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

• B
• Entertainment Unit
In summary: I would take the first card, and the number of spades for N and E can range from 0 to 2, which is 3 ways. Then look at the next card, and the number of spades for N and E can range from 0 to 2 again, which is 3 ways, but I would remove the card type from the previous card, if any. For 13 cards, I am expecting to get a 3^13 result. Then, for each of those 3^13 possibilities count the number of spades in N and E and if that is equal, increment a counter. The probability is the counter divided by 3^13.
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?

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.

Last edited:
Entertainment Unit said:
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%.

Entertainment Unit said:
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.

Entertainment Unit
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 and etotheipi
Thanks for the help! I finally got ##\frac {28035698399}{158753389900} \approx 0.177##.

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.

## 1. What is the probability that two bridge hands have the same number of spades?

The probability that two bridge hands have the same number of spades is approximately 0.25 or 25%. This means that there is a 25% chance that both hands will have the same number of spades.

## 2. How is the probability of two bridge hands having the same number of spades calculated?

The probability is calculated by taking the total number of ways two hands can have the same number of spades (which is 13 for each hand) and dividing it by the total number of possible bridge hands (which is 635,013,559,600). This gives us a probability of approximately 0.000000000002 or 0.0000000002%.

## 3. Does the order of the spades in the hand matter for the probability calculation?

No, the order of the spades does not matter for the probability calculation. As long as both hands have the same number of spades, they will be considered a match for this probability calculation.

## 4. What is the probability that two randomly dealt bridge hands will have the same number of spades?

The probability of two randomly dealt bridge hands having the same number of spades is approximately 0.0000000002% or 0.000000000002. This is a very small chance, but still possible.

## 5. Does the number of players in the bridge game affect the probability of two hands having the same number of spades?

No, the number of players in the bridge game does not affect the probability of two hands having the same number of spades. As long as there are two hands being dealt, the probability will remain the same.

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