{n})^{\lfloor x\rfloor}$. The corresponding rescaled boundary function, with the given scaling factor $a_{n}=(p_{n}q_{n})^{-1}$, is

[TABLE]

since $p_{n}q_{n}\to 0$ as $n\to\infty$.

Setting $m:=D+M$, and letting $\pi_{n}^{m}$ denote the probability distribution on $\mathcal{W}(n)$ for $\alpha^{(m)}$, we have

[TABLE]

for all $\varepsilon>0$ and all $x>0$, in accordance with (2). By virtue of Lemma 2, the same limit shape holds when configurations in the solitaire $\mathscr{B}(n,p_{n},q_{n})$ are represented by partitions $\mathcal{P}(n)$.

Since $\pi_{n,p_{n},q_{n}}$ is the stationary distribution of the Markov chain $(\lambda^{(0)},\lambda^{(1)},\dotsc)$, if we start with a partition $\lambda^{(0)}$ sampled from $\pi_{n,p_{n},q_{n}}$ and play $m$ moves, the resulting partition $\lambda^{(m)}$ will also be sampled from $\pi_{n,p_{n},q_{n}}$. Thus, the theorem follows by choosing $\lambda^{(0)}$ as a stochastic partition sampled from the stationary distribution. ∎

7 Conjectures

Recall that Theorem 1 was proved with $\mathscr{B}(n,p_{n},q_{n})$ being considered a process on $\mathcal{W}(n)$, and by virtue of Lemma 2 it also holds in $\mathcal{P}(n)$. We imposed the condition $\frac{p_{n}q_{n}^{2}n}{\log n}\to\infty$. Here we conjecture that the weaker condition $p_{n}q_{n}^{2}n\to\infty$ suffices in order for Theorem 1 to hold in $\mathcal{P}(n)$.

Conjecture 1**.**

Theorem 1 holds also when the condition $np_{n}q_{n}^{2}/{\log n}\to\infty$ is replaced by the weaker condition $np_{n}q_{n}^{2}\to\infty$.

The reason for this conjecture can be understood by considering the example $q_{n}=1$ and $p_{n}n=\log(\log n)$. For this example it is easy to prove that there is no limit shape when sorting is not performed. Since $q_{n}=1$, the number of picked cards in each move and thus the expected size of a new pile is $\textup{Bin}(n,p_{n})$ with expected value $np_{n}$ and standard deviation $\sigma\approx\sqrt{np_{n}}$. A pile of size $np_{n}$ will after $1/p_{n}$ moves have the expected size $np_{n}(1-p_{n})^{1/p_{n}}\to e^{-1}np_{n}$ as $n\to\infty$.

Thus, the probability for a “visible” deviation (i.e. greater than $d=\sqrt{np_{n}}$ standard deviations) is $P(\text{deviation}\geq d\sigma)=e^{-np_{n}}$, so for $1/p_{n}$ piles, the probability for a visible deviation anywhere is $\frac{e^{-np_{n}}}{p_{n}}=\frac{n}{\log n\log(\log n)}\to\infty$ as $n\to\infty$, i.e. the expected number of such large deviations tends to infinity as $n$ tends to infinity. This makes it impossible to achieve a convergence in probability towards a limit shape. However, from simulations we have reason to believe that the process converges towards a limit shape when sorting is performed.

Further, recall from Section 4 the other regimes $np_{n}q_{n}^{2}\to 0$ and $np_{n}q_{n}^{2}\to C$ from some constant $C>0$. We conjecture that the limit shapes in the $p_{n}$-random $q_{n}$-proportion Bulgarian solitaire in these regimes are the same as in the deterministic $q$-proportion Bulgarian solitaire developed in [3].

Conjecture 2**.**

If $p_{n}q_{n}^{2}n\to 0$ as $n\to\infty$, the limit shape of the $p_{n}$-random $q_{n}$-proportion Bulgarian solitaire is triangular.

Conjecture 3**.**

If $p_{n}q_{n}^{2}n\to C$ as $n\to\infty$ for some constant $C>0$, the limit shape of the $p_{n}$-random $q_{n}$-proportion Bulgarian solitaire is a piecewise linear shape that depends on the value of $C$.