Cactus Kev's Poker Hand Evaluator
A while ago, I decided to take a shot at writing a poker hand
evaluator in the programming language "C". There are already
numerous evaluators out there, but I had an idea for an algorithm
that might be faster than anything already out there. The basic
concept is to write a routine that would take a five card poker
hand and return it's overall "value". This is extremely valuable
in any poker-related software, since the code will constantly be
comparing various player's hand with each other to determine the
"winner". Here is my concept on how I thought I could write a
Okay, before we start digging into my algorithm, please read
this first. I usually get one or two emails every month from
somebody interested in my algorithm or poker code. And they
typically ask me if I happen to have a seven-card poker evaluator.
The answer is yes. I did indeed write a seven-card hand
evaluator a few years after writing the five-card one. It
used a completely new algorithm, completely unrelated to my
five-card version. I just never posted it on my web site because
(a) it was pretty lengthy, and (b) I was too lazy to write up
all that HTML.
However, one day, I got an email from an actual poker software
company from Canada called
Poker Academy. They also wanted to know if I had written
a seven-card evaluator, and if so, could they test it to see if it
was faster than their current code? We converted my code from 'C'
to Java, and they gave it a whirl. Turns out it was about three times
faster than what they were currently using, so they asked if I'd
be willing to sell/license it to their company for use in the next
version of their software. We worked out a contract and the deal
was inked. If you visit their site, download version 2.5, and select
Version History, you can see my name in the notes for the February 15,
2006 2.5.0 Release (b164). The downside of all this, is that I cannot
pass on my seven-card evaluator algorithm to any curious poker math-geeks
who stumble upon my site. So don't email me asking for the code or
algorithm, because I can't give it to you. Sorry, mate.
First off, any person who has studied combinatorics will know
that there are C(52,5),
or 2,598,960 possible unique poker hands.
I realized that even though there are nearly 2.6 million unique
hands, many of those hands actually have the same poker
hand value. In other words, somebody holding an AJ942 flush in
spades has the exact same value hand as somebody with an AJ942
flush in clubs. Even though both hands are unique, they still
hold the identical value, and would therefore "tie" in poker games.
Here's another way to look at it.
Suppose you were able to round up 2,598,960
of your friends on a football field, and you gave each of them one
of the unique 2,598,960 poker hands to hold. You then yell in a loud voice,
asking everyone to compare their hand with everybody else's
hand. (This will take some time, of course :)
Anyway, once they are done, you ask the person holding the best hand to step forward.
Of course, four people will step forward, each holding a Royal
Flush in each of the four suits. They all "tied" for having the
best hand. You group those people together, tie a rope around
them, label them with the number "1", and ask them to leave the
field. Now, you ask your friends to compare hands again and figure
out who has the best hand now. This time, four more people come
forward, each holding a King-High Straight Flush. You group them
together, label them with a "2", and escort them off the field.
You keep repeating this process of finding out who has the highest
hand and marking them with the next number. The next eight queries
Now, when you query for the eleventh time, four people should step
forward, each holding Four Aces with a King kicker. Then, Four
Aces with a Queen kicker. Then, Four Aces with a Jack kicker.
And so on, until we have Four Aces with a Deuce kicker. Next comes
Four Kings with an Ace kicker. This continues until we finally
get down to Four Deuces with a Trey kicker.
- 3: four people holding Queen-High Straight Flushes
- 4: four people holding Jack-High Straight Flushes
- 5: four people holding Ten-High Straight Flushes
- 6: four people holding Nine-High Straight Flushes
- 7: four people holding Eight-High Straight Flushes
- 8: four people holding Seven-High Straight Flushes
- 9: four people holding Six-High Straight Flushes
- 10: four people holding Five-High Straight Flushes
Next, for the 167th query, we find twenty-four people coming
forward, each holding a Full House of Aces over Kings. Then
24 people with Aces over Queens. And so on, for the remaining
Full House hands.
- 11: four people holding Four Aces with a King kicker
- 12: four people holding Four Aces with a Queen kicker
- 165: four people holding Four Deuces with a Four kicker
- 166: four people holding Four Deuces with a Trey kicker
Note that by using combinatorics, we can verify these totals. Assume
you have a Full House of Aces over Kings. There are
C(4,3) possible ways to select three Aces out of a
possible four, and C(4,2) possible
ways to choose two Kings out of a possible four. This yields 4 x 6,
or 24 possible combinations of such hands.
