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# Analysis of casino shelf shuffling machines

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Many casinos routinely use mechanical card shuffling machines. We were asked to evaluate a new product, a shelf shuffler. This leads to new probability, new combinatorics and to some practical advice which was adopted by the manufacturer. The interplay between theory, computing, and real-world application is developed.
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ANALYSIS OF CASINO SHELF SHUFFLING MACHINES
PERSI DIACONIS, JASON FULMAN, AND SUSAN HOLMES
Abstract. Many casinos routinely use mechanical card shuﬄing machines. We were
asked to evaluate a new product, a shelf shuﬄer. This leads to new probability, new
combinatorics, and to some practical advice which was adopted by the manufacturer. The
interplay between theory, computing, and real-world application is developed.
1. Introduction
We were contacted by a manufacturer of casino equipment to evaluate a new design for
a casino card-shuﬄing machine. The machine, already built, was a sophisticated “shelf
shuﬄer” consisting of an opaque box containing ten shelves. A deck of cards is dropped
into the top of the box. An internal elevator moves the deck up and down within the
box. Cards are sequentially dealt from the bottom of the deck onto the shelves; shelves are
chosen uniformly at random at the command of a random number generator. Each card is
randomly placed above or below previous cards on the shelf with probability 1/2. At the
end, each shelf contains about 1/10 of the deck. The ten piles are now assembled into one
pile, in random order. The manufacturer wanted to know if one pass through the machine
would yield a well-shuﬄed deck.
Testing for randomness is a basic task of statistics. A standard approach is to design
some ad hoc tests such as: Where do the original top and bottom cards wind up? What
is the distribution of cards that started out together? What is the distribution, after one
shuﬄe, of the relative order of groups of consecutive cards? Such tests had been carried
out by the engineers who designed the machine, and seemed satisfactory.
We ﬁnd closed-form expressions for the probability of being at a given permutation after
the shuﬄe. This gives exact expressions for various global distances to uniformity, e.g.,
total variation. These suggest that the machine has ﬂaws. The engineers (and their bosses)
needed further convincing; using our theory, we were able to show that a knowledgeable
player could guess about 9 1/2 cards correctly in a single run through a 52-card deck. For
a well-shuﬄed deck, the optimal strategy gets about 4 1/2 cards correct. This data did
convince the company. The theory also suggested a useful remedy. Journalist accounts of
our shuﬄing adventures can be found in Klarreich (2003, 2002); Mackenzie (2002).
Section 2 gives background on casino shuﬄers, needed probability, and the literature of
shuﬄing. Section 3 gives an analysis of a single shuﬄe; we give a closed formula for the
chance that a deck of n cards passed through a machine with m shelves is in ﬁnal order
w. This is used to compute several classical distances to randomness. In particular it is
shown that, for n cards, the l() distance is asymptotic to e
1/12c
2
1 if the number of
shelves m = cn
3/2
and n is large. The combinatorics of shelf shuﬄers turns out to have
connections to the “peak algebra” of algebraic combinatorics. This allows nice formulae for
Date: July 18, 2011.
The ﬁrst author is supported in part by National Science Foundation grant DMS 0804324.
The second author is supported in part by National Science Foundation grant DMS 0802082 and National
Security Agency grant H98230-08-1-0133.
The third author is supported in part by National Institutes of Health grant R01 GM086884-02.
1
arXiv:1107.2961v1 [math.CO] 14 Jul 2011
2 PERSI DIACONIS, JASON FULMAN, AND SUSAN HOLMES
the distribution of several classical test statistics: the cycle structure (e.g., the number of
ﬁxed points), the descent structure, and the length of the longest increasing subsequence.
Section 4 develops tools for analyzing repeated shelf shuﬄing. Section 5 develops our
“how many can be correctly guessed” tests. This section also contains our ﬁnal conclusions.
2. Background
This section gives background and a literature review. Section 2.1 treats shuﬄing ma-
chines; Section 2.2 gives probability background; Section 2.3 gives an overview of related
literature and results on the mathematics of shuﬄing cards.
2.1. Card shuﬄing machines. Casinos worldwide routinely employ mechanical card-
shuﬄing machines for games such as blackjack and poker. For example, for a single deck
game, two decks are used. While the dealer is using the ﬁrst deck in the usual way, the
shuﬄing machine mixes the second deck. When the ﬁrst deck is used up (or perhaps half-
used), the second deck is brought into play and the ﬁrst deck is inserted into the machine.
Two-, four-, and six-deck machines of various designs are also in active use.
The primary rationale seems to be that dealer shuﬄing takes time and use of a machine
results in approximately 20% more hands per hour. The machines may also limit dealer
cheating.
The machines in use are sophisticated, precision devices, rented to the casino (with service
contracts) for approximately \$500 per month per machine. One company told us they had
about 8,000 such machines in active use; this amounts to millions of dollars per year. The
companies involved are substantial businesses, listed on the New York Stock Exchange.
One widely used machine simulates an ordinary riﬄe shuﬄe by pushing two halves of a
single deck together using mechanical pressure to make the halves interlace. The randomness
comes from slight physical diﬀerences in alignment and pressure. In contrast, the shelf
shuﬄers we analyze here use computer-generated pseudo-random numbers as a source of
their randomness.
The pressure shuﬄers require multiple passes (perhaps seven to ten) to adequately mix
52 cards. Our manufacturer was keen to have a single pass through suﬃce.
2.2. Probability background. Let S
n
denote the group of permutations of n objects. Let
U(σ) = 1/n! denote the uniform distribution on S
n
. If P is a probability on S
n
, the total
variation, separation, and l
distances to uniformity are
(2.1)
kP Uk
TV
=
1
2
X
w
|P (w) U (w)| = max
AS
n
|P (A) U(A)| =
1
2
max
kfk
1
|P (f) U (f )|,
sep(P ) = max
w
1
P (w)
U(w)
, kP Uk
= max
w
1
P (w)
U(w)
.
Note that kP Uk
TV
sep(P ) kP Uk
. The ﬁrst two distances are less than 1; the
k k
norm can be as large as n! 1.
If one of these distances is suitably small then many test statistics evaluate to approx-
imately the same thing under P and U . This gives an alternative to ad hoc tests. The
methods developed below allow exact evaluation of these and many further distances (e.g.,
chi-square or entropy).
Repeated shuﬄing is modeled by convolution:
P P (w) =
X
v
P (v)P (wv
1
), P
k
(w) = P P
(k1)
(w).
ANALYSIS OF CASINO SHELF SHUFFLING MACHINES 3
All of the shelf shuﬄers generate ergodic Markov chains (even if only one shelf is involved),
and so P
k
(w) U (w) as k . One question of interest is the quantitative measurement
of this convergence using one of the metrics above.
2.3. Previous work on shuﬄing.
Early work. The careful analysis of repeated shuﬄes of a deck of cards has challenged
probabilists for over a century. The ﬁrst eﬀorts were made by Markov (1906) in his papers
on Markov chains. Later, Poincar´e (1912) studied the problem. These great mathematicians
proved that in principle repeated shuﬄing would mix cards at an exponential rate but gave
no examples or quantitative methods to get useful numbers in practical problems.
