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Best of Donald Catlin  # How Many Decks?

3 July 2009

The title of this article is the question every blackjack player needs to ask. Why? Well, if you study blackjack strategy literature you will find that, everything else being equal, the more decks that are in play the higher the house edge. Of course, it generally isn't that simple. Expected return in multiple-deck games can be improved by introducing other rules that favor the player and single-deck games can be made worse by introducing rules that go against the player. But if two games have exactly the same sets of rules but one is a single-deck and one is an eight-deck game, the eight-deck game is the worst choice of the two. In this article I want to explain why.

On any given deal the dealer is as likely as the player to be dealt a blackjack. However, since the player's BJ gets paid 3:2 and the dealer only gets even money this represents an advantage to the player. Thus the higher the probability of a BJ occurring the better it is for the player. What I want to show you below is that with fewer decks the probability for a player BJ is higher than with more decks.

Recall that the number of combinations of n things taken k at a time , written C(n, k), is given by

C(n, k) = n! / [(n-k)!k!]

where n! = n(n-1)(n-2) … 1. We will use this idea in the calculations below.

We will use one deck and eight decks to illustrate the idea. For one deck the probability of a player BJ, written P(PBJ), is given by

P(PBJ) = C(4,1)C(16,1) / C(52, 2) = 0.04826546

The rationale here is that there are 4 ways to choose the Ace, 16 ways to choose the ten card, and there are C(52, 2) ways to choose a two-card hand from 52 cards. Now some of these hands will occur with the dealer having a BJ as well, so let's compensate for that. The conditional probability of a DBJ given that the player already has a PBJ is given by

P(DBJ|PBJ) = C(3,1)C(15,1) / C(50,2) = 0.036734693

If we multiply this number by P(PBJ) the result is the probability that both the player and dealer have a BJ and this is

P(PBJ & DBJ) = 0.001773016

If we subtract this number from P(PBJ) the result will be the probability that the player has a BJ and there is no tie. That number is

P(PBJ & no tie) = 0.046492443 (one deck)

We can do the same calculations for eight decks and the resulting number is

P(PBJ & no tie) = 0.045266132 (eight decks)

The ratio here is approximately 1.028 so the one-deck situation is 2.8% better than the eight-deck situation. That might not seem very large but in a game as close as blackjack every little bit helps. In fact, simulations have shown that increasing the number of decks from one to eight increases the house edge approximately 0.57%. So look for single-deck games with good rules (there are still a few around) but stay away from those 6:5 games. See you next month.

Don Catlin can be reached at 711cat@comcast.net

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Best of Donald Catlin
Donald Catlin Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

#### Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers
Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

#### Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers