Solved – the chance that someone is dealt a suit in the game of Bridge

combinatoricsgamesprobability

In a game of Bridge, what is the probability that some player has a complete suit?

(There are four players in a Bridge card game. A deck of Bridge cards consists of 52 cards arranged in four suits of thirteen each. Playing Bridge begins by distributing the cards randomly to four players, namely North, South, East, West, so that each receives 13 cards.)

I am referring to various books on statistics to solve this problem.

Best Answer

The answer is a tiny number, but large enough to suggest someone has at sometime been dealt a suit in the US. This post shows how to find that chance (with a sequence of three simple calculations), provides an interpretation, and concludes by showing how to compute it accurately.

Let's begin with a small generalization, because it uncovers the essence of the problem. Let a "suit" consist of $k\ge 1$ cards. A "deck" is the union of $m\ge 1$ distinct suits. In the question, $k=13$ and $m=4$. To model a deal, suppose the deck is randomly shuffled and partitioned into $m$ groups of $k$ contiguous cards in the shuffle. Each of these groups is a "hand".

First let's find the chance that one pre-specified player is dealt a suit. There are $$\binom{mk}{k} = \frac{(mk)(mk-1)\cdots((m-1)k+1)}{(k)(k-1)\cdots(1)}$$ possible hands, all of them equally likely, and $m$ of them are suits. This chance therefore is

$$p^{(1)}(m,k) = \frac{m}{\binom{mk}{k}}.\tag{1}$$

If this is what the question was asking, we are done. However, the more likely interpretation is that it asks for the chance that one or more of the hands is a suit.

To do this, proceed to find the chance that two pre-designated players are dealt suits. Conditional on the first player being dealt a suit (the chance given by $(1)$), there remain $m-1$ suits. Result $(1)$ applies with $m$ replaced by $m-1$ to give the conditional probability. These two values multiply to give the joint probability

$$p^{(2)}(m,k) = p^{(1)}(m,k)p^{(1)}(m-1,k).$$

Continuing this reasoning inductively gives the chance that $s\ge 1$ pre-designated players each are dealt suits,

$$p^{(s)}(m,k) = \prod_{i=0}^{s-1} p^{(1)}(m-i,k).\tag{2}$$

The Principle of Inclusion-Exclusion ("PIE") supplies the chance that one or more players (not designated in advance) are dealt suits; it is

$$p(m,k) = \sum_{s=1}^m (-1)^{s-1} \binom{m}{s} p^{(s)}(m,k).\tag{3}$$

In particular, $$\eqalign{ p(4,13) &= \frac{(3181)(233437)(25281233)}{(2^6)(3)(5^4)(7^4)(17^3)(19^2)(23^2)(29)(31)(37)(41)(43)(47)} \\ &=\frac{18772910672458601}{745065802298455456100520000}\\ &\approx 2.519631234522642\times 10^{-11}. }$$

The small primes in the denominator were expected: they cannot exceed $mk=52$. The large primes in the numerator strongly suggest there exists no general closed form formula for $p(m,k)$.

What does this answer mean?

Wikipedia traces the modern version of Bridge to 1904, states that it used to be more popular in the US, and reports there are around 25 million players today in the US. Although it's difficult to know exactly what it means to be a Bridge player, we might expect each one on average to play between a few hands and a few hundred hands annually, with each hand involving four players. (In Duplicate Bridge some deals are played multiple times, but let's ignore that complication and just absorb it into the "few to a few hundred" estimate.) The expected number of Bridge deals annually in the US in which a suit is dealt therefore is on the order of ten to a hundred times the product

$$p(4,13) \times 25 \times 10^6 \approx \frac{1}{1588}.$$

Accounting for the $110$ or so years that have transpired since 1904, we might multiply this expectation by another two orders of magnitude. The result is somewhere between $1/10$ and $10$. Although $p(4,13)$ might seem "impossibly small," it is not negligible: depending on the assumptions about how active Bridge players have been, it's somewhere between plausible and highly likely a suit has already been dealt in the US.

Many people have reported such hands. The obvious explanation is that some (many?) decks are not randomly shuffled or dealt. See Peter Rowlett on Four Perfect Hands or Science News on Thirteen Spades.

Computing notes

Computing the answer is as straightforward and simple as it looks in formulas $(1)$, $(2)$, and $(3)$: see the R example below. When applying PIE it's usually best to avoid large values of $m$ due to the alternating addition and subtraction in the final formula: round-off error can accumulate rapidly when some individual terms in the sum are much greater in size than the result. This situation is nicer. Since in general the first term--based on the chance of just one particular player getting a suit--dominates the rest, this code performs the sum in reverse order to avoid that roundoff error.

# NB: `choose` computes the binomial coefficient
p <- function(m, k) {
  p.1 <- function(m, k) m / choose(m*k, k)            # Formula (1)
  p.s <- function(s, m, k) prod(p.1(m:(m-s+1), k))    # Formula (2)
  p.s <- Vectorize(p.s, "s")
  sum((-1)^(m:1-1) * choose(m, m:1) * p.s(m:1, m, k)) # Formula (3)
}
print(p(4, 13), digits=16)

[1] 2.519631234522642e-11

The result is correct to the full precision inherent in IEEE floating point arithmetic.

References

Gridgeman, N. T. "The Mystery of the Missing Deal." Amer. Stat. 18, 15-16, Feb. 1964.

Mosteller, F. "Perfect Bridge Hand." Problem 8 in Fifty Challenging Problems in Probability with Solutions. New York: Dover, pp. 2 and 22-24, 1987.

Wolfram Mathworld quotes the same rational value for $p(4,13)$. Its references are Mosteller and Gridgeman.

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