Card Shuffling
Much of this section is based upon an article by Brad Mann,[Notes 1] which is an exposition of an article by David Bayer and Persi Diaconis.[Notes 2]
Riffle Shuffles
Given a deck of [math]n[/math] cards, how many times must we shuffle it to make it “random”? Of course, the answer depends upon the method of shuffling which is used and what we mean by “random.” We shall begin the study of this question by considering a standard model for the riffle shuffle.
We begin with a deck of [math]n[/math] cards, which we will assume are labelled in
increasing order with the integers from 1 to [math]n[/math]. A riffle shuffle consists of a cut
of the deck into two stacks and an interleaving of the two
stacks. For example, if [math]n = 6[/math], the initial ordering is
[math](1, 2, 3, 4, 5, 6)[/math], and a cut might occur between cards 2 and 3. This gives rise to
two stacks, namely [math](1, 2)[/math] and [math](3, 4, 5, 6)[/math]. These are interleaved to form a new
ordering of the deck. For example, these two stacks might form the ordering [math](1, 3,
4, 2, 5, 6)[/math]. In order to discuss such shuffles, we need to assign a probability
distribution to the set of all possible shuffles. There are several reasonable ways in
which this can be done. We will give several different assignment strategies, and
show that they are equivalent. (This does not mean that this assignment is the only
reasonable one.) First, we assign the binomial probability [math]b(n, 1/2, k)[/math] to the
event that the cut occurs after the [math]k[/math]th card. Next, we assume that all possible
interleavings, given a cut, are equally likely. Thus, to complete the assignment of
probabilities, we need to determine the number of possible interleavings of two stacks
of cards, with [math]k[/math] and [math]n-k[/math] cards, respectively.
We begin by writing the second stack in a line, with spaces in between each pair
of consecutive cards, and with spaces at the beginning and end (so there are [math]n-k+1[/math]
spaces). We choose, with replacement, [math]k[/math] of these spaces, and place the cards from
the first stack in the chosen spaces. This can be done in
ways. Thus, the probability of a given interleaving should be
Next, we note that if the new ordering is not the identity ordering, it is the
result of a unique cut-interleaving pair. If the new ordering is the identity, it is
the result of any one of [math]n+1[/math] cut-interleaving pairs.
We define a rising sequence in an ordering to be a maximal
subsequence of consecutive integers in increasing order. For example, in the ordering
there are 4 rising sequences; they are [math](1)[/math], [math](2, 3, 4)[/math], [math](5, 6)[/math], and [math](7)[/math]. It is easy to see that an ordering is the result of a riffle shuffle applied to the identity ordering if and only if it has no more than two rising sequences. (If the ordering has two rising sequences, then these rising sequences correspond to the two stacks induced by the cut, and if the ordering has one rising sequence, then it is the identity ordering.) Thus, the sample space of orderings obtained by applying a riffle shuffle to the identity ordering is naturally described as the set of all orderings with at most two rising sequences.
It is now easy to assign a probability distribution to this sample space. Each
ordering with two rising sequences is assigned the value
and the identity ordering is assigned the value
There is another way to view a riffle shuffle. We can imagine starting with a
deck cut into two stacks as before, with the same probabilities assignment as before
i.e., the binomial distribution. Once we have the two stacks, we take cards, one by
one, off of the bottom of the two stacks, and place them onto one stack. If there are
[math]k_1[/math] and [math]k_2[/math] cards, respectively, in the two stacks at some point in this process,
then we make the assumption that the probabilities that the next card to be taken comes
from a given stack is proportional to the current stack size. This implies that the
probability that we take the next card from the first stack equals
and the corresponding probability for the second stack is
We shall now show that this process assigns the uniform probability to each of the possible interleavings of the two stacks.