For query #323, four people will step foward, each holding
a AKQJ9 Flush. Then, four people holding AKQJ8
Flushes, and so forth.
- 167: twenty-four people holding Full Houses of Aces over Kings
- 168: twenty-four people holding Full Houses of Aces over Queens
- 321: twenty-four people holding Full Houses of Deuces over Fours
- 322: twenty-four people holding Full Houses of Deuces over Treys
For our next query, we get a whopping 1020 people step forward with
Ace High Straights. This is followed by another 1020 people with
King High Straights, and so on down the line.
- 323: four people holding AKQJ9 Flushes
- 324: four people holding AKQJ8 Flushes
- 1598: four people holding 76432 Flushes
- 1599: four people holding 75432 Flushes
Next comes all the Three of a Kinds, 64 people each.
- 1600: 1,020 people holding Ace High Straights
- 1601: 1,020 people holding King High Straights
- 1608: 1,020 people holding Six High Straights
- 1609: 1,020 people holding Five High Straights
Then the people in groups of 144, each holding Two Pair.
- 1610: sixty-four people holding AAAKQ
- 1611: sixty-four people holding AAAKJ
- 2466: sixty-four people holding 22253
- 2467: sixty-four people holding 22243
Almost done! Next comes the One Pair hands, 384 people each.
- 2468: 144 people holding AAKKQ
- 2469: 144 people holding AAKKJ
- 3324: 144 people holding 33225
- 3325: 144 people holding 33224
And finally, we have the "dreck" High Card hands, with
1,020 people per hand. Notice they are identical to the
above "Flush" hands, except these are not flushes.
- 3326: 384 people holding AAKQJ
- 3327: 384 people holding AAKQT
- 6184: 384 people holding 22643
- 6185: 384 people holding 22543
There! After enumerating and collapsing all the 2.6 million
unique five-card poker hands, we wind up with just 7462 distinct
poker hand values. That was pretty surprising to me when I
saw the final total. If you are interested in seeing a table
of all 7462 values, you'll find it here.
This is how the numbers stack up:
- 6186: 1,020 people holding AKQJ9
- 6187: 1,020 people holding AKQJ8
- 7461: 1,020 people holding 76432
- 7462: 1,020 people holding 75432
Once I determined that there were only 7462 distinct values of
poker hands, I needed a way to quickly transform each of the
2,598,960 unique five-card poker hands into its actual value.
To complicate matters, the algorithm needed to be order independant.
In other words, if I pass the cards
Kd Qs Jc Th 9s
to my evaluator, it must generate the value 1601. However, if I change
the order of the cards in any fashion, it must still return
the value of 1601. Mixing up the five cards does not change the
overall value of the hand. At first, I thought that I could always
simply sort the hand first before passing it to the evaluator; but
sorting takes time, and I didn't want to waste any CPU cycles sorting
hands. I needed a method that didn't care what order the five cards
were given as.
After a lot of thought, I had a brainstorm to use prime numbers.
I would assign a prime number value to each of the thirteen card
ranks, in this manner:
|Four of a Kind
|Three of a Kind
The beauty of this system is that if you multiply the prime
values of the rank of each card in your hand, you get a unique
product, regardless of the order of the five cards. In my above
example, the King High Straight hand will always
generate a product value of 14,535,931. Since multiplication is one
of the fastest calculations a computer can make, we have shaved
hundreds of millseconds off our time had we been forced to sort
each hand before evaluation.
One last step is required, however, before multiplying our prime
rank values. We must first check to see if all five cards are of
the same suit. It is extremely important that our evaluator
realize that the value of a KQJT9 hand is much
higher if all the suits are the same.
Okay. At this point, I was ready to start writing some code.
I decided to use the following bit scheme for a single card,
where each card is of type integer (and therefore,
four bytes long).
p = prime number of rank (deuce=2,trey=3,four=5,...,ace=41)
r = rank of card (deuce=0,trey=1,four=2,five=3,...,ace=12)
cdhs = suit of card (bit turned on based on suit of card)
b = bit turned on depending on rank of card
Using such a scheme, here are some bit pattern examples:
xxxAKQJT 98765432 CDHSrrrr xxPPPPPP
00001000 00000000 01001011 00100101 King of Diamonds
00000000 00001000 00010011 00000111 Five of Spades
00000010 00000000 10001001 00011101 Jack of Clubs
Now, for my evaluator function, I would pass in five integers,
each representing one of the five cards in the hand to evaluate.
