5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @Hugh,You will get all information about bridge card game at www.rpbridge.net $\endgroup$ – Dhamnekar Winod Jun 30 '17 at 15:46
  • $\begingroup$ Describe the problem mathematically. Like: 52 cards are dealt to 4 people, what is the probability that any player got 13 cards of the same suit? $\endgroup$ – Hugh Jun 30 '17 at 15:55
5
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

Using the conditional probabilities, $P$ with a subscript $i$ $i=1, 2, \ldots, 13$ to denote the probability of getting the "right" card, we can use inductive reasoning to calculate the probability. We can assume WLOG that the player is dealt all 13 cards in sequence irrespective of what may be dealt to the other players. We care not what the suit of the first card is that is dealt, so the $P_1 = 1$, but the next card that is dealt must match that suit, which of the 51 remaining there are only 12 to draw, so $P_2 = 12/51$. Continuing in this fashion, the 13th card has probability $P_{13} = 1 / 40$ to be the correct card. Mathematically, this combinatoric reduces to (13! 39! / 52!) which is impossibly small.

$\endgroup$
  • $\begingroup$ This answer implicitly assumes one of several possible interpretations of the question: namely, what is the chance that this particular player is dealt a full suit? The harder version--the one that actually seems to be articulated in the original post--is what is the chance that some player (that is, one or more players) are dealt a full suit? $\endgroup$ – whuber Jun 30 '17 at 17:26
  • $\begingroup$ I remember to read (but not where) that sometimes, under a course, the four players each were dealt a full suit ... $\endgroup$ – kjetil b halvorsen Jul 24 '17 at 18:00
0
$\begingroup$

One player = $\frac {\binom{4}{1}} {\binom{52}{13}} = 6.29907808979643E-012$

One player not having it is 1 - 6.29907808979643E-012 so that to the 4th is none
Any player $1 - (1 - 6.29907808979643E-012)^4 = 2.5196289499263E-011$
One or more is one minus the chance of no player

This is a little odd because if 3 are suited then the 4th has to be suited

$\endgroup$
  • $\begingroup$ It is indeed odd: your formula appears to assume the chances of suits for each player are independent, but your remark shows why that cannot be the case. In fact, the formula doesn't quite give the correct number. It is an approximation and it works only because the chances of two or more players being dealt suits are so tiny. $\endgroup$ – whuber Jul 24 '17 at 15:06
  • $\begingroup$ @whuber I am used to the chance of no one having it done that way. Your answer comes out to 4 time the chance of one player. When it consumes the whole deck like that and only one hand can be left over I am not clear. $\endgroup$ – paparazzo Jul 24 '17 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.