Suppose, for example, that an interleaving came about as the result of choosing
cards from the two stacks in some order. The probability that this result occurred is
the product of the probabilities at each point in the process, since the choice of
card at each point is assumed to be independent of the previous choices. Each factor
of this product is of the form
where [math]i = 1[/math] or [math]2[/math], and the denominator of each factor equals the number of cards left to be chosen. Thus, the denominator of the probability is just [math]n![/math]. At the moment when a card is chosen from a stack that has [math]i[/math] cards in it, the numerator of the corresponding factor in the probability is [math]i[/math], and the number of cards in this stack decreases by 1. Thus, the numerator is seen to be [math]k!(n-k)![/math], since all cards in both stacks are eventually chosen. Therefore, this process assigns the probability
to each possible interleaving.
We now turn to the question of what happens when we riffle shuffle [math]s[/math] times. It
should be clear that if we start with the identity ordering, we obtain an ordering
with at most [math]2^s[/math] rising sequences, since a riffle shuffle creates at most two rising
sequences from every rising sequence in the starting ordering. In fact, it is not
hard to see that each such ordering is the result of [math]s[/math] riffle shuffles. The
question becomes, then, in how many ways can an ordering with [math]r[/math] rising sequences
come about by applying [math]s[/math] riffle shuffles to the identity ordering? In order to
answer this question, we turn to the idea of an [math]a[/math]-shuffle.
[math]a[/math]-Shuffles
There are several ways to visualize an [math]a[/math]-shuffle. One way is to imagine a creature with [math]a[/math] hands who is given a deck of cards to riffle shuffle. The creature naturally cuts the deck into [math]a[/math] stacks, and then riffles them together. (Imagine that!) Thus, the ordinary riffle shuffle is a 2-shuffle. As in the case of the ordinary 2-shuffle, we allow some of the stacks to have 0 cards. Another way to visualize an [math]a[/math]-shuffle is to think about its inverse, called an [math]a[/math]-unshuffle. This idea is described in the proof of the next theorem.
We will now show that an [math]a[/math]-shuffle followed by a [math]b[/math]-shuffle is equivalent to an
[math]ab[/math]-shuffle. This means, in particular, that [math]s[/math] riffle shuffles in succession are
equivalent to one
[math]2^s[/math]-shuffle. This equivalence is made precise by the following theorem.
Let [math]a[/math] and [math]b[/math] be two positive integers. Let [math]S_{a,b}[/math] be the set of all ordered pairs in which the first entry is an [math]a[/math]-shuffle and the second entry is a [math]b[/math]-shuffle. Let [math]S_{ab}[/math] be the set of all [math]ab[/math]-shuffles. Then there is a 1-1 correspondence between [math]S_{a,b}[/math] and [math]S_{ab}[/math] with the following property. Suppose that [math](T_1, T_2)[/math] corresponds to [math]T_3[/math]. If [math]T_1[/math] is applied to the identity ordering, and [math]T_2[/math] is applied to the resulting ordering, then the final ordering is the same as the ordering that is obtained by applying [math]T_3[/math] to the identity ordering.
Show ProofThe easiest way to describe the required correspondence is through the idea of an unshuffle. An [math]a[/math]-unshuffle begins with a deck of [math]n[/math] cards. One by one, cards are taken from the top of the deck and placed, with equal probability, on the bottom of any one of [math]a[/math] stacks, where the stacks are labelled from 0 to [math]a-1[/math]. After all of the cards have been distributed, we combine the stacks to form one stack by placing stack [math]i[/math] on top of stack [math]i+1[/math], for [math]0 \le i \le a-1[/math]. It is easy to see that if one starts with a deck, there is exactly one way to cut the deck to obtain the [math]a[/math] stacks generated by the [math]a[/math]-unshuffle, and with these [math]a[/math] stacks, there is exactly one way to interleave them to obtain the deck in the order that it was in before the unshuffle was performed. Thus, this [math]a[/math]-unshuffle corresponds to a unique [math]a[/math]-shuffle, and this [math]a[/math]-shuffle is the inverse of the original [math]a[/math]-unshuffle.