Borel and Ch´eron (1955) studied riﬄe shuﬄing and concluded heuristically that about
seven shuﬄes would be required to mix 52 cards. Emile Borel also reported joint work with
Paul Levy, one of the great probabilists of the twentieth century; they posed some problems
but were unable to make real progress.
Isolated but serious work on shuﬄing was reported in a 1955 Bell Laboratories report by
Edgar Gilbert. He used information theory to attack the problems and gave some tools for
riﬄe shuﬄing developed jointly with Claude Shannon.
They proposed what has come to be called the Gilbert–Shannon–Reeds model for riﬄe
shuﬄing; this presaged much later work. Thorp (1973) proposed a less realistic model and
showed how poor shuﬄing could be exploited in casino games. Thorp’s model is analyzed
in Morris (2009). Epstein (1977) reports practical studies of how casino dealers shuﬄe with
data gathered with a very precise microphone! The upshot of this work was a well-posed
mathematics problem and some heuristics; further early history appears in Chapter 4 of
Diaconis (1988).
The modern era. The modern era in quantitative analysis of shuﬄing begins with papers
of Diaconis and Shahshahani (1981) and Aldous (1983). They introduced rigorous meth-
ods, Fourier analysis on groups, and coupling. These gave sharp upper and lower bounds,
suitably close, for real problems. In particular, Aldous sketched out a proof that
3
2
log
2
n
riﬄe shuﬄes mixed n cards. A more careful argument for riﬄe shuﬄing was presented by
Aldous and Diaconis (1986). This introduced “strong stationary times,” a powerful method
of proof which has seen wide application. It is applied here in Section 4.
A deﬁnitive analysis of riﬄe shuﬄing was ﬁnally carried out in Bayer and Diaconis (1992)
and Diaconis et al. (1995). They were able to derive simple closed-form expressions for all
quantities involved and do exact computations for n = 52 (or 32 or 104 or . . . ). This results
in the “seven shuﬄes theorem” explained below. A clear elementary account of these ideas
is in Mann (1994, 1995) reprinted in Grinstead and Snell (1997). See Ethier (2010) for an
informative textbook account.
The successful analysis of shuﬄing led to a host of developments, the techniques reﬁned
and extended. For example, it is natural to want not only the order of the cards, but
also the “up-down pattern” of one-way backs to be randomized. Highlights include work
of Bidigare et al. (1999) and Brown and Diaconis (1998) who gave a geometric interpre-
tation of shuﬄing which had many extensions to which the same analysis applied. Lalley
(1996, 1999) studied less random methods of riﬄe shuﬄing. Fulman (2000a,b, 2001) showed
that interspersing cuts doesn’t materially eﬀect things and gave high level explanations for
miraculous accidents connecting shuﬄing and Lie theory. The work is active and ongoing.
Recent surveys are given by Diaconis (1996, 2003); Fulman (1998); Stark et al. (2002).
In recent work, Diaconis et al. (1995); Conger and Viswanath (2006) and Assaf et al.
(2011) have studied the number of shuﬄes required to have selected features randomized
(e.g., the original top card, or the values but not the suits). Here, fewer shuﬄes suﬃce.
4 PERSI DIACONIS, JASON FULMAN, AND SUSAN HOLMES
Conger and Howald (2010) shows that the way the cards are dealt out after shuﬄing aﬀects
things. The mathematics of shuﬄing is closely connected to modern algebraic combinatorics
through quasi-symmetric functions Stanley (2001). The descent theory underlying shuﬄing
makes equivalent appearances in the basic task of carries when adding integers Diaconis
and Fulman (2009a,b); Diaconis and Fulman (2011).
3. Analysis of one pass through a shelf shuffler
This section gives a fairly complete analysis of a single pass through a shelf shuﬄer.
Section 3.1 gives several equivalent descriptions of the shuﬄe. In Section 3.2, a closed-form
formula for the chance of any permutation w is given. This in turn depends only on the
number of “valleys” in w. The number of permutations with j valleys is easily calculated
and so exact computations for any of the distances above are available. Section 3.3 uses the
exact formulae to get asymptotic rates of convergence for l() and separation distances.
Section 3.4 gives the distribution of such permutations by cycle type. Section 3.5 gives the
distribution of the “shape” of such a permutation under the Robinson–Schensted–Knuth
map. Section 3.6 gives the distribution of the number of descents. We ﬁnd it surprising that
a real-world applied problem makes novel contact with elegant combinatorics. In Section 4,
iterations of a shelf shuﬄer are shown to be equivalent to shelf shuﬄing with more shelves.
Thus all of the formulae of this section apply.
3.1. Alternative descriptions. Consider two basic shelf shuﬄers: for the ﬁrst, a deck of
n cards is sequentially distributed on one of m shelves. (Here, n = 52, m = 10, are possible
choices.) Each time, the cards are taken from the bottom of the deck, a shelf is chosen at
random from one to m, and the bottom card is placed on top of any previous cards on the
shelf. At the end, the packets on the shelves are unloaded into a ﬁnal deck of n. This may
be done in order or at random; it turns out not to matter. Bayer and Diaconis (1992) called
this an inverse m-shuﬄe.
This shuﬄing scheme, that is the main object of study, is based on m shelves. At each
stage that a card is placed on a shelf, the choice of whether to put it on the top or the
bottom of the existing pile on that shelf is made at random (1/2 each side). This will be
called a shelf shuﬄe. There are several equivalent descriptions of shelf shuﬄes:
Description 1 (Shelf shuﬄes). A deck of cards is initially in order 1, 2, 3, . . . , n. Label the
back of each card with n random numbers chosen at random between one and 2m. Remove
all cards labeled 1 and place them on top, keeping them in the same relative order. Then
remove all cards labeled 2 and place them under the cards labeled 1, reversing their relative
order. This continues with the cards labeled 3, labeled 4, and so on, reversing the order in
each even labeled packet. If at any stage there are no cards with a given label, this empty
packet still counts in the alternating pattern.
For example, a twelve-card deck with 2m = 4,
Label 2 1 1 4 3 3 1 2 4 3 4 1
Card 1 2 3 4 5 6 7 8 9 10 11 12
is reordered as
2 3 7 12 8 1 5 6 10 11 9 4.
Description 2 (Inverse shelf shuﬄes). Cut a deck of n cards into 2m piles according to a
multinomial distribution; thus the number of cards cut oﬀ in pile i has the same description
as the number of balls in the ith box if n balls are dropped randomly into 2m boxes. Reverse
the order of the even-numbered packets. Finally, riﬄe shuﬄe the 2m packets together by
the Gilbert–Shannon–Reeds (GSR) distribution Bayer and Diaconis (1992) dropping each
ANALYSIS OF CASINO SHELF SHUFFLING MACHINES 5
card sequentially with probability proportional to packet size. This makes all possible
interleavings equally likely.
1
10
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
B
Figure 1. Two shelves in shelf shuﬄe.
Description 3 (Geometric description). Consider the function f
m
(x) from [0, 1] to [0, 1]
which has “tents,” each of slope ±2m centered at
1
2m
,
3
2m
,
5
2m
, . . . ,
2m1
2m
. Figure 1 illustrates
an example with m = 2. Place n labeled points uniformly at random into the unit interval.
