## introduction

Draw poker is played with a standard deck of 52 cards: 13 ranks (Ace through 10, and jack, queen, king) in 4 suits. (An ace usually counts as higher than a king, unless it is being used as “1″ in a run.) Each player is dealt 5 cards.

Before I ever played poker – at home – I was required to learn a mantra: one pair, two pair, three of a kind… straight, flush, full house… four of a kind, straight flush. From least to best, those are the possible winning hands. (In fact, I will distinguish a royal straight flush from other straight flushes.)

What is the probability of being dealt each one of them?

That’s the first question I’ll answer, in several parts.

After the cards have been dealt and bet on, each player may discard some and get replacements. In a second post I will ask and answer: what are the odds against improving any given hand?

There is an excellent wiki article out there. In fact, the nice way to calculate the probabilities comes from it (especially for one pair, and for nothing).

## number of hands

How many 5-card poker hands are there? If order mattered, the first card could be any of the 52, the second card could be any of the remaining 51, and so on, and the fifth card could be any of the remaining 48. We could write that, of course, as

52 x 51 x 50 x 49 x 48 = 311,875,200.

Or we could write it a little more compactly as

52! / 47! = 311,875,200

where n! (n factorial) means n times n-1 times n-2 … times 2 times 1. Call that the number of possible deals.

But that’s if order matters. In a poker hand, it doesn’t matter. Well, then, given one particular set of five cards, how many ways could it have been dealt?

5!

Any one of the five cards could have been dealt first, any one of the remaining four could have been dealt second, etc.

To compute the number of possible poker hands, we take the number of possible deals and divide by 5!

hands = 52! / 47! / 5! = 2,598,960.

A more formal way to write that is to let this symbol:

$\binom{52}{5}$

stand for the number 52! / 47! / 5!, where we infer the 47 as 52 – 5.

Furthermore, it is often pronounced as the problem it solves. We read it as “52 choose 5″ or “choose 5 out of 52″. That’s exactly what we did: we chose a subset of 5 cards out of a set of 52 cards (without replacement).

## flushes and straight flushes

A flush is 5 cards of the same suit. A straight flush is simultaneously a flush and a straight, i.e. a flush with five consecutive cards: i.e. 1,2,3,4,5 up to 10,J, Q,K,A (=1); equivalently, it’s a straight all in the same suit. A straight flush 10 through ace is called a royal straight flush.

How many all-spade hands are there? This many ordered hands of one suit… this many unordered… so the number of flushes is 4 times that:

(That is, allflush = 5148.)

There is a cleaner way of stating that calculation: choose 1 suit out of 4; and choose 5 cards out of the 13 in that suit.

$\binom{4}{1} \binom{13}{5} = 5148$

Yes, “4 choose 1″ = 4, so I’m typing extra characters by writing

$\binom{4}{1}$

instead of “4″. But I like emphasizing the conceptual commonality.

Of all those flushes, exactly 4 are royal straight flushes.

royal = 4.

How many flushes are straight flushes? A straight can begin with Ace thru 10, so there are precisely 40 straight flushes, including the royal… hence 36 straight flushes excluding the royal.

str8flush = 40 – royal = 36.

Then the number of flushes which are not straights is:

flush = allflush – royal – str8flush = 5108.

## straights

Any straight has 40 choices for the lowest card, and 4 for each of the remaining, so there are

allstr8 = 40*4^4 = 10240.

How many of those are straight flushes or royal straight flushes? We know that: 40 (which I saved as 4 and 36).Thus the number of straights which are not straight flushes is:

str8 = allstr8 – royal – str8flush = 10200.

Now let’s rephrase that nicely. To get the total number of straights, choose 1 out of 10 for the lowest rank in the straight, and then choose 1 out of 4 suits for each of the 5 cards.

$\binom{10}{1} \binom{4}{1}^5 = 10240$

## four of a kind

How many 4-of-a-kind hands? 13 choices for the kind, 48 possibilities for the 5th card:

fours
= 13 * 48 = 624.

Put nicely, choose 1 out of 13 for the kind, then choose 1 out of 48 for the fifth card.

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

## full house

A full house is three of a kind and two of another kind.

How many full-houses? 13 choices for the triple, 12 for the pair, but there is an extra card for the triple, and two extra cards for the double.

full=(4*13) (6*12) = 3744.

It’s so much better to do it nicely: choose 1 of 13 for the value of the triple, then choose 3 out of 4 suits to fill out the triple… then choose 1 out of 12 for the value of the pair, and choose 2 out of 4 suits to get a specific pair:

$\binom{13}{1} \binom{4}{3} \binom{12}{1} \binom{4}{2} = 3744$

## three of a kind

How many exactly 3-of-a-kind? 13 choices for the kind, but 4 ways to get each one… 4*12 choices for the 2nd rank and 4*11 choices for the 3rd rank… but order doesn’t matter, so divide by 2.

threes = (4*13) (4*12) (4*11) / 2 = 54912

Really? We know how to break this down. Choose 1 of 13 for the rank of the triple, then 3 out of 4 suits; choose 2 ranks out of 12 for the other two ranks, and for each, chose 1 of 4 suits:

$\binom{13}{1} \binom{4}{3} \binom{12}{2} \binom{4}{1}^2 = 54912$

## two pair

The first pair is one of 13, with 6 distinct pairs and second pair is one of 12, with 6 distinct pairs… but order doesn’t matter (first pair, second pair)… and the 5th card can be any of the remaining 44.

(That is, twopair = 123,552.)

Do I believe that reasoning? Sometimes!

So let me do it cleanly. Choose two ranks out of 13 for the ranks of the two pairs… for each pair, choose 2 out of 4 suits… then pick one of the 44 cards other than these 8 to fill out the hand:

$\binom{13}{2} \binom{4}{2}^2 \binom{44}{1} = 123552$

The numbers are the same — but I trust the second calculation. The first one is too ad hoc.