If we apply an [math]ab[/math]-unshuffle [math]U_3[/math] to a deck, we obtain a set of [math]ab[/math] stacks,
which are then combined, in order, to form one stack. We label these stacks with
ordered pairs of integers, where the first coordinate is between 0 and [math]a-1[/math], and the
second coordinate is between 0 and [math]b-1[/math]. Then we label each card with the label of
its stack. The number of possible labels is [math]ab[/math], as required. Using this labelling,
we can describe how to find a [math]b[/math]-unshuffle and an
[math]a[/math]-unshuffle, such that if these two unshuffles are applied in this order to the
deck, we obtain the same set of [math]ab[/math] stacks as were obtained by the [math]ab[/math]-unshuffle.
To obtain the [math]b[/math]-unshuffle [math]U_2[/math], we sort the deck into [math]b[/math] stacks, with the
[math]i[/math]th stack containing all of the cards with second coordinate [math]i[/math], for [math]0 \le i \le
b-1[/math]. Then these stacks are combined to form one stack. The [math]a[/math]-unshuffle [math]U_1[/math]
proceeds in the same manner, except that the first coordinates of the labels are
used. The resulting [math]a[/math] stacks are then combined to form one stack.
The above description shows that the cards ending up on top are all those
labelled [math](0, 0)[/math]. These are followed by those labelled [math](0, 1),\ (0, 2),[/math] [math]\
\ldots,\ (0, b - 1),\ (1, 0),\ (1,1),\ldots,\ (a-1, b-1)[/math]. Furthermore, the relative order of any
pair of cards with the same labels is never altered. But this is exactly the same as an
[math]ab[/math]-unshuffle, if, at the beginning of such an unshuffle, we label each of the cards with one of
the labels [math](0, 0),\ (0, 1),\ \ldots,\ (0, b-1),\ (1, 0),\ (1,1),\ \ldots,\ (a-1, b-1)[/math]. This
completes the proof.
In Figure, we show the labels for a 2-unshuffle of a deck with 10 cards. There
are 4 cards with the label 0 and 6 cards with the label 1, so if the 2-unshuffle is performed,
the first stack will have 4 cards and the second stack will have 6 cards. When this unshuffle is
performed, the deck ends up in the identity ordering.
In Figure, we show the labels for a 4-unshuffle of the same deck (because there
are four labels being used). This figure can also be regarded as an example of a pair of
2-unshuffles, as described in the proof above. The first 2-unshuffle will use the second coordinate
of the labels to determine the stacks. In this case, the two stacks contain the cards whose values
are
After this 2-unshuffle has been performed, the deck is in the order shown in Figure, as the reader should check. If we wish to perform a 4-unshuffle on the deck, using the labels shown, we sort the cards lexicographically, obtaining the four stacks
When these stacks are combined, we once again obtain the identity ordering of the deck. The point of the above theorem is that both sorting procedures always lead to the same initial ordering.
If [math]D[/math] is any ordering that is the result of applying an [math]a[/math]-shuffle and then a [math]b[/math]-shuffle to the identity ordering, then the probability assigned to [math]D[/math] by this pair of operations is the same as the probability assigned to [math]D[/math] by the process of applying an [math]ab[/math]-shuffle to the identity ordering.
Show ProofCall the sample space of [math]a[/math]-shuffles [math]S_a[/math]. If we label the stacks by the integers from [math]0[/math] to [math]a-1[/math], then each cut-interleaving pair, i.e., shuffle, corresponds to exactly one [math]n[/math]-digit base [math]a[/math] integer, where the [math]i[/math]th digit in the integer is the stack of which the [math]i[/math]th card is a member. Thus, the number of cut-interleaving pairs is equal to the number of [math]n[/math]-digit base [math]a[/math] integers, which is [math]a^n[/math]. Of course, not all of these pairs leads to different orderings. The number of pairs leading to a given ordering will be discussed later. For our purposes it is enough to point out that it is the cut-interleaving pairs that determine the probability assignment.