Label them, from left to right, x
1
, x
2
, . . . , x
n
. Applying f
m
gives y
i
= f
m
(x
i
). This gives
the permutation
1 2 ··· n
π
1
π
2
··· π
n
with π
1
the relative position from the bottom of y
1
, . . . , π
i
the relative position from the
bottom of y
i
among the other y
j
. This permutation has the distribution of an inverse shelf
shuﬄe. It is important to note that the natural distances to uniformity (total variation,
separation, l
) are the same for inverse shuﬄes and forward shuﬄes. In Section 4, this
description is used to show that repeated shelf shuﬄing results in a shelf shuﬄe with more
shelves.
3.2. Formula for the chance of a permutation produced by a shelf shuﬄer. To
describe the main result, we call i a valley of the permutation w S
n
if 1 < i < n and
w(i 1) > w(i) < w(i + 1). Thus w = 5136724 has two valleys. The number of valleys is
classically used as a test of randomness for time series. See Warren and Seneta (1996) and
their references. If v(n, k) denotes the number of permutations on n symbols with k valleys,
then Warren and Seneta (1996) v(1, 0) = 1, v(n, k) = (2k + 2)v(n 1, k) + (n 2k)v(n
1, k 1). So v(n, k) is easy to compute for numbers of practical interest. Asymptotics are
in Warren and Seneta (1996) which also shows the close connections between valleys and
descents.
Theorem 3.1. The chance that a shelf shuﬄer with m shelves and n cards outputs a
permutation w is
4
v(w)+1
2(2m)
n
m1
X
a=0
n + m a 1
n

n 1 2v(w)
a v(w)
where v(w) is the number of valleys of w. This can be seen to be the coeﬃcient of t
m
in
1
2(2m)
n
(1 + t)
n+1
(1 t)
n+1
4t
(1 + t)
2
v(w)+1
.
6 PERSI DIACONIS, JASON FULMAN, AND SUSAN HOLMES
Example. Suppose that m = 1. Then the theorem yields the uniform distribution on the
2
n1
permutations with no valleys. Permutations with no valleys are also sometimes called
unimodal permutations. These arise in social choice theory through Coombs’ “unfolding”
hypothesis (Diaconis, 1988, Chap. 6).
Remark. By considering the cases m n and n m we see that, in the formula of Theorem
3.1, the range of summation can be taken up to n 1 instead of m 1. This will be useful
later.
Theorem 3.1 makes it easy to compute the distance to stationarity for any of the metrics
in Section 2.2. Indeed, the separation and l() distance is attained at either permutations
with a maximum number of valleys (when n = 52, this maximum is 25) or for permutations
with 0 valleys. For the total variation distance, with P
m
(v) denoting the probability in
Theorem 3.1,
kP
m
Uk
TV
=
1
2
b
n1
2
c
X
a=0
v(n, a)
P
m
(a)
1
n!
.
Table 1. Distances for various numbers of shelves m.
m 10 15 20 25 30 35 50 100 150 200 250 300
kP
m
Uk
TV
1 .943 .720 .544 .391 .299 .159 .041 .018 .010 .007 .005
sep(P
m
) 1 1 1 1 1 .996 .910 .431 .219 .130 .085 .060
kP
m
Uk
45118 3961 716 39 1.9 .615 .313 .192 .130
Table 1 gives these distances when n = 52 for various numbers of shelves m. Larger values
of m are of interest because of the convolution results explained in Section 4. These numbers
show that ten shelves are woefully insuﬃcient. Indeed, 50 shelves are hardly suﬃcient.
To prove Theorem 3.1, we will relate it to the following 2m-shuﬄe on the hyperoctahedral
group B
n
: cut the deck multinomially into 2m piles. Then ﬂip over the odd numbered stacks,
and riﬄe the piles together, by dropping one card at a time from one of the stacks (at each
stage with probability proportional to stack size). When m = 1 this shuﬄe was studied in
Bayer and Diaconis (1992), and for larger m it was studied in Fulman (2001).
It will be helpful to have a description of the inverse of this 2m-shuﬄe. To each of the
numbers {1, . . . , n} is assigned independently and uniformly at random one of 1, 1, 2, 2,
. . . , m, m. Then a signed permutation is formed by starting with numbers mapped to 1
(in decreasing order and with negative signs), continuing with the numbers mapped to 1
(in increasing order and with positive signs), then continuing to the numbers mapped to
2 (in decreasing order and with negative signs), and so on. For example the assignment
{1, 3, 8} 7→ 1, {5} 7→ 1, {2, 7} 7→ 2, {6} 7→ 3, {4} 7→ 3
8 3 1 5 2 7 6 4.
The proof of Theorem 3.1 depends on an interesting relation with shuﬄes for signed
permutations (hyperoctahedral group). This is given next followed by the proof of Theorem
3.1.
Theorem 3.2 gives a formula for the probability for w after a hyperoctahedral 2m-shuﬄe,
when one forgets signs. Here p(w) is the number of peaks of w, where i is said to be a peak
of w if 1 < i < n and w(i 1) < w(i) > w(i + 1). Also Λ(w) denotes the peak set of w
and D(w) denotes the descent set of w (i.e., the set of points i such that w(i) > w(i + 1)).
Finally, let [n] = {1, . . . , n}.
ANALYSIS OF CASINO SHELF SHUFFLING MACHINES 7
Theorem 3.2. The chance of a permutation w obtained by performing a 2m shuﬄe on the
hyperoctahedral group and then forgetting signs is
4
p(w
1
)+1
2(2m)
n
m1
X
a=0
n + m a 1
n

n 1 2p(w
1
)
a p(w
1
)
where p(w
1
) is the number of peaks of w
1
.
Proof. Let P
0
(m) denote the set of nonzero integers of absolute value at most m, totally
ordered so that
1 1 2 2 ··· m m.
Then given a permutation w = (w
1
, ··· , w
n
), page 768 of Stembridge (1997) deﬁnes a
quantity ∆(w). (Stembridge calls it ∆(w, γ), but throughout we always choose γ to be the
identity map on [n], and so suppress the symbol γ whenever he uses it). By deﬁnition,
∆(w) enumerates the number of maps f : [n] 7→ P
0
(m) such that
f (w
1
) ··· f(w
n
)
f (w
i
) = f(w
i+1
) > 0 i / D(w)
f (w
i
) = f(w
i+1
) < 0 i D(w)
We claim that the number of maps f : [n] 7→ P
0
(m) with the above three properties is
equal to (2m)
n
multiplied by the chance that a hyperoctahedral 2m-shuﬄe results in the
permutation w
1
. This is most clearly explained by example:
w = 8 3 1 5 2 7 6 4
f = 1 1 1 1 2 2 3 3
¿From the inverse description of the hyperoctahedral 2m-shuﬄe stated before the proof,
the assignment yields w. This proves the claim.
Let Λ(w) denote the set of peaks of w. From Proposition 3.5 of Stembridge (1997),
∆(w) = 2
p(w)+1
X
E[n1]:Λ(w)E4(E+1)
L
E
.
