[Math] How many hands (of size 5) of a cards from a standard deck with 4 of a kind are there

combinatorics

I was trying to count how many cards with 4 of a kind I could get from a standard deck of 52 cards in a hand of length 5.

i.e. 4 of the same value (rank) and a card with a different value.

I was told the answer was $13 \times 12 \times 4 = 624$ but I was a little confused why.

The explanation given to me was because a four of a kind can be completely specified by a sequence saying

The rank of the four cards
The rank of the extra card
The suit of the extra card

However, there is a specific reason I am confused. For me this reasoning seems to be undercounting. Why do I think that, well, I believe that the suit of the cards with same rank should be important too because a 4 of a kind hand with {8 spades, 8 spades, 8 spades, 8 spades, Q hearts} is a different 4 of a kind than {8 clubs, 8 clubs, 8 clubs, 8 clubs, Q hearts}

i.e. does the suit of the cards with the same kind not matter?

Why is it not $13 \times 4 \times 12 \times 4$ i.e.

The rank of the four cards
The suit of the four cards
The rank of the extra card
The suit of the extra card

I just don't understand why that is wrong.

Actually, that is wrong too, why is it not:

The rank of the four cards
The suit of each of the four cards in a combination way
The rank of the extra card
The suit of the extra card

As in why is the answer not:

$$ 13 \times C^4_1 \times C^4_1 \times C^4_1 \times C^4_1 \times 12 \times 4$$

or something that takes into account the different suits for the 4 of a kind cards?

Best Answer

Note first that in a standard 52-card deck, there is only one copy of each specific rank-suit card in the deck (i.e., only one Ace of spades, only one Ace of clubs, only one Ace of hearts, etc...)

When determining how many four-of-a-kind hands there are, you may choose to consider what suits the 4-tuple consists of as it helps to generalize the problem further. After the first step of picking what rank the 4-tuple is in, decide which 4 of the available 4 suits the cards will be, that is to say, there are $\binom{4}{4}$ "different" ways of having the suits selected for the 4-tuple. Due to the fact that $\binom{4}{4}=1$, many people choose to leave it out of the calculations because it is readily apparent to them that to have four of the same rank, you will necessarily use all four available suits and there exist no other configurations.

Down here instead, consider the related problem of asking to find how many full houses exist (again from a standard deck) (a full house consists of three cards of one rank and two cards of another)

Breaking it into steps: 1. Choose the rank of the triple: $\binom {13}{1} = 13 $

Choose the suits of the triple: $\binom {4}{3} = 4 $
Choose the rank of the double: $\binom{12}{1} = 12$
Choose the suits of the double : $\binom{4}{2} = 6$

Thus, the total number of full house hands in a standard deck is $13\cdot 4\cdot 12\cdot 6$

Note that in this case we can identify the triple from the double. If we asked how many two - pair there are you will need to divide by symmetry

Consider another similar situation in order to highlight the differences of using a nonstandard deck consisting of 65 cards: thirteen ranks available as usual (2,3,4,5,6,7,8,9,10,J,Q,K,A), but this time consisting of five different suits: (hearts, diamonds, clubs, spades, and swords)

How many four-of-a-kind hands exist for this nonstandard deck then?

Breaking it up into steps as before:

Choose the rank of the 4-tuple: $\binom{13}{1} = 13$
Choose the suits of the 4-tuple: $\binom{5}{4} = 5$
Choose the rank of the singleton: $\binom{12}{1} = 12$
Choose the suit of the singleton: $\binom{5}{1} = 5$

So, all together, there will be $13\cdot 5\cdot 12\cdot 5$ number of 4ofakind hands in this nonstandard deck.

Related Solutions

[Math] Solution verification: picking $5$-card hands from standard deck of $52$, with conditions

Extended Comment: This is a simulation mixed with some exact computations. It may help as a reality check for a couple of the parts of your question.

The simulation in R deals a million 5-card hands and counts Clubs and Spades. The deck consists numbers 1, 2, 3, 4, for H, D, C, S, respectively; 13 of each. (The sample function distinguishes among elements labeled with the same number.)

With a million hands, probabilities should be accurate to about two or three places. Exact values of $13{39 \choose 4}/{52 \choose 5}$ and the corresponding hypergeometric probability are also shown. It may be of interest for you to look at the hypergeometric distribution in your textbook or on Wikipedia.

deck = rep(1:4, times=13)
deck
# [1] 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2
#[27] 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4

m = 10^6;  nr.cl = nr.sp = numeric(m)
for (i in 1:m) {
  hand = sample(deck,5)
  nr.cl[i] = sum(hand==3)
  nr.sp[i] = sum(hand==4)  }

mean(nr.cl==1);  mean(nr.sp==1)
## 0.411229      # aprx P(exactly 1 Club)
## 0.412112      # aprx P(exactly 1 Spade)
13*choose(39,4)/choose(52,5)
## 0.4114196     # exact value of above from combinatorics
dhyper(1, 13, 39, 5)
## 0.4114196     # exact value of above from hypergeometric dist'n

mean((nr.cl==1) | (nr.sp==1))
## 0.654191      # aprx P(ex 1 Club OR ex 1 Spade)
mean((nr.cl==1) & (nr.sp==1))
## 0.16915       # aprx P(ex 1 Club AND ex 1 Spade)

[Math] How many different hands of 7 cards…

You do not distinguish hand by permutation of cards.

Firstly, you wish to count ways to: select four from four suits for one from thirteen rank, and three from four suits for one from the twelve remainining ranks.

$$\binom 44\binom {13}1~\binom 43\binom {12}1$$

That is, there are $\mathrm C(4,4)C(13,1)C(4,3)C(12,1)$ ways to select seven cards that contain 4 cards of one rank and 3 cards of a second rank .

Secondly, you wish to count ways to select two from four suits each for three from thirteen ranks, and one from four suits for one from the ten remaining ranks. $$...$$

That is, there are so many ways to select seven cards that contain 3 pairs of 3 different ranks and a single card of a fourth rank.

Best Answer

Related Solutions

[Math] Solution verification: picking $5$-card hands from standard deck of $52$, with conditions

[Math] How many different hands of 7 cards…

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