count method in 'if' doesn't work - python -

i don't it, i'm trying count 2 in list , when this:

hand=['d2', 'h5', 's2', 'sk', 'cj', 'h7', 'cq', 'h9', 'd10', 'ck'] f=''.join(hand) count2=f.count('2') print count2 

it works , prints me 2 number of times 2 in list. when i'm putting in if doesn't work:

def same_rank(hand, n):     if hand.count('2')>n:         print hand.count('2')     else:         print 'bite me'    hand=['d2', 'h5', 's2', 'sk', 'cj', 'h7', 'cq', 'h9', 'd10', 'ck'] f=''.join(hand) n=raw_input('give n ') print same_rank(hand,n) 

if user gives n=1 supposed print 2 because number 2 twice in list , want more 1 is! why doesn't return that?

raw_input() returns string; strings sorted after numbers, 2 > '1' false:

>>> 2 > '1' false 

convert input integer first:

n = int(raw_input('give n ')) 

had used python 3, you'd have gotten exception instead:

>>> 2 > '1' traceback (most recent call last):   file "<stdin>", line 1, in <module> typeerror: unorderable types: int() > str() 

because python 3 has done away giving arbitrary types relative ordering.

next, don't pass in f, passing in hand, list:

>>> hand.count('2') 0 >>> f 'd2h5s2skcjh7cqh9d10ck' >>> f.count('2') 2 

you wanted pass in latter, function doesn't work otherwise.