Here
L
E
=
X
1i
1
≤···≤i
n
m
kEi
k
<i
k+1
1,
and 4 denotes symmetric diﬀerence, i.e., A4B = (A B) (B A). Now a simple
combinatorial argument shows that L
E
=
n+m−|E|−1
n
. Indeed, L
E
is equal to the number
of integral i
1
, . . . , i
n
with 1 i
1
··· i
n
m |E|, which by a “stars and bars” argument
is
n+m−|E|−1
n
. Thus
∆(w) = 2
p(w)+1
X
E[n1]:Λ(w)E4(E+1)
n + m |E| 1
n
.
Now let us count the number of E of size a appearing in this sum. For each j Λ(w),
exactly one of j or j 1 must belong to E, and the remaining n 1 2p(w) elements of
[n 1] can be independently and arbitrarily included in E. Thus the number of sets E of
size a appearing in the sum is 2
p(w)
n12p(w)
ap(w)
. Hence
∆(w) =
4
p(w)+1
2
n1
X
a=0
n + m a 1
n

n 1 2p(w)
a p(w)
,
which completes the proof.
8 PERSI DIACONIS, JASON FULMAN, AND SUSAN HOLMES
Proof of Theorem 3.1. To deduce Theorem 3.1 from Theorem 3.2, it is not hard to see that a
shelf shuﬄe with m shelves is equivalent to taking w
0
to be the inverse of a permutation after
a hyperoctahedral 2m-shuﬄe, then taking a permutation w deﬁned by w(i) = n w
0
(i) + 1.
Thus the shelf shuﬄe formula is obtained from the hyperoctahedral 2m-shuﬄe formula by
replacing peaks by valleys.
Remarks.
The paper Fulman (2001) gives an explicit formula for the chance of a signed per-
mutation after a 2m-shuﬄe on B
n
in terms of cyclic descents. Namely it shows this
probability to be
m+ncd(w
1
)
n
(2m)
n
where cd(w) is the number of cyclic descents of w, deﬁned as follows: Ordering the
integers 1 < 2 < 3 < ··· < ··· < 3 < 2 < 1,
w has a cyclic descent at position i for 1 i n 1 if w(i) > w(i + 1).
w has a cyclic descent at position n if w(n) < 0.
w has a cyclic descent at position 1 if w(1) > 0.
For example the permutation 3 1 2 4 5 has two cyclic descents at position 1 and
a cyclic descent at position 3, so cd(w) = 3.
This allows one to study aspects of shelf shuﬄers by lifting the problem to B
n
,
using cyclic descents (where calculations are often easier), then forgetting about
signs. This idea was used in Fulman (2001) to study the cycle structure of unimodal
permutations, and in Aguiar et al. (2004) to study peak algebras of types B and D.
The appearance of peaks in the study of shelf shuﬄers is interesting, as peak algebras
have appeared in various parts of mathematics. Nyman (2003) proves that the
peak algebra is a subalgebra of the symmetric group algebra, and connections with
geometry of polytopes can be found in Aguiar et al. (2006) and Billera et al. (2003).
There are also close connections with the theory of P -partitions Petersen (2005,
2007); Stembridge (1997).
The following corollary shows that for a shelf shuﬄer of n cards with m shelves, the
chance of a permutation w with v valleys is monotone decreasing in v. Thus, the identity
(or any other unimodal permutation) is most likely and an alternating permutation (down,
up, down, up, . . . ) is least likely. From Theorem 3.1, the chance of a ﬁxed permutation
with v valleys is
(3.1) P (v) =
4
v+1
2(2m)
n
n1
X
a=0
n + m 1 a
n

n 1 2v
a v
.
Corollary 3.3. For P (v) deﬁned at (3.1), P (v) P (v + 1), 0 v (n 1)/2.
Proof. Canceling common terms, and setting a v = j (so a = j + v) in (3.1), we
have
2(2m)
n
4
v+1
P (v) =
P
n12v
j=0
f(j + v)
n12v
j
= 2
n12v
E(f(S
n12v
+ v)) with f(a) =
n+m1a
n
and S
n12v
distributed as Binomial(n 1 2v,
1
2
). The proposed inequality is
equivalent to
(3.2) E (f (S
n12v
+ v)) E (f (S
n12v2
+ v + 1)) .
To prove this, represent S
n12v
= S
n12v2
+ Y
1
+ Y
2
, with Y
i
independent taking values
in {0, 1}, with probability 1/2. Then (3.2) is equivalent to
(3.3)
X
j
1
4
f(j + v) +
1
2
f(j + v + 1) +
1
4
f(j + v + 2) f(j + v + 1)
P {S
n12v2
= j} 0.
ANALYSIS OF CASINO SHELF SHUFFLING MACHINES 9
Thus if
1
2
f(j + v) +
1
2
f(j + v + 2) f(j + v + 1), e.g., f(a) is convex, we are done. Writing
out the expression f(a) + f(a + 2) 2f(a + 1) and canceling common terms, it must be
shown that
(3.4) (m + n 1 a)(m + n 2 a) + (m 1 a)(m 2 a) 2(m + n 2 a)(m 1 a)
for all 0 a n 1. Subtracting the right side from the left, the coeﬃcients of a
2
and a
cancel, leaving n(n 1) 0.
3.3. Asymptotics for the kP Uk
and separation distances. Recall the distances
kP Uk
= max
w
1
P (w)
U(w)
and sep(P ) = max
w
1
P (w)
U(w)
.
Theorem 3.4. Consider the shelf shuﬄing measure P
m
with n cards and m shelves. Sup-
pose that m = cn
3/2
. Then, as n tends to inﬁnity with 0 < c < ﬁxed,
kP
m
Uk
e
1/(12c
2
)
1,
sep(P
m
) 1 e
1/(24c
2
)
.
Remark. We ﬁnd it surprising that this many shelves are needed. For example, when n = 52,
to make the distance less than 1/100, m
.
= 1, 085 shelves are required for kP
m
U k
and
m
.
= 764 are required for sep(P
m
).
Proof. Using Corollary 3.3, the distance is achieved at the identity permutation or a per-
mutation with b(n 1)/2c valleys. For the identity, consider n!P
m
(id). Using Theorem
3.1,
(3.5) n!P
m
(id) =
2(n!)
(2m)
n
n1
X
a=0
m + n a 1
n

n 1
a
.
To bound this sum, observe that
n1
a
/2
n1
is the binomial probability density. To keep
the bookkeeping simple, assume throughout that n is odd. The argument for even n is
similar.
For a =
n1
2
+ j, the local central limit theorem as in Feller (1968, Chap. VII.2), shows
(3.6)
n1
n1
2
+j
2
n1
e
2j
2
/n
p
πn/2
for j = o(n
2/3
).
In the following, we show further that
(3.7)
n!
m
n
m +
n1
2
j
n
e
1
24c
2
+
j
c
n
uniformly for j = o(n).
