# How many hands (of size 5) of a cards from a standard deck with 4 of a kind are there?

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

1. The rank of the four cards
2. The rank of the extra card
3. 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.

1. The rank of the four cards
2. The suit of the four cards
3. The rank of the extra card
4. The suit of the extra card

I just don't understand why that is wrong.

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

1. The rank of the four cards
2. The suit of each of the four cards in a combination way
3. The rank of the extra card
4. 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?

• There is only one copy of each specific rank-suit card in the deck. In other words, once you know that the four of a kind is in aces, you know it consists of the ace of hearts, clubs, spades, and diamonds respectively. It cannot consist of multiple copies of the ace of clubs (unless either someone is cheating or you combine multiple decks) – JMoravitz Nov 5 '14 at 1:10
• @JMoravitz Oh, I see. That makes sense. Please provide your answer, I will be happy to accept it. You explained it really well. – Pinocchio Nov 5 '14 at 1:11
• You could use your final thought and consider the suits important (for the sake of making the jump to the more general problem) you will then multiply by $\binom {4}{4}$, the number of ways to choose which four of the four suits you use. Note that it simplifies to 1, hence it's omission. – JMoravitz Nov 5 '14 at 1:17
• From your question in the comment two above, it would be the second interpretation – JMoravitz Nov 5 '14 at 1:29
• @Pinocchio: It's all right, I see it has been improved now. Actually I have generously dealt you an additional 8 of spades (from a spare deck), so there is a true four-of-a-card now. – Marc van Leeuwen Nov 5 '14 at 14:45

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: $\binom {13}{1} = 13$

1. Choose the suits of the triple: $\binom {4}{3} = 4$

2. Choose the rank of the double: $\binom{12}{1} = 12$

3. 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:

1. Choose the rank of the 4-tuple: $\binom{13}{1} = 13$

2. Choose the suits of the 4-tuple: $\binom{5}{4} = 5$

3. Choose the rank of the singleton: $\binom{12}{1} = 12$

4. 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.

• The six of course comes from 4 choose 2 – JMoravitz Nov 5 '14 at 1:30
• Since I feel I learned the answer to my question and more from you, if you were to add (copy/paste) the comment you put under my question and just correct that 6 with a 4 choose 2, I feel that you should get my acceptance. Don't worry about it now, you can do it later/tomorrow. When you have a computer. Thanks so much!! :D – Pinocchio Nov 5 '14 at 1:33
• Updated with an additional example. For practice, in the setting of a 5-card hand from a standard deck, I recommend also trying to calculate the number of 2-pair hands (consisting of three distinct ranks of cards, two of which appearing twice), the number of royal flushes(10,J,Q,K,A of a single suit), the number of straight flushes (five of same suit of adjacent rank, i.e. 4,5,6,7,8 of hearts but not a royal flush), and using those previous two, the number of flushes (five cards of same suit which are neither royal nor straight flushes). These are the more challenging examples. – JMoravitz Nov 5 '14 at 18:32

The four cards include all suits, and there is only one way to include all four suits. It might be 8 hearts, 8 spades, 8 clubs, 8 diamonds, Q clubs.

• I am sure you are right, but there are not enough details in your answer for me to understand the fundamental flaw in my reasoning. – Pinocchio Nov 5 '14 at 1:10
• You wrote '8 clubs,8 clubs,8 clubs,Q hearts' In four of a kind, you don't repeat the card '8 clubs', you find different cards with the same rank (8) and different suits (Clubs,Hearts,Diamonds,Spades). – Empy2 Nov 5 '14 at 1:12
• yes you are right. You should have said that in your answer! Its hard to be a good teacher unless you specify explicitly what was the students misunderstanding. Just stating the answer sometimes is not enough. You should have said why my example was ridiculous! Which it was. Thanks. :) – Pinocchio Nov 5 '14 at 1:15

To solve your confusion, we will not be able to create any four of a kind hands with {8 spades, 8 spades, 8 spades, Q hearts} or {8 clubs, 8 clubs, 8 clubs, Q hearts}. The reason for this is in a standard deck of 52 cards there are 4 suits with 13 distinct ranks in each suit. This means that once the 8 of clubs of Queen of hearts is chosen, we can no longer choose this card.

A solution to this problem can be broken down into two stages

Stage 1: Choose the rank of the card for the 4 cards -> $\binom {13}{1}$ since there are 13 ranks

Stage 2: Choose the remaining card -> $\binom {48}{1}$ since we need one card and we chose 4 so there are only 48 left in the deck

Since the outcomes of each stage are independent of each other we can multiply the results of each stage to find the total number of hands with a four of a kind and one other card.

Total = $\binom {13}{1}$ x $\binom {48}{1}$ = 13 x 48 = 624

The other persons answer is the same thing in disguise, by choosing the remaining ranks and then suit you are simply choosing the remaining cards in the deck, notice that 12 x 4 = 48.

Here is another way to explain.

In the case that "kind" means rank. There are $13$ ranks in the $52$ cards. Each rank has exactly $4$ cards. To select $5$-card hand with $4$ cards of one rank:

• Firstly, select $4$ cards of one rank: $13$ ways to do this since there are $13$ or $\binom{13}{1}$ ways
• Secondly, select the last card: after selecting $4$ cards, we know have $48$ cards left. -> $48$ or $\binom{48}1$ or $12\times 4$ (remainedRanks*cardsPerRank) ways to select the last cards

Hence, there are $13 \times 48$ ways to select $5$-card hand of which $4$ cards are of one kind.

I hope this helps.