B The Probability that N and E bridge hands have the same number of spades?

[Entertainment Unit]

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

• $2^{13}$ 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.

By the fundamental principle of counting, $|A_0| = 2^{13} \binom {39}{13} \frac {26!}{13!13!}$.

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 the fundamental principle of counting, $|A_1| = \binom {13}{2} \binom {50}{24} \frac {26!}{13!13!}$.

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

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

Since the events $A_i, i \in \{0,1,2,3,4,5,6\}$ are disjoint,

​

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

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.

 

[PeroK]

By the Discrete Uniform Probability Law,

$\mathbb {P}(A) =\frac {|A|}{|S|} = \frac {|A|}{\frac {52!}{13!13!13!13!}} = \frac {91957}{1508157204050}$​

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

That looks like a very small number. I make it about 18%.

 

[willem2]

In the case of event A0A0A_0, there are

• 2132132^{13} ways to deal the 13 spades into hands S and W,
• (3913)(3913)\binom {39} {13} ways to "top up" hands S and W, and

You only count, the number of ways 13 cards can be selected to "top up" S and W, but not the number of ways these can be divided between S and W.

 

[PeroK]

Here's a rough estimate:

Let's say that North gets two, three or four spades with roughly equal probabilty of $1/4$, leaving $1/4$ for everything else. And, assume that East gets the same independent of what North has.

That gives us an approximate probability of $1/16$ for North and East to have two spades each and the same for three spades each and four spades each. That's a total of $3/16$. That's a first guess.

To calculate the number accurately, we take the six cases where E and N share zero to six spades.

The number of ways that N has $n$ spades is:
$$N(n)= \binom{13}{n}\binom{39}{13-n}$$
The total number of N hands is:
$$N = \binom{52}{13}$$
Then for East we have (given that N has $n$ spades):
$$E(n) = \binom{13-n}{n}\binom{26+n}{13-n}$$
And the total number of East hands is:
$$E = \binom{39}{13}$$
Then the probability that N and E both have $n$ spades is:
$$P(n) = \frac{N(n)}{N} \frac{E(n)}{E}$$
And you can sum the six resulting probabilities. The most likely, for example, is three spades each, which has a probability of about $8 \%$. And the total probability is about $18\%$, which is a bit less than the rough estimate above.

 

[Entertainment Unit]

Thanks for the help! I finally got $\frac {28035698399}{158753389900} \approx 0.177$.

 

[Vanadium 50]

I didn't solve the problem, but I would be inclined to attack it via the general case: 4N cards in a deck, and work it out by hand for N=1, 2 and maybe 3. Once I get it right for N, plug in N=13.

For example, I would not be trying to partition cards between S and W.