Combining (3.6), (3.7), gives a Riemann sum for the integral
e
1
24c
2
p
π/2
Z
−∞
e
2x
2
+x/c
dx = e
1/(12c
2
)
,
the claimed result. This part of the argument follows Feller (1968, Chap. VII.2) and we
suppress further details. To complete the argument the tails of the sum in (3.5) must be
bounded.
We ﬁrst prove (3.7). From the deﬁnitions
n!
m
n
m j +
n1
2
n
=
n1
2
Y
i=
n1
2
1
j
m
+
i
m
10 PERSI DIACONIS, JASON FULMAN, AND SUSAN HOLMES
using log(1 x) = x
x
2
2
+ O(x
3
),
n1
2
X
i=
n1
2
log
1
j
m
+
i
m
=
X
i
j
m
+
i
m
1
2
X
i
j
m
+
i
m
2
+ nO
n
m
3
=
nj
m
1
2
nj
2
m
2
+
1
12
n(n
2
1)
m
2
+ O
1
n
=
j
c
n
j
2
2c
2
n
2
1
24c
2
+ O
1
n
.
(3.8)
The error term in (3.8) is uniform in j. For j = o(n), j
2
/n
2
= o(1) and (3.7) follows.
To bound the tails of the sum, ﬁrst observe that (3.8) implies that
n!
m
n
mj+
n1
2
n
= e
O
(
n
)
for all j. From Bernstein’s inequality, if X
i
= ±1 with probability 1/2, P (|X
1
+···+X
n1
| >
a) 2e
a
2
/(n1)
. Using this, the sum over |j| An
3/4
is negligible for A suﬃciently large.
The Gaussian approximation to the binomial works for j n
2/3
. To bound the sum for
|j| between n
2/3
and n
3/4
, observe from (3.8) that in this range,
n!
m
n
mj+
n1
2
n
= O
e
n
1/4
.
Then, Feller (1968, p. 195) shows
n1
n1
2
+j
2
n1
1
p
πn/2
e
1
2
(j)
2
/(n/4)f
j/
n/4
with f(x) =
P
a=3
(
1
2
)
a1
+
(
1
2
)
a1
a(a1)
1
n/4
a2
x
a
= c
1
x
4
n
+ c
2
x
6
n
2
+ . . . for explicit constants
c
1
, c
2
, . . . . For θ
1
n
2/3
|j| θ
2
n
3/4
, the sum under study is dominated by A
P
jn
2/3
e
Bj
1/6
which tends to zero.
The separation distance is achieved at permutations with
n1
2
valleys (recall we are
assuming that n is odd). From (3.1),
1 n!P
m
n 1
2
= 1
n!
m
n
m +
n1
2
n
.
The result now follows from (3.7) with j = 0.
Remark. A similar argument allows asymptotic evaluation of total variation. We have not
carried out the details.
3.4. Distribution of cycle type. The number of ﬁxed points and the number of cycles
are classic descriptive statistics of a permutation. More generally, the number of i-cycles for
1 i n has been intensively studied Shepp and Lloyd (1966); Diaconis et al. (1995). This
section investigates the distribution of cycle type of a permutation w produced from a shelf
shuﬄer with m shelves and n cards. Similar results for ordinary riﬄe shuﬄes appeared in
Diaconis et al. (1995), and closely related results in the type B case (not in the language
of shelf-shuﬄing) appear in Fulman (2001, 2002). Recall also that in the case of one shelf,
the shelf shuﬄer generates one of the 2
n1
unimodal permutations uniformly at random.
The cycle structure of unimodal permutations has been studied in several papers in the
literature: see Fulman (2001, 2002); Thibon (2001) for algebraic/combinatorial approaches
and Gannon (2001); Rogers (1981) for approaches using dynamical systems.
ANALYSIS OF CASINO SHELF SHUFFLING MACHINES 11
For what follows, we deﬁne
f
i,m
=
1
2i
X
d|i
d odd
µ(d)(2m)
i/d
where µ is the oebius function of elementary number theory.
Theorem 3.5. Let P
m
(w) denote the probability that a shelf shuﬄer with m shelves produces
a permutation w. Let N
i
(w) denote the number of i-cycles of a permutation w in S
n
. Then
(3.9) 1 +
X
n1
u
n
X
wS
n
P
m
(w)
Y
i1
x
N
i
(w)
i
=
Y
i1
1 + x
i
(u/2m)
i
1 x
i
(u/2m)
i
f
i,m
.
Proof. By the proof of Theorem 3.1, a permutation produced by a shelf shuﬄer with m
shelves is equivalent to forgetting signs after the inverse of a type B riﬄe shuﬄe with 2m
piles, then conjugating by the longest element n, n 1, . . . , 1. Since a permutation and
its inverse have the same cycle type and conjugation leaves cycle type invariant, the result
follows from either (Fulman, 2001, Thm. 7) or (Fulman, 2002, Thm. 9), both of which
derived the generating function for cycle type after type B shuﬄes.
Theorem 3.5 leads to several corollaries. We say that a random variable X is Bino-
mial(n,p) if P(X = j) =
n
j
p
j
(1 p)
nj
, 0 j n, and that X is negative binomial with
parameters (f, p) if P(X = j) =
f+j1
j
p
j
(1 p)
f
, 0 j < .
Corollary 3.6. Let N
i
(w) be the number of i-cycles of a permutation w.
(1) Fix u such that 0 < u < 1. Then choose a random number N of cards so that
P(N = n) = (1 u)u
n
. Let w be produced by a shelf shuﬄer with m shelves and N
cards. Then any ﬁnite number of the random variables {N
i
} are independent, and
N
i
is distributed as the convolution of a Binomial
f
i,m
,
(u/2m)
i
1+(u/2m)
i
and a negative
binomial with parameters (f
i,m
, (u/2m)
i
).
(2) Let w be produced by a shelf shuﬄer with m shelves and n cards. Then in the n
limit, any ﬁnite number of the random variables {N
i
} are independent. The N
i
are
distributed as the convolution of a Binomial
f
i,m
,
1
(2m)
i
+1
and a negative binomial
with parameters (f
i,m
, (1/2m)
i
).
Proof. Setting all x
i
= 1 in equation (3.9) yields the equation
(3.10) (1 u)
1
=
Y
i1
1 + (u/2m)
i
1 (u/2m)
i
f
i,m
.
Taking reciprocals of equation (3.10) and multiplying by equation (3.9) gives the equality
(3.11) (1 u) +
X
n1
(1 u)u
n
X
wS
n
P
m
(w)
Y
i1
x
n
i
(w)
i
=
Y
i1
1 + x
i
(u/2m)
i
1 + (u/2m)
i
f
i,m
·
1 (u/2m)
i
1 x
i
(u/2m)
i
f
i,m
.
This proves part 1 of the theorem, the ﬁrst term on the right corresponding to the convo-
lution of binomials, and the second term to the convolution of negative binomials.
The second part follows from the claim that if a generating function f(u) has a Taylor
series which converges at u = 1, then the n limit of the coeﬃcient of u
n
in f(u)/(1
12 PERSI DIACONIS, JASON FULMAN, AND SUSAN HOLMES
u) is f (1). Indeed, write the Taylor expansion f(u) =
P
n=0
a
n
u
n
and observe that the
coeﬃcient of u
n
in f(u)/(1 u) is
P
n
i=0
a
i
. Now apply the claim to equation (3.11) with
all but ﬁnitely many x
i
equal to 1.
