Craps In Python
2021年9月7日Register here: http://gg.gg/vx7yx
R/learnpython: Subreddit for posting questions and asking for general advice about your python code. Basic Knowledge Python, Craps Game.
*Blackjack In Python
*Craps Game In Python
*Craps In Python
*Craps Game In Python
*Craps In Pythonewoks4life
*Simulate the game of Craps using Python. Write a program that plays 10 rounds of Craps, tracking how many times the player wins and the dealer wins. The rules of craps are as follows: The player rolls a pair of 6-sided dice and the value of the 2 dice are displayed to the user.
*If you think coding a dice in Python is hard you are complete wrong! You don’t need 800 lines to do it, 5 is more than enough. TAKE A LOOK AT THE MORE COMPLETE VERSION OF THIS POST OVER HERE: Handle your 1st Python project and make a Dice Simulator Web App. Welcome everybody! Today we will make a simple dice simulator from scratch.
*I don’t know how to play craps but i’ve looked up the rules and the first one says that if the come out roll is 2,3, or 12 the shooter craps out and they lose their money. I’m not sure how to implement this in code though. Also the way it is explained is a bit convoluted. If anyone knows what’s supposed to happen I’d appreciate the help.Hi there,
I had an idea for a new Craps prop bet and was trying to write some Python code that would simulate the bet, but ran into an issue.
My assumption was that if you flipped the rules of a bet and you flipped the payouts of the bet the house edge on that new hypothetical bet could be expressed by simply changing the sign of the house edge on the original bet.
Say you turned the hard 8 bet into a lay the hard 8 bet. You win on 7 or easy 8 before the next hard 8. You’re essentially acting like the casino at that point and the payouts go from 9:1 to 1:9. Shouldn’t the house edge on this new hypothetical bet be -9.09% since the house edge on hard 8 is 9.09%?
To test my assumption, I ran a 10 million round simulation of the traditional hard 8 bet and laying the hard 8 and was surprised to find that while the original hardway bet did show a house edge of ~ 9%, laying the hard 8 showed a house edge of ~ -1%.
I’m trying to determine if the mistake I’ve made is in the assumption that a house edge flips if you flip the payouts and the rules of the bet or if the mistake I’ve made is in the Python code I’ve written.
Any thoughts?
Returns:
Hard eight player edge: -9.11%
Lay hard eight player edge: 0.98%ThatDonGuyYour code appears correct.
Your mistake is, while you are ’acting like the casino’ when you lay the bet, the casino is betting 1/9 against your 1, rather than 1 against your 9, but your house edge is based on your bet rather than the casino’s.ewoks4lifeI’m sorry but I really don’t understand what you mean. Are you saying that this bet really would only be an advantage of 1% if the player could make it? I don’t see how that could be.
If the casino has a 9% advantage on a hard 8, surely the player would have a 9% advantage if they could book the same exact bet as the casino, right? Putting my $9 up against the casino’s $1 (which is the same scenario that plays out on a standard hard 8 bet, just reversed) is the same as putting my $1 up against the casino’s $(1/9).odiousgambitfair odds are 10:1 .. I’d guess if they ever let you play darkside you’d put up something like $15 to win $1
if you only had to put up $9 I agree that the player would get the casino’s exact edge for himself
in any case that has to be the most unheard of bet in a casino - to take the large end of the bet. I can’t think of anything like that for the moment
I take it back, the most unheard of thing is to intentionally allow a bet with a player edgethe next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. .. Arnold SnyderRSThanks for this post from:
I’m sorry but I really don’t understand what you mean. Are you saying that this bet really would only be an advantage of 1% if the player could make it? I don’t see how that could be.
If the casino has a 9% advantage on a hard 8, surely the player would have a 9% advantage if they could book the same exact bet as the casino, right? Putting my $9 up against the casino’s $1 (which is the same scenario that plays out on a standard hard 8 bet, just reversed) is the same as putting my $1 up against the casino’s $(1/9).
’House Edge’ is actually a bit of a misnomer.
The casino isn’t betting $1 against your $1. The casino is betting $9 against your $1. You just have a 9% disadvantage.
The casino has a 1% or so advantage on their $9 bet. You have a 9% (or so) disadvantage on your $1 bet.
IE: Player $1 bet: has 10 ways to lose $1 and 1 way to win $9. The total money wagered is $11. The sum of outcomes (10*-1 + 1*9) is -1.
