Python: Negative result when subtracting two positive values

[Bazooka9T]

Hey group,
I am new on here, and also to Python. I have been setting myself various tasks to try to learn more about this language and have a small query.
I set myself a "Cash Machine" task, and although I am still yet to add multiples of 10, overdraft etc.... I keep getting a negative value as a result which is literally "bugging" me *sticks fork in leg for the sad pun.
Any ideas on where exactly right under my nose I have booboo'd? Even with a positive result, the script outputs a negative value.

Thanks.

######################################
#!/usr/bin/python -tt
# -*- coding: cp1252 -*-

from random import choice
import sys, math

class BreakLoop:
    pin_attempts = "PIN failed to authorise on three attempts"
    card_returned = "Card returned to account holder"

class Account_Detail(object):
    def __init__(self, name, surname, balance, account, PIN):
        self.name = name
        self.surname = surname
        self.balance = balance
        self.account = account
        self.PIN = PIN

### Populate database
acc1 = Account_Detail('Jeff', 'Minter', 200.18, 789321, 1234)
acc2 = Account_Detail('Tim', 'Lambesis', 182.63, 147852, 7896)
acc3 = Account_Detail('Matt', 'Heafy', 350.58, 963258, 9874)

### Prep and setup random choice for entering card into machine
account1 = [acc1.name, acc1.surname, acc1.balance, acc1.account, acc1.PIN]
account2 = [acc2.name, acc2.surname, acc2.balance, acc2.account, acc2.PIN]
account3 = [acc3.name, acc3.surname, acc3.balance, acc3.account, acc3.PIN]

rand_acc = [account1, account2, account3]
card_choice = choice(rand_acc)

print card_choice[4]

def ins_card():
    count = 0
    while count < 3:
        print "Your account number is %i" % (card_choice[3])
        card_in = input ("Please enter your PIN: ")
        if card_in == (card_choice[4]):
            print "Hello, %s %s" %(card_choice[0], card_choice[1])
            print "Your account balance is £%.2f " % (card_choice[2])
            count = 4
        elif card_in == (''):
            print "No PIN entered"
        else:
            print "Incorrect PIN"
            count = count + 1
            
        if count == 3:
            print; print "PIN entry has exceeded three attempts" 
            sys.exit(BreakLoop.pin_attempts)  
       
def withdraw():        
    amount = input ("How much would you like to withdraw?: ")
    if amount <= card_choice[2]:
        #print amount
        #print card_choice[2]
        tot_bal = amount - card_choice[2]
        print; print "Dispensing £%i" % (amount)
        print "Your new balance is £%.2f" % (tot_bal)
        print "Have a nice day"
    else:
        funds = "You do not have sufficient funds available"
        print; print funds.upper() 
        sys.exit(BreakLoop.card_returned)
                
if __name__ == "__main__":
    ins_card()
    withdraw()
####################################

[bvdet]

Change this line:

tot_bal = amount - card_choice[2]

to this:

tot_bal = card_choice[2] - amount

[dwblas]

i.e. your logic is bass-ackwards

    if amount <= card_choice[2]:
         #print amount
         #print card_choice[2]
         tot_bal = amount - card_choice[2]

[Bazooka9T]

My mind told me to do that, but I'm fiddling at work and thought "nah! That's not it"
A serious lesson learnt here, and a pathetically stupid schoolboy error.

Thank you, I feel like a fool (because I am)

[bvdet]

You shouldn't feel that way. Sometimes the simplest errors are the hardest to find. The best way to learn is to make mistakes.

[dwblas]

I prefer to avoid the top if statement as it can be confusing sometimes and use instead

tot_bal = amount - card_choice[2] 
if tot_bal >= 0:

[Bazooka9T]

@bvdet I'm cool, I just kick myself when I miss the obvious. But as you say, these errors make us stronger.
@dwblas I will look at that, good tip!

I'm still expanding with this, I have found it a good project to test the things I have thus far learnt.
Thank you both for your input, I hope one day I will be assisting other newbies with such issues.