Remark. For example, when i = 1, f
i,m
= m; the number of ﬁxed points are distributed
as a sum of Binomial
m,
1
2m+1
and negative binomial
m,
1
2m
. Each of these converges
to Poisson(1/2) and so the number of ﬁxed points is approximately Poisson(1). A similar
analysis holds for the other cycle counts. Corollary 3.6 could also be proved by the method
of moments, along the lines of the arguments of Diaconis et al. (1995) for the case of ordinary
riﬄe shuﬄes.
For the next result, recall that the limiting distribution of the large cycles of a uniformly
chosen permutation in S
n
has been determined by Goncharov Gontcharoﬀ (1942); Gon-
charov (1944), Shepp and Lloyd (1966), Vershik and Shmidt (1977, 1978), and others. For
instance the average length of the longest cycle L
1
is approximately .63n and L
1
/n has a
known limiting distribution. The next result shows that even with a ﬁxed number of shelves,
the distribution of the large cycles approaches that of a uniform random permutation, as
long as the number of cards is growing. We omit the proof, which goes exactly along the
lines of the corresponding result for riﬄe shuﬄes in Diaconis et al. (1995).
Corollary 3.7. Fix k and let L
1
(w), L
2
(w), . . . , L
k
(w) be the lengths of the k longest cycles
of w S
n
produced by a shelf shuﬄer with m shelves. Then for m ﬁxed, or growing with n,
as n ,
|P
m
{L
1
/n t
1
, . . . , L
k
/n t
n
} P
{L
1
/n t
1
, . . . , L
k
/n t
n
}| 0
uniformly in t
1
, t
2
, . . . , t
k
.
As a ﬁnal corollary, we note that Theorems 3.1 and 3.5 give the following generating
function for the joint distribution of permutations by valleys and cycle type. Note that
this gives the joint generating function for the distribution of permutations by peaks and
cycle type, since conjugating by the permutation n, n1, . . . , 1 preserves the cycle type and
swaps valleys and peaks.
Corollary 3.8. Let v(w) denote the number of valleys of a permutation w. Then
t
1 t
+
X
n1
u
n
X
wS
n
1
2
(1 + t)
n+1
(1 t)
n+1
4t
(1 + t)
2
v(w)+1
Y
i1
x
N
i
(w)
i
=
X
m1
t
m
Y
i1
(
1 + x
i
u
i
1 x
i
u
i
)
f
i,m
.
The same result holds with v(w) replaced by p(w), the number of peaks of w.
Remark. There is a large literature on the joint distribution of permutations by cycles
and descents Gessel and Reutenauer (1993); Diaconis et al. (1995); Reiner (1993); Fulman
(2000b); Blessenohl et al. (2005); Poirier (1998) and by cycles and cyclic descents Fulman
(2001, 2002, 2000a), but Corollary 3.8 seems to be the ﬁrst result on the joint distribution
by cycles and peaks.
3.5. Distribution of RSK shape. In this section we obtain the distribution of the Rob-
inson–Schensted–Knuth (RSK) shape of a permutation w produced from a shelf shuﬄer
with m shelves and n cards. For background on the RSK algorithm, see Stanley (1999).
The RSK bijection associates to a permutation w S
n
a pair of standard Young tableaux
(P (w), Q(w)) of the same shape and size n. Q(w) is called the recording tableau of w.
To state our main result, we use a symmetric function S
λ
studied in Stembridge (1997) (a
special case of the extended Schur functions in Kerov and Vershik (1986)). One deﬁnition
ANALYSIS OF CASINO SHELF SHUFFLING MACHINES 13
of the S
λ
is as the determinant
S
λ
(y) = det(q
λ
i
i+j
)
where q
r
= 0 for r > 0 and for r 0, q
r
is deﬁned by setting
X
n0
q
n
t
n
=
Y
i1
1 + y
i
t
1 y
i
t
.
We also let f
λ
denote the number of standard Young tableaux of shape λ.
Theorem 3.9. The probability that a shelf shuﬄer with m shelves and n cards produces a
permutation with recording tableau T is equal to
1
2
n
S
λ
1
m
, . . . ,
1
m
for any T of shape λ, where S
λ
has m variables. Thus the probability that w has RSK shape
λ is equal to
f
λ
2
n
S
λ
1
m
, . . . ,
1
m
.
Proof. By the proof of Theorem 3.1, a permutation produced by a shelf shuﬄer with m
shelves is equivalent to forgetting signs after the inverse of a type B 2m-shuﬄe, and then
conjugating by the permutation n n 1 . . . 1. Since a permutation and its inverse have the
same RSK shape (Stanley, 1999, Sect. 7.13), and conjugation by n, n 1, . . . , 1 leaves the
RSK shape unchanged (Stanley, 1999, Thm. A1.2.10), the result follows from Theorem 8
of Fulman (2002), which studied RSK shape after type B riﬄe shuﬄes.
3.6. Distribution of descents. A permutation w is said to have a descent at position i
(1 i n 1) if w(i) > w(i + 1). We let d(w) denote the total number of descents of π.
For example the permutation 3 1 5 4 2 has d(w) = 3 and descent set 1, 3, 4. The purpose of
this section is to derive a generating function for the number of descents in a permutation
w produced by a shelf shuﬄer with m shelves and n cards. More precisely, we prove the
following result.
Theorem 3.10. Let P
m
(w) denote the probability that a shelf shuﬄer with m shelves and
n cards produces a permutation w. Letting [u
n
]f(u) denote the coeﬃcient of u
n
in a power
series f (u), one has that
(3.12)
X
wS
n
P
m
(w)t
d(w)+1
=
(1 t)
n+1
2
n
X
k1
t
k
[u
n
]
(1 + u/m)
km
(1 u/m)
km
.
The proof uses the result about RSK shape mentioned in Section 3.5, and symmetric
function theory; background on these topics can be found in the texts Stanley (1999) and
Macdonald (1995) respectively.
Proof. Let w be a permutation produced by a shelf shuﬄer with m shelves and n cards.
The RSK correspondence associates to w a pair of standard Young tableaux (P (w), Q(w))
of the same shape. Moreover, there is a notion of descent set for standard Young tableaux,
and by Lemma 7.23.1 of Stanley (1999), the descent set of w is equal to the descent set of
Q(w). Let f
λ
(r) denote the number of standard Young tableaux of shape λ with r descents.
Then Theorem 3.9 implies that
P(d(w) = r) =
X
|λ|=n
f
λ
(r)
2
n
S
λ
1
m
, . . . ,
1
m
.
