Algorithm to determine the winner of a Texas Hold'em Hand

Question

Ok, so I am making a Texas Hold\'em AI for my senior project. I\'ve created the gui and betting/dealing procedures, but I have reached the part where I need to determine who

Answer 1

An example of a ready made Texas Hold'em 7- and 5-card evaluator can be found here and further explained here. This might help you with performance. All feedback welcome at the e-mail address found therein.

Answer 2

The method used in ralu's post is by far the best alternative I've seen. I used this method in my own project, and its very fast.

Cliffs:

Do some preprocessing, to generate a table, containing one value for each distinct poker-hand. Make sure the table is sorted by hand-strength.

Each card-value has a corresponding prime-value. The table is indexed by the multiplication of each card-value in the hand. So to find the value of the hand AAAAK, you calculate the prime multiplication and use this as index for the table:

int prime = getPrime(hand); // Calculates A.getPrime()...*K.getPrime();
int value = table[prime];

(Sorry for the java syntax).

This way, AAAAK is the same hand as KAAAA, and you dont need a 5-dim table.

Note that you need to go through all combinations of the best 5 card hand, with the 7 cards you can choose from, to find the largest value, which is the real value of the hand.

You use a different table for flushes.

The table gets pretty beefy, as there are lots of wasted cells by this implementation. To counter this, you can create a map during preprocessing, which maps the large prime values to integer values, and use this as you source instead.

Answer 3

Monte Carlo? That's the first suggestion I see here. It's another senior project. Simple and slow, but otherwise you're probably looking at some complicated combinatorics that I won't pretend to know much about.

Answer 4

http://www.codingthewheel.com/archives/poker-hand-evaluator-roundup

Best algorithm you will get is 7 looks in lookup table of size 100 MB (if I remember correctly)

Answer 5

import itertools
from collections import Counter

# gets the most common element from a list
def Most_Common(lst):
    data = Counter(lst)
    return data.most_common(1)[0]



# gets card value from  a hand. converts A to 14,  is_seq function will convert the 14 to a 1 when necessary to evaluate A 2 3 4 5 straights
def convert_tonums(h, nums = {'T':10, 'J':11, 'Q':12, 'K':13, "A": 14}):
    for x in xrange(len(h)):

        if (h[x][0]) in nums.keys():

            h[x] = str(nums[h[x][0]]) + h[x][1]

    return h


# is royal flush
# if a hand is a straight and a flush and the lowest value is a 10 then it is a royal flush
def is_royal(h):
    nh = convert_tonums(h)
    if is_seq(h):
        if is_flush(h):
            nn = [int(x[:-1]) for x in nh]
            if min(nn) == 10:
                return True

    else:
        return False


# converts hand to number valeus and then evaluates if they are sequential  AKA a straight  
def is_seq(h):
    ace = False
    r = h[:]

    h = [x[:-1] for x in convert_tonums(h)]


    h = [int(x) for x in h]
    h = list(sorted(h))
    ref = True
    for x in xrange(0,len(h)-1):
        if not h[x]+1 == h[x+1]:
            ref =  False
            break

    if ref:
        return True, r

    aces = [i for i in h if str(i) == "14"]
    if len(aces) == 1:
        for x in xrange(len(h)):
            if str(h[x]) == "14":
                h[x] = 1

    h = list(sorted(h))
    for x in xrange(0,len(h)-1):
        if not h[x]+1 == h[x+1]:

            return False
    return True, r

# call set() on the suite values of the hand and if it is 1 then they are all the same suit
def is_flush(h):
    suits = [x[-1] for x in h]
    if len(set(suits)) == 1:
        return True, h
    else:
        return False


# if the most common element occurs 4 times then it is a four of a kind
def is_fourofakind(h):
    h = [a[:-1] for a in h]
    i = Most_Common(h)
    if i[1] == 4:
        return True, i[0]
    else:
        return False


# if the most common element occurs 3 times then it is a three of a kind
def is_threeofakind(h):
    h = [a[:-1] for a in h]
    i = Most_Common(h)
    if i[1] == 3:
        return True, i[0]
    else:
        return False


# if the first 2 most common elements have counts of 3 and 2, then it is a full house
def is_fullhouse(h):
    h = [a[:-1] for a in h]
    data = Counter(h)
    a, b = data.most_common(1)[0], data.most_common(2)[-1]
    if str(a[1]) == '3' and str(b[1]) == '2':
        return True, (a, b)
    return False

# if the first 2 most common elements have counts of 2 and 2 then it is a two pair
def is_twopair(h):
    h = [a[:-1] for a in h]
    data = Counter(h)
    a, b = data.most_common(1)[0], data.most_common(2)[-1]
    if str(a[1]) == '2' and str(b[1]) == '2':
        return True, (a[0], b[0])
    return False


#if the first most common element is 2 then it is a pair
# DISCLAIMER: this will return true if the hand is a two pair, but this should not be a conflict because is_twopair is always evaluated and returned first 
def is_pair(h):
    h = [a[:-1] for a in h]
    data = Counter(h)
    a = data.most_common(1)[0]

    if str(a[1]) == '2':
        return True, (a[0]) 
    else:
        return False

#get the high card 
def get_high(h):
    return list(sorted([int(x[:-1]) for x in convert_tonums(h)], reverse =True))[0]

# FOR HIGH CARD or ties, this function compares two hands by ordering the hands from highest to lowest and comparing each card and returning when one is higher then the other
def compare(xs, ys):
  xs, ys = list(sorted(xs, reverse =True)), list(sorted(ys, reverse = True))

  for i, c in enumerate(xs):
    if ys[i] > c:
        return 'RIGHT'
    elif ys[i] < c:
        return 'LEFT'

  return "TIE"


