# Bridge - Probability of 2 persons owning 1 full suit

Consider a game of bridge with a standard deck of 52 cards and 4 players $$A, B, C, D$$, each dealt 13 cards.

What is the probability that any 2 players are dealt a full suit? That means, any 2 players $$AB, BC, CD, ...$$ get all 13 cards of 1 suit?

A full suit means any of the 4 suits. Player $$A$$ can have $$9 \heartsuit$$ and $$C$$ can have $$4 \heartsuit$$ and this counts. Similarly, $$A$$ can have $$6 \clubsuit$$ and D can have $$7 \clubsuit$$ and this counts.

Any 2 players can own any number of the suit, but both combined will form the complete suit. Hence, Player $$A$$ can have $$x$$ number of $$\spadesuit$$, and if Player $$B$$ has $$13-x$$ of $$\spadesuit$$, this is a valid combination.

• (i) Is the suit specified, or can it be anything? (ii) Is your condition satisfied if between them A and D have all the spades, like $10$ for A and $3$ for $D$? Commented Feb 3, 2013 at 5:27
• I am sorry for not being specific enough. Edited the question Commented Feb 3, 2013 at 5:28

Let's say $AB$ are going to get all the spades. Then $CD$ will have no spades between them, so their 26 cards are going to come from the 39 non-spades. We can choose the pair getting one suit in 6 ways, and the suit they get in 4 ways, so the probability is approximately $$\frac{\binom{39}{26}}{\binom{52}{26}} \cdot 6 \cdot 4 \approx 3.9 \times 10^{-4}.$$ This is only approximate because it's possible for more than one pair to have an entire suit, but that should be a small probability compared to the above. It's even possible for there to be a situation such as: $AB$ own all the clubs, $BC$ own all the diamonds, $CD$ own all the hearts, and $DA$ own all the spades. So the exact calculation will be messy.