14 PERSI DIACONIS, JASON FULMAN, AND SUSAN HOLMES
By equation 7.96 of Stanley (1999), one has that
X
r0
f
λ
(r)t
r+1
= (1 t)
n+1
X
k1
s
λ
(1, . . . , 1)t
k
where in the kth summand, s
λ
(1, . . . , 1) denotes the Schur function with k variables spe-
cialized to 1. Thus
X
r0
P(d(w) = r) · t
r+1
=
X
r0
X
|λ|=n
f
λ
(r)
2
n
S
λ
1
m
, . . . ,
1
m
· t
r+1
=
(1 t)
n+1
2
n
X
k1
t
k
X
|λ|=n
S
λ
1
m
, . . . ,
1
m
s
λ
(1, . . . , 1)
=
(1 t)
n+1
2
n
X
k1
t
k
[u
n
]
X
n0
X
|λ|=n
S
λ
1
m
, . . . ,
1
m
s
λ
(1, . . . , 1) · u
n
.
¿From Appendix A.4 of Stembridge (1997), if λ ranges over all partitions of all natural
numbers, then
X
λ
s
λ
(x)S
λ
(y) =
Y
i,j1
1 + x
i
y
j
1 x
i
y
j
.
Setting x
1
= ··· = x
k
= u and y
1
= ··· = y
m
=
1
m
completes the proof of the theorem.
For what follows we let A
n
(t) =
P
wS
n
t
d(w)+1
be the generating function of elements in
S
n
by descents. This is known as the Eulerian polynomial and from page 245 of Comtet
(1974), one has that
(3.13) A
n
(t) = (1 t)
n+1
X
k1
t
k
k
n
.
This also follows by letting m in equation (3.12).
The following corollary derives the mean and variance of the number of descents of a
permutation produced by a shelf shuﬄer.
Corollary 3.11. Let w be a permutation produced by a shelf shuﬄer with m shelves and
n 2 cards.
(1) The expected value of d(w) is
n1
2
.
(2) The variance of d(w) is
n+1
12
+
n2
6m
2
.
Proof. The ﬁrst step is to expand [u
n
]
(1+u/m)
km
(1u/m)
km
as a series in k. One calculates that
[u
n
]
(1 + u/m)
km
(1 u/m)
km
=
1
m
n
X
a0
km
a

km + n a 1
n a
=
1
m
n
X
a0
(km) . . . (km a + 1)
a!
(km + n a 1) . . . (km)
(n a)!
=
1
n!
2
n
k
n
+
2
n
n(n 1)(n 2)
12m
2
k
n2
+ . . .
ANALYSIS OF CASINO SHELF SHUFFLING MACHINES 15
where the . . . in the last equation denote terms of lower order in k. Thus Theorem 3.10
gives
X
w
P
m
(w)t
d(w)+1
=
(1 t)
n+1
n!
X
k1
t
k
k
n
+
n 2
12m
2
(1 t)
2
(1 t)
n1
(n 2)!
X
k1
t
k
k
n2
+ (1 t)
3
C(t)
where C(t) is a polynomial in t. By equation (3.13), it follows that
X
w
P
m
(w)t
d(w)+1
=
A
n
(t)
n!
+ (1 t)
2
n 2
12m
2
A
n2
(t)
(n 2)!
+ (1 t)
3
C(t).
Since the number of descents of a random permutation has mean (n 1)/2 and variance
(n + 1)/12 for n 2, it follows that
A
0
n
(1)
n!
=
(n+1)
2
and also that
A
00
n
(1)
n!
= (3n
2
+ n 2)/12.
Thus
X
w
P
m
(w)d(w) =
n 1
2
and
X
w
P
m
(w)d(w)[d(w) + 1] =
3n
2
+ n 2
12
+
n 2
6m
2
and the result follows.
Remarks.
Part 1 of Corollary 3.11 can be proved without generating functions simply by noting
that by the way the shelf shuﬄer works, w and its reversal are equally likely to be
produced.
Theorem 3.10 has an analog for ordinary riﬄe shuﬄes which is useful in the study
of carries in addition. See Diaconis and Fulman (2009a) for details.
4. Iterated shuffling
This section shows how to analyze repeated shuﬄes. Section 4.1 shows how to combine
shuﬄes. Section 4.2 gives a clean bound for the separation distance.
4.1. Combining shuﬄes. To describe what happens to various combinations of shuﬄes,
we need the notion of a signed mshuﬄe. This has the following geometric description:
divide the unit interval into sub-intervals of length
1
m
; each sub-interval contains the graph
of a straight line of slope ±m. The left-to-right pattern of signs ±s is indicated by a vector
x of length m. Thus if m = 4 and x = + + ++, an xshuﬄe is generated as shown on the
left side of Figure 2. If m = 4 and x = + +, the graph becomes that of the right side
of Figure 2. Call this function f
x
.
The shuﬄe proceeds as in the ﬁgure with n points dropped at random into the unit
interval, labeled left to right, y
1
, y
2
, . . . , y
n
and then permuted by f
x
. In each case there is
a simple forward description: the deck is cut into m piles by a multinomial distribution and
piles corresponding to negative coordinates are reversed. Finally, all packets are shuﬄed
together by the GSR procedure. Call the associated measure on permutations P
x
.
Remark. Thus, ordinary riﬄe shuﬄes are ++ shuﬄes. The shelf shuﬄe with 10 shelves is
an inverse + + ··· + (length 20) shuﬄe in this notation.
16 PERSI DIACONIS, JASON FULMAN, AND SUSAN HOLMES
0 1
1
1
4
1
2
3
4
B
B
B
B
B
B
B
B
B
B
B
0 1
1
1
4
1
2
3
4
B
B
B
B
B
B
B
B
B
B
B
Figure 2. Left: Four peaks; right: m = 4, x = + +.
The following theorem reduces repeated shuﬄes to a single shuﬄe. To state it, one piece
of notation is needed. Let x = (x
1
, x
2
, . . . , x
a
) and y = (y
1
, y
2
, . . . , y
b
) be two sequences
of ± signs. Deﬁne a sequence of length ab as x y = y
x
1
, y
x
2
, . . . , y
x
a
with (y
1
, . . . , y
b
)
1
=
(y
1
, . . . , y
b
) and (y
1
, . . . , y
b
)
1
= (y
b
, y
b1
, . . . , y
1
). This is an associative product on
strings; it is not commutative. Let P
x
be the measure induced on S
n
(forward shuﬄes).
Example.
(+ + +) (++) = + + + + ++
(+) (+) = + +
(+) (+ + +) = + + + + −−
Theorem 4.1. If x and y are ±1 sequences of length a and b, respectively, then
P
x
P
y
= P
xy
.
Proof. This follows most easily from the geometric description underlying Figure 1 and
Figure 2. If a uniformly chosen point in [0, 1] is expressed base a, the “digits” are uniform
and independently distributed in {0, 1, . . . , a 1}. Because of this, iterating the maps on
the same uniform points gives the convolution. The iterated maps have the claimed pattern
of slopes by a simple geometric argument.
Corollary 4.2. The convolution of k + shuﬄes is a + + ···+ (2
k
terms) shuﬄe.
Further, the convolution of a shelf shuﬄer with m
1
and then m
2
shelves is the same as a
shelf shuﬄer with 2m
1
m
2
shelves.
4.2. Bounds for separation distance. The following theorem gives a bound for separa-
tion (and so for total variation) for a general P
x
shuﬄe on S
n
.
Theorem 4.3. For any ±1 sequence x of length a, with P
x
the associated measure on S
n
,
and sep(P
x
) from (2.1),
(4.1) sep(P
x
) 1
n1
Y
i=1
1
i
a
.
