bjstrat letters

Overview - possible shuffles in the game of blackjack

In determining the possible ways a deck of cards can be dealt, there are 2 terms that we need to familiarize ourselves with: permutations, which take into consideration the order in which the cards can be arranged and combinations, which count any equivalent permutations only once. In order to illustrate the difference, let's create a very simple deck. Let this deck consist of only 3 cards - the K of spades, the Q of hearts, and the ace of diamonds. The number of permutations of this deck is equal to 6. How is this figure computed? We want to know all of unique orders the deck can be arranged in. If we randomly deal 1 card, there will be 3 possiblities, with 2 cards left. Therefore, for the second card, we will have 2 remaining possiblities. After dealing the second card, there will be only 1 remaining possibility. The number of permutations of a 3 card deck = (1st card possibilities times 2nd card possiblities times 3rd card possiblities) = 3*2*1 = 6. 3*2*1 is also known in mathematics as 3 factorial and is denoted 3!. In general for any number, n, the factorial of n = n*(n-1)*(n-2)*.......*1 and is denoted n!. For a deck containing n cards, the number of permutations is n!. For our sample 3 card deck the permutations are -

     K of spades, Q of hearts, ace of diamonds
     Q of hearts, K of spades, ace of diamonds
     K of spades, ace of diamonds, Q of hearts
     Q of hearts, ace of diamonds, K of spades
     ace of diamonds, K of spades, Q of hearts
     ace of diamonds, Q of hearts, K of spades

Now in blackjack, any picture card or ten are equivalent and known as ten value cards. Similarly any 2 cards of the same rank of 1 to 9 are equivalent, regardless of suit. Therefore, in our simple sample deck the K of spades is equivalent to the Q of hearts when it comes to evaluating our blackjack hand. Let's now just call any ten value card T. Any other card is simply known as its rank. K of spades is now known as T. Q of hearts is also known as T. Ace of diamonds is known as 1. We can now rewrite our permutations as -


As can be seen, there are 3 duplicates. When we count the number of equivalent permutations only once, we arrive at the number of combinations, which = 3 -


In order to show in general how to mathematically compute combinations, let's define r1,r2,r3,r4,r5,r6,r7,r8,r9,r10 as the number of cards of each rank, 1 to 10 respectively, present in an n-card deck. What needs to be done is to count the equivalent permutations of each rank only once, so we need to divide overall permutations by the permutations of each rank.

n = r1+r2+r3+r4+r5+r6+r7+r8+r9+r10 = total number cards in deck
and combinations = n!/r1!/r2!/r3!/r4!/r5!/r6!/r7!/r8!/r9!r10!

In other words, combinations = overall permutations / (product of permutations of each individual rank)

For our simple sample deck, there are only 2 ranks, which are tens and aces. The sample deck consists of 3 cards, 2 tens (r10 = 2) and 1 ace (r1 = 1). (r2 through r9 = 0 for the sample deck.) (Note: 0! = 1). Number of cards in the deck = n = (r10 + r1) = (2 + 1) = 3.

Then sample deck combinations = n!/r10!/r1! = 3!/2!/1! = 3, which agrees with what was enumerated above.

Permutations and combinations for a deck consisting of a larger number of cards can be computed similarly to the above.

Calculations for Permutations and Combinations for 1, 2, 4, 6 and 8 Decks

1 Deck
Permutations = 52! = 80658175170943878571660636856403766975289505440883277824000000000000
Combinations = 52! / 16! / (4!)9 = 1459243014608741105832394919396033250000000

2 Decks
Permutations = 104! = 1029901674514562762384858386476504428305377245499907218232549177688
Combinations = 104! / 32! / (8!)9 = 13897583953886915916997900322245344374849047779518415

4 Decks
Permutations = 208! = 2411110054505276003287178952912926704433581813332627467813915186196
Combinations = 208! / 64! / (16!)9 = 2472998157208866354450720400155283825622865530393258

6 Decks
Permutations = 312! = 2102026605126378359351726442119571336270439087200001675367568427286
Combinations = 312! / 96! / (24!)9 = 1555666597724897046676164210446162756169596852699926

8 Decks
Permutations = 416! = 3846313387719957490284353898010643560199216596359526228022425377205
Combinations = 416! / 128! / (32!)9 = 164944883791287598411943116655903865118824068362983

What can this information be used for?

The number of equivalent ways a deck can be shuffled for a game of blackjack is given by the Combinations figure in the above. As you can see, there are an awful lot of possibilities. One use of this information is to give some context to a blackjack computer simulation. A billion round sim simulates far less than a billion shuffles and even a billion shuffles is a very small portion of the possible shuffles. Therefore, we may ask, "What is the statistical significance of a computer sim?" or "What does a specific computer simulation program do to ensure that it produces statistically significant results given the number of possibilities that exists?" It can be noted that each shuffle resets the shoe to its full state so the starting point for each shoe that is simmed is repeated. At the start of a freshly shuffled shoe there is just one possible shoe consistency. After the first round is dealt from a randomly shuffled shoe there are relatively few combinations possible for the dealt cards and therefore relatively few consistencies for the cards remaining to be dealt. As more cards are dealt more combinations become possible. The maximum number of possible combinations for dealt/undealt cards occurs exactly at mid-shoe so that is where the maximum number of possible shoe consistencies occurs, after which the number of possible combinations starts to decrease. When there are only a few cards remaining to be dealt there are relatively few possible shoe consistencies. Therefore the beginning and end of a shoe would be the portions of a shoe that would be less variable relative to shoe composition than the middle of the shoe in a simulation. This would be true for any number of decks and would be more pronounced as number of decks increases. It can also be noted that any given shoe consistency (shoe composition) could be computed exactly by combinatorial analysis. This would require more computing time (for each trial) than simulating by employing a "brute force" number of trials but would require far less trials. The smaller the number of possible shoe consistencies is, the more accurate the result of simulating in conjunction with combinatorial analysis would be. At the beginning of a shoe there is exactly one shoe composition so in that case combinatorial analysis is 100% accurate. After 1 round is simmed there are more than 1 shoe compositions possible but there still aren't that many so a relatively small number of trials would be highly accurate, (although not 100% as in the case of a full shoe.) As more rounds are simmed accuracy decreases but at least a trend may be evident from the more accurate earlier rounds. Also, another consideration is that as number of cards dwindles, there is a possiblity of running out of cards.

Beyond that, it seems that the number of possible shuffles is just something to know and keep in mind.


Copyright 2010 (
All rights reserved.