## one pair

Let me confess that I have to do this one cleanly. I get it wrong any other way.

Choose 1 out of 13 for the rank, then choose 2 out of the 4 suits; choose 3 ranks out of the remaining 12, and for each of those, choose 1 out of 4 suits:

pair = $\binom{13}{1} \binom{4}{2} \binom{12}{3} \binom{4}{1}^3 = 1,098,240$

## nothing

This is slick, in my opinion. It comes from the wiki link at the beginning of the post. Pick 5 ranks out of 13… but exclude the 10 straights of any one suit; for each of the chosen 5 ranks, choose 1 out of 4 suits… but exclude the 4 flushes.

none = $(\binom{13}{5}-10) \left(\binom{4}{1}^5-4\right) = 1,302,540$

## summary

Here’s what we’ve computed:

$\left( \begin{array}{cc} \text{ONE PAIR} & 1098240 \\ \text{TWO PAIR} & 123552 \\ \text{THREE OF A KIND} & 54912 \\ \text{STRAIGHT} & 10200 \\ \text{FLUSH} & 5108 \\ \text{FULL HOUSE} & 3744 \\ \text{FOUR OF A KIND} & 624 \\ \text{STRAIGHT FLUSH} & 36 \\ \text{ROYAL STRAIGHT FLUSH} & 4 \end{array} \right)$

It’s nice to see that hands do get rarer as their value rises.

Now that we have collected the number of hands… how many valuable poker hands are there? Here are the individual numbers, so add them up:

Oh, while we have that list, let’s divide every entry by the total number of hands; this is the probability of being dealt each of these poker hands:

{0.422569, 0.047539, 0.0211285, 0.00392465, 0.0019654, 0.00144058, 0.000240096, 0.0000138517, 1.53908*10^-6}

There is, for example, a 42% chance of being a dealt exactly one pair; almost 5% chance of being dealt two pair; 2% chance of being dealt three of a kind.

How many hands are of no value? We have computed

none = 1,302,540.

and what the sum of none and poker?

none + poker = 2,598,960.

And how many possible hands did we compute that there were?

hands = 2,598,960.

Good. By independently computing the valueless hands, we got a check on our computation of poker hands.

Finally, the ratio none/hands is:

none / hands = 0.501177

… so the chances of being dealt 5 useless cards is just over 50%, but not by much. You could safely remember it as 50-50, for being dealt at least a pair.

In another post, I’ll look at the odds of improving the hand you were dealt.

### 11 Responses to “Poker Hands – 5 card draw”

1. John McDowel Says:

I’ve been playing cards most of my life and poker, mostly Texas Hold Em, for the last several years at Ultimate Bet an online casino. Like chess, it’s a simple to learn the mechanics of the game, but difficult to master. Take care,
John Mc

2. kevin Says:

Thank you for an informative discussion – this is exactly what I needed.

(However, I have other mathematical discussions pertaining to poker
which I have not seen and that is the relationship of probabilities as
the poker game progresses and the number of players at that specific
turn. My verbal discussions are that this is complex but I am certain
it is boundable as it is finite.)

Thank you again for a rare quality webpage.

Kevin

• rip Says:

You’re welcome. And it is very true that the probabilities of being dealt various hands, and the probabilities of improving various hands, are a far cry from a thorough model of the dynamics of a round of poker.

Jeff Rubens “Win at Poker” discusses the dynamics from a practical but not mathematical point of view.

In my case, my posture changes significantly when I have a good hand – so there’s not much point in my playing poker, because I can be read too easily.

3. dead mans hand poker Says:

wow. nice discussion. i’ve been playing online poker for years without knowing the probability to get a good hand. my judgement in game is based on my instincts and mood since its a gamble.

4. Anutr Says:

Great site and great article. Although maths is not everything in poker, it is extremely important. In 5-card draw, knowing the odds is essential just like other poker games. I remember play 5-card draw in my first poker game, I knew nothing about odds but it was still very fun! Thank you for these calculations.

5. Chrystal Says:

Thank you for the math. Curious though, how does the option to discard upto all five cards and receive replacements factor in?

• rip Says:

You’re welcome.

Improving the hand was the subject of the second post on draw poker , although the weakest deal I considered was two pair and drawing three cards. Since that case took up half the post, a precise analysis of improving a garbage hand would involve a lot of computation.

Maybe the right answer is: if you’re not dealt at least one pair, throw it in.

6. Math Hater Says:

I came to this line “where n! (n factorial) means n times n-1 times n-2 … times 2 times 1.” and then you lost me. I dont see see n! anywhere in the text above that line. This also reminded me how much I hate math. And I mean hate, I absolutetly hate it, have for over 20 years, its satans favourite game, if I could I would kill all math everywhere, its the most annoying thing on this planet, I prefer war over math, it sucks the fun out of every little thing I try to do.

• rip Says:

Did you mean this?

52! / 47! = 311,875,200

where n! (n factorial) means n times n-1 times n-2 … times 2 times 1. Call that the number of possible deals.

That’s not five hundred twenty one and four hundred seventy one, it’s 52 factorial divided by 47 factorial… or, in British schools I think, “52 bang divided by 47 bang”.

I’m sorry math is such a trial for you.

7. Willy dal Says:

Hi Rip..I have a question for you….if you have a stardard deck of cards, but the six of hearts is missing, what are the odds of being dealt a straight in a game of 5 card stud?…..Thanks!!!!

• rip Says:

Well, Willy, ask yourself how many straights contain a 6? Then how many contain the 6 of hearts? Subtract that from the total number of straights possible with a full deck.