The previous theorem shows that there is a 1-1 correspondence between [math]S_{a,b}[/math]
and [math]S_{ab}[/math]. Furthermore, corresponding elements give the same ordering when applied to
the identity ordering. Given any ordering [math]D[/math], let [math]m_1[/math] be the number of elements of
[math]S_{a,b}[/math] which, when applied to the identity ordering, result in [math]D[/math]. Let [math]m_2[/math] be
the number of elements of [math]S_{ab}[/math] which, when applied to the identity ordering,
result in [math]D[/math]. The previous theorem implies that [math]m_1 = m_2[/math]. Thus, both sets assign
the probability
Connection with the Birthday Problem
There is another point that can be made concerning the labels given to the cards by the successive unshuffles. Suppose that we 2-unshuffle an [math]n[/math]-card deck until the labels on the cards are all different. It is easy to see that this process produces each permutation with the same probability, i.e., this is a random process. To see this, note that if the labels become distinct on the [math]s[/math]th 2-unshuffle, then one can think of this sequence of 2-unshuffles as one [math]2^s[/math]-unshuffle, in which all of the stacks determined by the unshuffle have at most one card in them (remember, the stacks correspond to the labels). If each stack has at most one card in it, then given any two cards in the deck, it is equally likely that the first card has a lower or a higher label than the second card. Thus, each possible ordering is equally likely to result from this [math]2^s[/math]-unshuffle.
Let [math]T[/math] be the random variable that counts the number of 2-unshuffles until all labels are
distinct. One can think of [math]T[/math] as giving a measure of how long it takes in the unshuffling process
until randomness is reached. Since shuffling and unshuffling are inverse processes, [math]T[/math] also
measures the number of shuffles necessary to achieve randomness. Suppose that we have an [math]n[/math]-card
deck, and we ask for [math]P(T \le s)[/math]. This equals [math]1 - P(T \gt s)[/math]. But [math]T \gt s[/math] if and only if it is
the case that not all of the labels after [math]s[/math] 2-unshuffles are distinct. This is just the birthday
problem; we are asking for the probability that at least two people have the same birthday, given
that we have [math]n[/math] people and there are [math]2^s[/math] possible birthdays. Using our formula from
Example, we find that
In Chapter Expected Value and Variance, we will define the average value of a random variable. Using this idea, and
the above equation, one can calculate the average value of the random variable [math]T[/math] (see Exercise). For example, if
[math]n = 52[/math], then the average value of [math]T[/math] is about 11.7. This means that, on the average, about 12
riffle shuffles are needed for the process to be considered random.
Cut-Interleaving Pairs and Orderings
As was noted in the proof of Theorem, not all of the cut-interleaving pairs lead to different orderings. However, there is an easy formula which gives the number of such pairs that lead to a given ordering.
If an ordering of length [math]n[/math] has [math]r[/math] rising sequences, then the number of cut-interleaving pairs under an [math]a[/math]-shuffle of the identity ordering which lead to the ordering is
To see why this is true, we need to count the number of ways in which the cut in an [math]a[/math]-shuffle can be performed which will lead to a given ordering with [math]r[/math] rising sequences. We can disregard the interleavings, since once a cut has been made, at most one interleaving will lead to a given ordering. Since the given ordering has [math]r[/math] rising sequences, [math]r-1[/math] of the division points in the cut are determined. The remaining [math]a - 1 - (r - 1) = a - r[/math] division points can be placed anywhere. The number of places to put these remaining division points is [math]n+1[/math] (which is the number of spaces between the consecutive pairs of cards, including the positions at the beginning and the end of the deck). These places are chosen with repetition allowed, so the number of ways to make these choices is
In particular, this means that if [math]D[/math] is an ordering that is the result of
applying an
[math]a[/math]-shuffle to the identity ordering, and if [math]D[/math] has [math]r[/math] rising sequences, then the
probability assigned to [math]D[/math] by this process is
The above theorem shows that the essential information about the
probability assigned to an ordering under an [math]a[/math]-shuffle is just the number of rising
sequences in the ordering. Thus, if we determine the number of orderings which
contain exactly [math]r[/math] rising sequences, for each [math]r[/math] between 1 and [math]n[/math], then we will
have determined the distribution function of the random variable which consists of
applying a random [math]a[/math]-shuffle to the identity ordering.
