- #1

Entertainment Unit

- 16

- 1

- TL;DR Summary
- The question: suppose we deal four 13-card bridge hands from an ordinary 52-card deck. What is the probability that the North and East hands each have exactly the same number of spades?

I have an answer, but I can't find a solution anywhere to confirm it, and it's a little beyond my programming ability at the moment to do a brute force solution.

Is my answer correct? If not, where did I go wrong?

My answer is as follows:

Let ##S## be the set of all outcomes of dealing four labelled 13-card hands from a standard 52-card deck.

Let ##A## be event "N and E have exactly the same number of spades."

Let ##A_i## be event "N and E have exactly ##i## spades each."

Note that when ##i > 6##, ##\mathbb {P}(A_i) = 0## since there are only 13 spades in a standard deck of cards.

In the case of event ##A_0##, there are

In the case of event ##A_1##, there are

By similar logic to that used to find the cardinality of event ##A_1##:

By the Discrete Uniform Probability Law,

Is this correct? If not, where did I go wrong?

Edit: updated answer, but it's still likely wrong.

Let ##S## be the set of all outcomes of dealing four labelled 13-card hands from a standard 52-card deck.

Let ##A## be event "N and E have exactly the same number of spades."

Let ##A_i## be event "N and E have exactly ##i## spades each."

Note that when ##i > 6##, ##\mathbb {P}(A_i) = 0## since there are only 13 spades in a standard deck of cards.

In the case of event ##A_0##, there are

- ##2^{13}## ways to deal the 13 spades into hands S and W,
- ##\binom {39} {13}## ways to "top up" hands S and W, and
- ##\frac {26!}{13!13!}## ways to partition the remaining cards into hands N and E.

In the case of event ##A_1##, there are

- ##\binom {13}{2}## ways to assign 1 spade each to hands N and E,
- ##\binom {50}{24}## ways to "top up" hands N and E, and
- ##\frac {26!}{13!13!}## ways to partition the remaining cards into hands S and W.

By similar logic to that used to find the cardinality of event ##A_1##:

- ##|A_2| = \binom {13}{4} \binom {48}{22} \frac {26!}{13!13!}##
- ##|A_3| = \binom {13}{6} \binom {46}{20} \frac {26!}{13!13!}##
- ##|A_4| = \binom {13}{8} \binom {44}{18} \frac {26!}{13!13!}##
- ##|A_5| = \binom {13}{10} \binom {42}{16} \frac {26!}{13!13!}##
- ##|A_6| = \binom {13}{12} \binom {40}{14} \frac {26!}{13!13!}##

##|A| = |\bigcup_{i \in \{0,1,2,3,4,5,6\}} A_i| = \sum_{i=0}^6 |A_i|##

By the Discrete Uniform Probability Law,

##\mathbb {P}(A) =\frac {|A|}{|S|} = \frac {|A|}{\frac {52!}{13!13!13!13!}} = \frac {507696703}{31260638524067527680000}##

Is this correct? If not, where did I go wrong?

Edit: updated answer, but it's still likely wrong.

Last edited: