combinations - Python Poker hand single pair counter -


i wrote program below iterate through every possible poker hand , count how many of these hands single pair

a hand 5 cards.
single pair when 2 cards of same rank (number) , other 3 cards of different ranks e.g. (1,2,1,3,4)

i representing deck of cards list of numbers e.g.
- 1 = ace
- 2 = two
- 3 = three
...
- 11 = jack
- 12 = queen...

the program seems work find however, number of single pair hands finds = 1101984

but according multiple sources correct answer 1098240.

can see error in code is?

from itertools import combinations # generating deck deck = [] in range(52):     deck.append(i%13 + 1)  def paircount(hand):     paircount = 0     in hand:         count = 0         x in hand:             if x == i:                 count += 1         if count == 2:             paircount += .5 #adding 0.5 because each pair counted twice      return paircount  count = 0 in combinations(deck, 5): # loop through combinations of 5     if paircount(i) == 1:         count += 1  print(count)

the issue hand can contain following type of cards -

a 3 of kind , single pair

you calculating single pair well.

i modified code count number of hands such contains 3 of kind single pair together. code -

deck = [] in range(52):     deck.append((i//13 + 1, i%13 + 1))  def paircount(hand):     paircount = 0     threecount = 0     in hand:         count = 0         x in hand:             if x[1] == i[1]:                 count += 1         if count == 2:             paircount += .5 #adding 0.5 because each pair counted twice         if count == 3:             threecount += 0.33333333     return (round(paircount, 0) , round(threecount, 0))  count = 0 in combinations(deck, 5):     if paircount(i) == (1.0, 1.0):         count += 1

this counted number - 3744.

now, if subtract number number got - 1101984 - number expecting - 1098240 .


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