The number of orderings of [math]\{1, 2, \ldots, n\}[/math] with [math]r[/math] rising sequences is
denoted by [math]A(n, r)[/math], and is called an Eulerian number. There are
many ways to calculate the values of these numbers; the following theorem gives one recursive
method which follows immediately from what we already know about [math]a[/math]-shuffles.
Let [math]a[/math] and [math]n[/math] be positive integers. Then
The second equation can be used to calculate the values of the Eulerian numbers, and follows immediately from the Equation. The last equation is a consequence of the fact that the only ordering of [math]\{1, 2, \ldots, n\}[/math] with one rising sequence is the identity ordering. Thus, it remains to prove Equation. We will count the set of [math]a[/math]-shuffles of a deck with [math]n[/math] cards in two ways. First, we know that there are [math]a^n[/math] such shuffles (this was noted in the proof of Theorem). But there are [math]A(n, r)[/math] orderings of [math]\{1, 2, \ldots, n\}[/math] with [math]r[/math] rising sequences, and Theorem states that for each such ordering, there are exactly
cut-interleaving pairs that lead to the ordering. Therefore, the right-hand side of Equation counts the set of [math]a[/math]-shuffles of an [math]n[/math]-card deck. This completes the\linebreak[4] proof.
Random Orderings and Random Processes
We now turn to the second question that was asked at the beginning of this section: What do we mean by a “random” ordering? It is somewhat misleading to think about a given ordering as being random or not random. If we want to choose a random ordering from the set of all orderings of [math]\{1, 2, \ldots, n\}[/math], we mean that we want every ordering to be chosen with the same probability, i.e., any ordering is as “random” as any other.
The word “random” should really be used to describe a process. We will say that
a process that produces an object from a (finite) set of objects is a random
process if each object in the set is produced with the
same probability by the process. In the present situation, the objects are the orderings, and the
process which produces these objects is the shuffling process. It is easy to see that no [math]a[/math]-shuffle
is really a random process, since if [math]T_1[/math] and [math]T_2[/math] are two orderings with a different
number of rising sequences, then they are produced by an [math]a[/math]-shuffle, applied to the
identity ordering, with different probabilities.
Variation Distance
Instead of requiring that a sequence of shuffles yield a process which is random, we will define a measure that describes how far away a given process is from a random process. Let [math]X[/math] be any process which produces an ordering of [math]\{1, 2, \ldots, n\}[/math]. Define [math]f_X(\pi)[/math] be the probability that [math]X[/math] produces the ordering [math]\pi[/math]. (Thus, [math]X[/math] can be thought of as a random variable with distribution function [math]f[/math].) Let [math]\Omega_n[/math] be the set of all orderings of [math]\{1, 2, \ldots, n\}[/math]. Finally, let [math]u(\pi) = 1/|\Omega_n|[/math] for all [math]\pi \in \Omega_n[/math]. The function [math]u[/math] is the distribution function of a process which produces orderings and which is random. For each ordering [math]\pi \in \Omega_n[/math], the quantity
is the difference between the actual and desired probabilities that [math]X[/math] produces [math]\pi[/math]. If we sum this over all orderings [math]\pi[/math] and call this sum [math]S[/math], we see that [math]S = 0[/math] if and only if [math]X[/math] is random, and otherwise [math]S[/math] is positive. It is easy to show that the maximum value of [math]S[/math] is 2, so we will multiply the sum by [math]1/2[/math] so that the value falls in the interval [math][0, 1][/math]. Thus, we obtain the following sum as the formula for the variation distance between the two processes:
Now we apply this idea to the case of shuffling. We let [math]X[/math] be the process of
[math]s[/math] successive riffle shuffles applied to the identity ordering. We know that it is
also possible to think of [math]X[/math] as one [math]2^s[/math]-shuffle. We also know that [math]f_X[/math] is
constant on the set of all orderings with [math]r[/math] rising sequences, where [math]r[/math] is any
positive integer. Finally, we know the value of [math]f_X[/math] on an ordering with [math]r[/math] rising
sequences, and we know how many such orderings there are. Thus, in this specific
case, we have
Since this sum has only [math]n[/math] summands, it is easy to compute this for moderate sized values of [math]n[/math]. For [math]n = 52[/math], we obtain the list of values given in Table.