-1/11 ~ -9%
The casino, on the other hand, is betting $9. 1 way to lose $9 and 10 ways to win $1. Total money wagered is $99. The sum of outcomes (-9*1 + 1*10) is +1.
+1/99 ~ +1%.ThatDonGuyThanks for this post from:
’House Edge’ is actually a bit of a misnomer.
The casino isn’t betting $1 against your $1. The casino is betting $9 against your $1. You just have a 9% disadvantage.
The casino has a 1% or so advantage on their $9 bet. You have a 9% (or so) disadvantage on your $1 bet.
This. +1. Whatever the term for ’exactly’ is this week.
The amount of the house edge is the same on both sides of the place bet because you are putting up 9x what the casino is.
The percentage, on the other hand, is different because each side is betting a different amount.ewoks4lifeThank you both for the explanation. It’s much clearer now. I looked over the code 10x and knew it was making the bet and payouts in the exact opposite way so figured it must have just been my lack of understanding the math.
in any case that has to be the most unheard of bet in a casino - to take the large end of the bet. I can’t think of anything like that for the moment
I take it back, the most unheard of thing is to intentionally allow a bet with a player edge
When it comes table games I think Craps has the largest lay bet available at 19 to 41 when laying the 4 or 10 (when factoring in the vig). I can’t think of anything with a smaller payout compared to the original bet.
You can lay much higher odds on sports. The infamous No Safety SB bet comes to mind.
This is just a thought experiment for me, not a serious idea that I’m going to try to sell to casinos or anything. They wouldn’t want this bet. Obviously if a casino did implement this it would not be at 1 to 9. It would be at 1 to 10 or 1 to 11. Much like how 99% of sports betters don’t lay -900 on No Safety in the SB, 99% of players would never lay 1 to 9 on a bet at a Craps table even if they had an advantage so no players would ever lay 1 to 10 or 1 to 11 on a bet where the casino had an advantage so it would just never be implemented.
I’m just messing around trying to intersect my interest in (basic) programming with my interest in gambling.
In this chapter we will develop two programs involving loops that are slightly more involved than the simple examples from the last chapter.1.1. Example: the game of craps¶
For the first example, we’ll write a program that allows a user to play the game of “craps”. It is played by rolling a pair of dice. Normally, the player (and spectators) bet on the outcome. We assume that a player starts with a fixed amount of money, called the “bankroll.”
Craps is played like this. Each round of the game has two “phases”:
In the first phase, you roll the dice once. If the result is 7 or 11, you win immediately and the round ends. If the result is 2, 3, or 12, you lose immediately, and the round ends. In all other cases the number you roll (which has to be a 4, 5, 6, 8, 9, or 10) is called the “point”, and you go on to the second phase.
In the second phase you keep rolling the dice until you roll a 7 or you roll the value of the “point” from the previous phase. If you roll a 7 first, you lose, and if you roll the point first, you win.
The payoff for the bets is always one-to-one: if you bet $10 and win, the bank gives you $10.
We can simulate rolling the dice using the randint function. To make it interactive, we’ll use an input statement to pause until the player presses the “Enter” key. Then we just generate two random values 1 through 6 and add them together.import random# Returns the value of a simulated roll of two dicedef roll_dice(): input(’Press ENTER to roll the dice..’) a = random.randint(1, 6) b = random.randint(1, 6) return a + bprint(roll_dice())
We’ll start by writing a function to play a single round. We can assume that the amount of the bet is a parameter to the function, and the return value of the function is the amount won, where a negative number means a loss. The logic for the first phase just needs a conditional statement.import randomdef play_one_round(bet): # Returns the value of a simulated roll of two dice def roll_dice(): input(’Press ENTER to roll the dice..’) a = random.randint(1, 6) b = random.randint(1, 6) return a + b roll = roll_dice() # First phase: 7 or 11 wins, 2, 3, or 12 loses if roll 7 or roll 11: print(’You win!’) return bet elif roll 2 or roll 3 or roll 12: print (’Sorry, you lose.’) return -bet else: point = roll print() print (’The point is now’, point, ’.’)play_one_round(100000000000)
For the second phase, the player has to keep rolling the dice until she rolls a 7 or rolls the point. There is no way to predict how long that might take, right? So we’ll need a while-loop.Blackjack In Python
Step 1: What is the repeated action?
Step 2: How do we start out?Rolling the dice the first time.
Step 3: How do we know whether to keep doing it?