We will label these c1 through c5.
First, we check to see if all the suits are the same. If so,
then we have either a Flush (or Straight Flush).
The quickest way to calculate this is by evaluating this:
c1 AND c2 AND c3 AND c4 AND c5 AND 0xF000
This is bit-wise AND, not a boolean AND, by the way. If the
above expression yields a value of zero, then we don't have
a Flush. If it is non-zero, then we do. Assuming we have
a Flush, I needed a fast way to calculate the hand's actual
value (i.e. convert it to a value from 1 to 7462). Table lookups
are one of the fastest ways to generate values based on index keys,
so I used the following coding technique.
First, I do another bitwise OR of all five cards, and then bit shift
that value 16 bits to the right:
q = (c1 OR c2 OR c3 OR c4 OR c5) >> 16
Note that if we have a Flush, then all five card ranks are
guaranteed to be unique. We should have exactly five bits
set in our shifted value. The smallest bit pattern would be
0x001F (decimal 31), and
the highest would be
0x1F00 (decimal 7936).
I created a lookup array containing 7937 elements, and then populated
it with the correct hand values based on each valid bit pattern.
Let's call this array flushes.
If you checked the value of
flushes, you should find
the value 9, because 9
is the distinct value for a Six-High Straight Flush. As another
example, let's say you were holding an AKQJ9 Flush. The
bit pattern for that hand would be
0x1E80, or decimal 7808.
This means that
flushes should hold
the value 323, which is the distinct value for AKQJ9 Flushes.
Obviously, there are a lot of entries in this array that will never
be accessed. That is the price of lookup tables. You gain speed
in favor of "wasted" space. However, since the highest distinct
hand value is 7462, we can make our array up of type short.
That means, our array will take up only 15874 bytes. Not very large
If we should determine that we do not have Flush or
Straight Flush hand, we move on to tackle Straight and
High Card hands. Again, we use a lookup table for speed.
We create another array, which will we will call
unique5. We use the
same index q as the index
into this new array. If the value found there is zero, then we
do not have a Straight or High Card hand. If it is
non-zero, then it holds the actual hand's distinct value. For
example, a King-High Straight has a bit pattern of
0x0F80, or decimal
3968. This means that
hold the value 1601. The absolute worst
poker hand of 75432 (unsuited) would yield
unique5[0x002F] = unique5,
With just two easy calculations and two quick lookups, we have
eliminated 2574 out of the 7462 possible distinct hand values.
That's a little over a third of all the hands. Now it's time
to use our trick with prime numbers. We take each card,
extract its prime number value, and then multiply them all
q = (c1 AND 0xFF) * (c2 AND 0xFF) * ... * (c5 AND 0xFF)
Again, these are bitwise ANDs. The resulting values will range
from the small to the very large. The smallest value is obtained
if you hold the hand 22223
(four Deuces with a Trey kicker).
Multiplying its prime values yields the number 48. The largest value
comes with holding AAAAK.
Doing the math, we get 104,553,157
(41 x 41 x 41 x 41 x 37). These numbers are way too large to use
a lookup table, so we must find some other method to convert these
values to their proper distinct hand value. Since there are only
4888 remaining hand values (7462 minus 2574), I decided to use
search algorithm. It's fast, with a complexity of O(log n).
I took the remaining 4888 hands, calculated the prime multiplication
value for each, placed them in an array, and then sorted it. When I
need to find a distinct hand value, after all previous efforts fail,
I do a binary search, which returns an index. That index is used
in a regular array, which holds the remaining distinct values.
That's it! I coded this algorithm, and ran it against some of the
other poker evaluators out there, and it beat them all.
I'm pretty proud of it, and kudos to you if you've actually read this far.
For your prize, you get to download the actual source code to
LATE BREAKING NEWS!!! Paul Senzee of Florida decided that he could
speed up my evaluator by using a pre-computed perfect hash function
instead of a binary search for those final 4888 hand values. He
says he obtained a speedup factor of 2.7x! Not too shabby! Anyway,
you can read about his optimization and download his new evaluator