'Python blackjack: How do I structure an if statement to determine whether an Ace should be turned into 1?

I'm trying to solve a simple coding exercise: Given three integers between 1 and 11, if their sum is less than or equal to 21, return their sum. If their sum exceeds 21 and there are elevens, reduce the 11's. Finally, if the sum (even after adjustment) exceeds 21, return 'BUST'

def blackjack(a,b,c):
    for i in range(0,12):
        if sum((a,b,c)) <= 21:
            return sum((a,b,c))
        elif sum ((a,b,c)) > 21 and 11 in (a,b,c):
            if a == 11:
                a -= 10
                continue
                if b == 11 and sum((a,b,c)) > 21:
                    b -= 10
                    continue
                    if c == 11 and sum((a,b,c)) > 21:
                        c -=10
                        break
                        return sum((a,b,c))

        else:
            return 'Bust'

print(blackjack(11,11,11))

I expected the output of print(blackjack(11,11,11)) to be 13, but I get none.

python python-2.7

Solution 1:^[1]

def blackjack(a,b,c):

    Ace_count = [a,b,c].count(11)

    total = sum([a,b,c])

    if total <= 21:
        return ("Not busted", total)

    while total > 21:
        if Ace_count > 0:
            Ace_count -= 1
            total -= 10
        else:
            return "Bust"

    return ("Not busted", total)

print(blackjack(11,11,11))

I think this should work, returns:

('Not busted', 13)

Solution 2:^[2]

def blackjack(a,b,c):  
    total = a+b+c
    if total <= 21:     
        return total
    elif 11 in [a,b,c] and total > 21:   
        new_total = total-10     
        if new_total > 21:    
            return 'Bust' 
        else:         
            return new_total   
    else: 
        return 'Bust'

Solution 3:^[3]

def func(x,y,z):
    if sum([x+y+z]) <= 21:
        return sum([x,y,z])
    elif 11 in (x, y, z) and sum((x, y, z)) <= 31:
        i = 0
        while 11 in [x,y,z] and sum([x,y,z]) <= 31:
            i -= 10
        return sum([x,y,z]) - i
    else:
        return 'bust'

print(func(9,9,9)) # bust

print(func(11,11,11)) # 13

print(func(11,10,9,9)) # bust

Solution 4:^[4]

def blackjack(x,y,z):
    tot=int(x+y+z)
    if tot < 21:
        return tot
    
    elif tot > 21 and x == 11 or y == 11 or z == 11:
        tot2=tot-10
        return tot2
    
    else:
        return "BUST"
        
    
    
print(blackjack(5, 6, 7))
print(blackjack(9, 9, 9))
print(blackjack(9, 9, 11))

Solution 5:^[5]

def blackjack(numbers):
    if len(numbers) == 3 and max(numbers) <= 11 and min(numbers)>=1:
        sum_n = sum(numbers)
        res =[]
        if (sum_n  <= 21):
            res = sum_n 
        else:
            if 11 in numbers:
                res =sum_n-10
            else:
                res = "BUST"
        return res
    else:
        return "numbers length should be 3 and max is 11 and min 1"

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution	Source
Solution 1	DarkLeader
Solution 2	David Buck
Solution 3	Eric Jin
Solution 4	faisal almashharawi
Solution 5	prashanth

'Python blackjack: How do I structure an if statement to determine whether an Ace should be turned into 1?

Solution 1:[1]

Solution 2:[2]

Solution 3:[3]

Solution 4:[4]

Solution 5:[5]