| Number of Riffle Shuffles | Variation Distance |
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 0.9999995334 |
| 5 | 0.9237329294 |
| 6 | 0.6135495966 |
| 7 | 0.3340609995 |
| 8 | 0.1671586419 |
| 9 | 0.0854201934 |
| 10 | 0.0429455489 |
| 11 | 0.0215023760 |
| 12 | 0.0107548935 |
| 13 | 0.0053779101 |
| 14 | 0.0026890130 |
To help in understanding these data, they are shown in graphical form in Figure. The program VariationList produces the data shown in both Table and Figure. One sees that until 5 shuffles have occurred, the output of [math]X[/math] is very far from random. After 5 shuffles, the distance from the random process is essentially halved each time a shuffle occurs.
Given the distribution functions [math]f_X(\pi)[/math] and [math]u(\pi)[/math] as above, there is
another way to view the variation distance [math]\parallel f_X - u\parallel[/math]. Given any
event [math]T[/math] (which is a subset of
[math]S_n[/math]), we can calculate its probability under the process [math]X[/math] and under the uniform
process. For example, we can imagine that [math]T[/math] represents the set of all permutations
in which the first player in a 7-player poker game is dealt a straight flush (five
consecutive cards in the same suit). It is interesting to consider how much the
probability of this event after a certain number of shuffles differs from the
probability of this event if all permutations are equally likely. This difference can
be thought of as describing how close the process [math]X[/math] is to the random process with
respect to the event [math]T[/math].
Now consider the event [math]T[/math] such that the absolute value of the difference between
these two probabilities is as large as possible. It can be shown that this absolute
value is the variation distance between the process [math]X[/math] and the uniform process. (The
reader is asked to prove this fact in Exercise.)
We have just seen that, for a deck of 52 cards, the variation distance between
the 7-riffle shuffle process and the random process is about [math].334[/math]. It is of
interest to find an event [math]T[/math] such that the difference between the probabilities
that the two processes produce [math]T[/math] is close to
[math].334[/math]. An event with this property can be described in terms of the game called
New-Age Solitaire.
New-Age Solitaire
This game was invented by Peter Doyle. It is played with a standard 52-card deck. We deal the cards face up, one at a time, onto a discard pile. If an ace is encountered, say the ace of Hearts, we use it to start a Heart pile. Each suit pile must be built up in order, from ace to king, using only subsequently dealt cards. Once we have dealt all of the cards, we pick up the discard pile and continue. We define the Yin suits to be Hearts and Clubs, and the Yang suits to be Diamonds and Spades. The game ends when either both Yin suit piles have been completed, or both Yang suit piles have been completed. It is clear that if the ordering of the deck is produced by the random process, then the probability that the Yin suit piles are completed first is exactly 1/2.
Now suppose that we buy a new deck of cards, break the seal on the package, and
riffle shuffle the deck 7 times. If one tries this, one finds that the Yin suits win
about 75% of the time. This is 25% more than we would get if the deck were in truly
random order. This deviation is reasonably close to the theoretical maximum of
[math]33.4[/math]% obtained above.
Why do the Yin suits win so often? In a brand new deck of cards, the suits are
in the following order, from top to bottom: ace through king of Hearts, ace through
king of Clubs, king through ace of Diamonds, and king through ace of Spades. Note
that if the cards were not shuffled at all, then the Yin suit piles would be completed
on the first pass, before any Yang suit cards are even seen. If we were to continue
playing the game until the Yang suit piles are completed, it would take 13 passes
through the deck to do this. Thus, one can see that in a new deck, the Yin suits are
in the most advantageous order and the Yang suits are in the least advantageous
order. Under 7 riffle shuffles, the relative advantage of the Yin suits over the Yang
suits is preserved to a certain extent.
General references
Doyle, Peter G. (2006). "Grinstead and Snell's Introduction to Probability" (PDF). Retrieved June 6, 2024.