Proof. It is easiest to argue using shuﬄes as in Description 1. There, the backs of cards are
labeled, independently and uniformly, with symbols 1, 2, . . . , a. For the inverse shuﬄe, all
cards labeled 1 are removed, keeping them in their same relative order, and placed on top
followed by the cards labeled 2 (placed under the 1s) and so on, with the following proviso:
ANALYSIS OF CASINO SHELF SHUFFLING MACHINES 17
if the ith coordinate of x is 1, the cards labeled i have their order reversed; so if they are
1, 5, 17 from top down, they are placed in order 17, 5, 1. All of this results in a single
permutation drawn from P
x
. Repeated shuﬄes are modeled by labeling each card with a
vector of symbols. The kth shuﬄe is determined by the kth coordinate of this vector. The
ﬁrst time t that the ﬁrst k coordinates of those n vectors are all distinct forms a strong
stationary time. See Aldous and Diaconis (1986) or Fulman (1998) for further details. The
usual bound for separation yields
sep(P
x
) P {all n labels are distinct}.
The bound (4.1) now follows from the classical birthday problem.
Remarks.
For a large with respect to n, the right side is well-approximated by 1 e
n(n1)
2a
.
This is small when n
2
a.
The theorem gives a clean upper bound on the distance to uniformity. For example,
when n = 52, after 8 ordinary riﬄe shuﬄes (so x = + + ··· + +, length 256), the
bound (4.1) is sep(P
x
) 0.997, in agreement with Table 1 of Assaf et al. (2011).
For the actual shelf shuﬄe with x = + + ··· + (length 20), the bound gives
sep(P
x
) = 1 but sep(P
x
P
x
) 0.969 and sep(P
x
P
x
P
x
) 0.153.
The bound in Theorem 4.3 is simple and general. However, it is not sharp for the
original shelf shuﬄer. The results of Section 3.3 show that m = cn
3/2
shelves suﬃce
to make sep(P
m
) small when c is large. Theorem 4.3 shows that m = cn
2
steps
suﬃce.
5. Practical tests and conclusions
The engineers and executives who consulted us found it hard to understand the total
variation distance. They asked for more down-to-earth notions of discrepancy. This section
reports some ad hoc tests which convinced them that the machine had to be used diﬀerently.
Section 5.1 describes the number of cards guessed correctly. Section 5.2 brieﬂy describes
three other tests. Section 5.3 describes conclusions and recommendations.
5.1. Card guessing with feedback. Suppose, after a shuﬄe, cards are dealt face-up, one
at a time, onto the table. Before each card is shown, a guess is made at the value of the
card. Let X
i
, 1 i n, be one or zero as the ith guess is correct and T
n
= X
1
+ ··· + X
n
the total number of correct guesses. If the cards were perfectly mixed, the chance that
X
1
= 1 is 1/n, the chance that X
2
= 1 is 1/(n 1), . . . , that X
i
= 1 is 1/(n i + 1).
Further, the X
i
are independent. Thus elementary arguments give the following.
Proposition 5.1. Under the uniform distribution, the number of cards guessed correctly
T
n
satisﬁes
E(T
n
) =
1
n
+
1
n1
+ ··· + 1 log n + γ + O
1
n
with γ
.
= 0.577 Euler’s constant.
var(T
n
) =
1
n
1
1
n
+
1
n1
1
1
n1
+ ··· +
1
2
1
1
2
log n + γ
π
2
6
+ O
1
n
.
Normalized by its mean and variance, T
n
has an approximate normal distribution.
When n = 52, T
n
has mean approximately 4.5, standard deviation approximately
2.9,
and the number of correct guesses is between 2.7 and 6.3, 70% of the time.
Based on the theory developed in Section 3 we constructed a guessing strategy con-
jectured to be optimal for use after a shelf shuﬄe.
Strategy
To begin, guess card 1.
18 PERSI DIACONIS, JASON FULMAN, AND SUSAN HOLMES
If guess is correct, remove card 1 from the list of available cards. Then guess card
2, card 3, . . . .
If guess is incorrect and card i is shown, remove card i from the list of available
cards and guess card i + 1, card i + 2, . . . .
Continue until a descent is observed (order reversal with the value of the current
card smaller than the value of the previously seen card). Then change the guessing
strategy to guess the next-smallest available card.
Continue until an ascent is observed, then guess the next-largest available card, and
so on.
Table 2. Mean and variance for n = 52 after a shelf shuﬄe with m shelves under
the conjectured optimal strategy.
m 1 2 4 10 20 64
mean 39 27 17.6 9.3 6.2 4.7
variance 3.2 5.6 6.0 4.7 3.8 3.1
A Monte Carlo experiment was run to determine the distribution of T
n
for n = 52 with
various values of m (10,000 runs for each value). Table 2 shows the mean and variance for
various numbers of shelves. Thus for the actual shuﬄer, m = 10 gives about 9.3 correct
guesses versus 4.5 for a well-shuﬄed deck. A closely related study of optimal strategy for
the GSR measure (without feedback) is carried out by Ciucu (1998).
5.2. Three other tests. For the shelf shuﬄer with m shelves, an easy argument shows
that the chance that the original top card is still on top is at least 1/2m instead of 1/n.
When n = 52, this is 1/20 versus 1/52. The chance that card 2 is on top is approximately
1
2m
1
1
2m
while the chance that card 2 is second from the top is roughly
1
(2m)
2
. The
same probabilities hold for the bottom cards. While not as striking as the guessing test of
Section 5.1, this still suggests that the machine is “oﬀ.”
Our second test supposed that the deck was originally arranged with all the red cards on
top and all the black cards at the bottom. The test statistic is the number of changes of
color going through the shuﬄed deck. Under uniformity, simulations show this has mean
26 and standard deviation 3.6. With a 10-shelf machine, simulations showed 17 ± 1.83, a
noticeable deviation.
The third test is based on the spacings between cards originally near the top of the
deck. Let w
j
denote the position of the card originally at position j from the top. Let
D
j
= |w
j
w
j+1
|. Figure 3 shows a histogram of D
j
for 1 j 9, from a simulation with
n = 52 based on a 10-shelf shuﬄer. Figure 4 shows histograms for the same statistics for a
well-shuﬄed deck; there are striking discrepancies.
5.3. Conclusions and recommendations. The study above shows that a single iteration
of a 10-shelf shuﬄer is not suﬃciently random. The president of the company responded
“We are not pleased with your conclusions, but we believe them and that’s what we hired
you for.”
We suggested a simple alternative: use the machine twice. This results in a shuﬄe
equivalent to a 200-shelf machine. Our mathematical analysis and further tests, not reported
here, show that this is adequately random. Indeed, Table 1 shows, for total variation, this
is equivalent to 8-to-9 ordinary riﬄe shuﬄes.
ANALYSIS OF CASINO SHELF SHUFFLING MACHINES 19
Figure 3. 9 spacings from a 10-shelf shuﬄe; j varies from top left to bottom right,
1 j 9.
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Departments of Mathematics and Statistics, Stanford University