Blackjack, what are the chances?

Warning: I have just used brute-force. It is very inefficient and slow, but it was the best I could do. Also, the answer is not precise, and keeps on changing every time the code is run. The precise answer can be obtained by running this code several times and taking an average.

I would like to see a better solution.

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from random import shuffle

# define the card ranks, and suits
ranks = [_ for _ in range(2, 11)] + ['JACK', 'QUEEN', 'KING', 'ACE']
suits = ['SPADE', 'HEART ', 'DIAMOND', 'CLUB']

def get_deck():
    """Return a new deck of cards."""
    return [[rank, suit] for rank in ranks for suit in suits]

# get a deck of cards, and randomly shuffle it
deck = get_deck()
shuffle(deck)

# issue the player and dealer their first two cards
player_hand = [deck.pop(), deck.pop()]


def isBJ(player_hand):
    if player_hand[0][0] == "ACE":
        if player_hand[1][0] == 10 or player_hand[1][0] == "JACK" or player_hand[1][0] == "QUEEN" or player_hand[1][0] == "KING":
            return True
        else:
            return False
    elif  player_hand[0][0] == 10 or player_hand[0][0] == "JACK" or player_hand[0][0] == "QUEEN" or player_hand[0][0] == "KING":
        if player_hand[1][0] == "ACE":
            return True
        else:
            return False
    else:
        return False

count = 0 # this is total number of times hand is dealt
count_True = 0 # number of hands that are Blackjack
count_False = 0 # number of hands that are not Blackjack

while count < 10**6:
    print player_hand
    print isBJ(player_hand)
    if isBJ(player_hand):
        count_True += 1
    else:
        count_False += 1
    deck = get_deck()
    shuffle(deck)
    player_hand = [deck.pop(), deck.pop()]
    count += 1

print count
print count_True
print count_False

To get $\hat{f} _{\text{blackjack}}$ , we divide count_True by count .

When I ran this code, I got count_True $= 48215$ , so the fraction $\hat{f} _{\text{blackjack}} \approx 0.0482$