When the loop finishes, we know that the value of ?roll? Poker table casino near me. is either equal to 7 or equal to the point. We just have to check which one, in order to determine whether the player won or lost.
Here is the complete function:import random# Returns the value of a simulated roll of two dicedef roll_dice(): input(’Press ENTER to roll the dice..’) a = random.randint(1, 6) b = random.randint(1, 6) return a + b# Plays one round of craps using the given amount as the bet. Returns the# amount won as a positive number or the amount lost as a negative number.def play_one_round(bet): roll = roll_dice() # First phase: 7 or 11 wins, 2, 3, or 12 loses if roll 7 or roll 11: print (’You win!’) return bet elif roll 2 or roll 3 or roll 12: print (’Sorry, you lose.’) return -bet else: point = roll print() print (’The point is now’, point, ’.’) # Second phase roll = roll_dice() print (’You rolled ’, roll) while roll != 7 and roll != point: roll = roll_dice() print (’You rolled ’, roll) # after loop, roll is 7 or is equal to point if roll 7: print (’Sorry, you lose.’) return -bet else: print (’You win!’) return betplay_one_round(100000000000)
To make this into a complete application, we need a user interface. The interface should allow the user to play again or quit, it should allow her to choose how much to bet, and should keep track of how much she’s won. This could be similar to the user interface for the number-guessing game in the last chapter. This part is left as an exercise.1.2. Example: a loan table¶
In this section, the problem we are solving is based on the following scenario. Suppose you buy a car and you are making monthly payments. Two years later, you discover you need to trade it in for a minivan. How much do you still owe on your car?
The answer to a question like this is found by creating something called an “amortization table” for the loan. This is just a table that shows for each payment, how much went to pay interest, how much went towards paying off what you owe, called the principal, and how much you still owe, called the balance.
As always, we’ll start with a small concrete example. Suppose you borrow $100 at an interest rate of 10% per month, and you make monthly payments of $30. A month goes by, and you make your first payment. After that, how much do you owe?
The interest you owed for the month is 10% of 100 dollars, or 10 dollars.
The amount paid on your loan is 30 - 10 , or 20 dollars.
The balance you owe is now 100 - 20, or 80 dollars.
We can summarize all this by writing one line in a table:
Then, you make your second payment.
Next month you make another payment.
The process is similar for the next payment.
But now something different happens.
The interest you owe is 10% of 7.18, or 72 cents.The total amount you still owe is the 7.18 balance + 72 cents interest, which is a total of 7.90.Another 30 dollar payment would be too much! So you make a final payment of 7.90, and then the balance is zero
When will online gambling be legal in pa. We’ll put the code into a function. The function needs three pieces of information: the amount of the loan, the payment amount, and the interest rate, so our function will have three parameters.
defprint_loan_table(amount,payment,annual_rate):
Let?s analyze what we have done and try to write the code.
Step 1: what is the repeated action?
Look at the calculations we did over and over again:
Introducing some appropriate variable names, we can write these steps by following what we did for the examples. For instance, in the first step, we had a balance of 100 dollars. We found the interest by multiplying the balance by the monthly rate of 10 percent. We found the amount paid on the principal by subracting the interest from the payment amount. We found the new balance by subtracting the principal from the balance.
To print a line of the table, we?ll use a format string with 12 spaces per column. Using a format specifier of ?percent 12 point 2 f? will round everything to two decimal places.print(’%12.2f%12.2f%12.2f’%(interest,principal,balance))
That takes care of the loop body.
Step 2: how do we start out?
Initially the balance is the original amount borrowed. The monthly rate is the annual rate, divided by 12. At this point we have this much written. The statements representing the “repeated action” become the loop body.Craps Game In Python
Step 3: How do we know whether to keep doing it?
Looking back at our example, we knew it was the end when the amount we owed was only 7.90, which was less than the payment. So the loop should look something like this:Craps In Python
How did we know the amount owed? It was the balance, plus the interest for the month, that is:
balance+balance*monthly_rate
So we could write the condition asCraps Game In Python
while(balance+balance*monthly_rate)<=payment:Craps In Python
That takes care of the loop, but we’re not quite done. We still have to write the very last line of the table showing the final payment.