# categorized a hand based on previous functions
def evaluate_hand(h):

    if is_royal(h):
        return "ROYAL FLUSH", h, 10
    elif is_seq(h) and is_flush(h) :
        return "STRAIGHT FLUSH", h, 9 
    elif is_fourofakind(h):
        _, fourofakind = is_fourofakind(h)
        return "FOUR OF A KIND", fourofakind, 8
    elif is_fullhouse(h):
        return "FULL HOUSE", h, 7
    elif is_flush(h):
        _, flush = is_flush(h)
        return "FLUSH", h, 6
    elif is_seq(h):
        _, seq = is_seq(h)
        return "STRAIGHT", h, 5
    elif is_threeofakind(h):
        _, threeofakind = is_threeofakind(h)
        return "THREE OF A KIND", threeofakind, 4
    elif is_twopair(h):
        _, two_pair = is_twopair(h)
        return "TWO PAIR", two_pair, 3
    elif is_pair(h):
        _, pair = is_pair(h)
        return "PAIR", pair, 2 
    else:
        return "HIGH CARD", h, 1



#this monster function evaluates two hands and also deals with ties and edge cases
# this probably should be broken up into separate functions but aint no body got time for that
def compare_hands(h1,h2):
    one, two = evaluate_hand(h1), evaluate_hand(h2)
    if one[0] == two[0]:

        if one[0] =="STRAIGHT FLUSH":

            sett1, sett2 = convert_tonums(h1), convert_tonums(h2)
            sett1, sett2 = [int(x[:-1]) for x in sett1], [int(x[:-1]) for x in sett2]
            com = compare(sett1, sett2)

            if com == "TIE":
                return "none", one[1], two[1]
            elif com == "RIGHT":
                return "right", two[0], two[1]
            else:
                return "left", one[0], one[1]

        elif one[0] == "TWO PAIR":

            leftover1, leftover2 = is_twopair(h1), is_twopair(h2)
            twm1, twm2 = max([int(x) for x in list(leftover1[1])]), max([int(x) for x in list(leftover2[1])])
            if twm1 > twm2:
                return "left", one[0], one[1]
            elif twm1 < twm2:
                return "right", two[0], two[1]


            if compare(list(leftover1[1]), list(leftover2[1])) == "TIE":
                l1 = [x[:-1] for x in h1 if x[:-1] not in leftover1[1]]
                l2 = [x[:-1] for x in h2 if x[:-1] not in leftover2[1]]
                if int(l1[0]) == int(l2[0]):
                    return "none", one[1], two[1]
                elif int(l1[0]) > int(l2[0]):
                    return "left", one[0], one[1]
                else:
                    return "right", two[0], two[1]
            elif compare(list(leftover1[1]), list(leftover2[1]))  == "RIGHT":
                return "right", two[0], two[1]
            elif  compare(list(leftover1[1]), list(leftover2[1]))  == "LEFT":
                return "left", one[0], one[1]


        elif one[0] == "PAIR":
            sh1, sh2 = int(is_pair(h1)[1]), int(is_pair(h2)[1])
            if sh1 == sh2:

                c1 = [int(x[:-1]) for x in convert_tonums(h1) if not int(sh1) == int(x[:-1])]
                c2 = [int(x[:-1]) for x in convert_tonums(h2) if not int(sh1) == int(x[:-1])]
                if compare(c1, c2) == "TIE":
                    return "none", one[1], two[1]
                elif compare(c1, c2) == "RIGHT":
                    return "right", two[0], two[1]
                else:
                    return "left", one[0], one[1]




            elif h1 > h2:
                return "right", two[0], two[1]
            else:
                return "left", one[0], one[1]

        elif one[0] == 'FULL HOUSE':

            fh1, fh2 =  int(is_fullhouse(h1)[1][0][0]), int(is_fullhouse(h2)[1][0][0])
            if fh1 > fh2:
                return "left", one[0], one[1]
            else:
                return "right", two[0], two[1]
        elif one[0] == "HIGH CARD":
            sett1, sett2 = convert_tonums(h1), convert_tonums(h2)
            sett1, sett2 = [int(x[:-1]) for x in sett1], [int(x[:-1]) for x in sett2]
            com = compare(sett1, sett2)
            if com == "TIE":
                return "none", one[1], two[1]
            elif com == "RIGHT":
                return "right", two[0], two[1]
            else:
                return "left", one[0], one[1]



        elif len(one[1]) < 5:
            if max(one[1])  == max(two[1]):
                return "none", one[1], two[1]
            elif max(one[1]) > max(two[1]):
                return "left", one[0], one[1]
            else:
                return "right", two[0], two[1]
        else:
            n_one, n_two = convert_tonums(one[1]), convert_tonums(two[1])
            n_one, n_two = [int(x[:-1]) for x in n_one], [int(x[:-1]) for x in n_two]

            if max(n_one)  == max(n_two):
                return "none", one[1], two[1]
            elif max(n_one) > max(n_two):
                return "left", one[0], one[1]
            else:
                return "right", two[0], two[1]
    elif one[2] > two[2]:
        return "left", one[0], one[1]
    else:
        return "right", two[0], two[1]



'''
a = ['QD', 'KD', '9D', 'JD', 'TD'] 
b = ['JS', '8S', 'KS', 'AS', 'QS']
print compare_hands(a,b)
'''