The complete function is shown below.# Prints out an amortization tabledef print_loan_table(amount, payment, annual_rate): balance = amount monthly_rate = annual_rate / 12.0 while payment <= balance + balance * monthly_rate: interest = balance * monthly_rate principal = payment - interest balance = balance - principal print (’%12.2f%12.2f%12.2f’ % (interest, principal, balance)) # print last line interest = balance * monthly_rate print (’%12.2f%12.2f%12.2f’ % (interest, balance, 0))print_loan_table(100, 20, 1.2)
Register here: http://gg.gg/vx7yx
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R/learnpython: Subreddit for posting questions and asking for general advice about your python code. Basic Knowledge Python, Craps Game.
*Blackjack In Python
*Craps Game In Python
*Craps In Python
*Craps Game In Python
*Craps In Pythonewoks4life
*Simulate the game of Craps using Python. Write a program that plays 10 rounds of Craps, tracking how many times the player wins and the dealer wins. The rules of craps are as follows: The player rolls a pair of 6-sided dice and the value of the 2 dice are displayed to the user.
*If you think coding a dice in Python is hard you are complete wrong! You don’t need 800 lines to do it, 5 is more than enough. TAKE A LOOK AT THE MORE COMPLETE VERSION OF THIS POST OVER HERE: Handle your 1st Python project and make a Dice Simulator Web App. Welcome everybody! Today we will make a simple dice simulator from scratch.
*I don’t know how to play craps but i’ve looked up the rules and the first one says that if the come out roll is 2,3, or 12 the shooter craps out and they lose their money. I’m not sure how to implement this in code though. Also the way it is explained is a bit convoluted. If anyone knows what’s supposed to happen I’d appreciate the help.Hi there,
I had an idea for a new Craps prop bet and was trying to write some Python code that would simulate the bet, but ran into an issue.
My assumption was that if you flipped the rules of a bet and you flipped the payouts of the bet the house edge on that new hypothetical bet could be expressed by simply changing the sign of the house edge on the original bet.
Say you turned the hard 8 bet into a lay the hard 8 bet. You win on 7 or easy 8 before the next hard 8. You’re essentially acting like the casino at that point and the payouts go from 9:1 to 1:9. Shouldn’t the house edge on this new hypothetical bet be -9.09% since the house edge on hard 8 is 9.09%?
To test my assumption, I ran a 10 million round simulation of the traditional hard 8 bet and laying the hard 8 and was surprised to find that while the original hardway bet did show a house edge of ~ 9%, laying the hard 8 showed a house edge of ~ -1%.
I’m trying to determine if the mistake I’ve made is in the assumption that a house edge flips if you flip the payouts and the rules of the bet or if the mistake I’ve made is in the Python code I’ve written.
Any thoughts?
Returns:
Hard eight player edge: -9.11%
Lay hard eight player edge: 0.98%ThatDonGuyYour code appears correct.
Your mistake is, while you are ’acting like the casino’ when you lay the bet, the casino is betting 1/9 against your 1, rather than 1 against your 9, but your house edge is based on your bet rather than the casino’s.ewoks4lifeI’m sorry but I really don’t understand what you mean. Are you saying that this bet really would only be an advantage of 1% if the player could make it? I don’t see how that could be.
If the casino has a 9% advantage on a hard 8, surely the player would have a 9% advantage if they could book the same exact bet as the casino, right? Putting my $9 up against the casino’s $1 (which is the same scenario that plays out on a standard hard 8 bet, just reversed) is the same as putting my $1 up against the casino’s $(1/9).odiousgambitfair odds are 10:1 .. I’d guess if they ever let you play darkside you’d put up something like $15 to win $1
if you only had to put up $9 I agree that the player would get the casino’s exact edge for himself
in any case that has to be the most unheard of bet in a casino - to take the large end of the bet. I can’t think of anything like that for the moment
I take it back, the most unheard of thing is to intentionally allow a bet with a player edgethe next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. .. Arnold SnyderRSThanks for this post from:
I’m sorry but I really don’t understand what you mean. Are you saying that this bet really would only be an advantage of 1% if the player could make it? I don’t see how that could be.
If the casino has a 9% advantage on a hard 8, surely the player would have a 9% advantage if they could book the same exact bet as the casino, right? Putting my $9 up against the casino’s $1 (which is the same scenario that plays out on a standard hard 8 bet, just reversed) is the same as putting my $1 up against the casino’s $(1/9).
’House Edge’ is actually a bit of a misnomer.
The casino isn’t betting $1 against your $1. The casino is betting $9 against your $1. You just have a 9% disadvantage.
The casino has a 1% or so advantage on their $9 bet. You have a 9% (or so) disadvantage on your $1 bet.
IE: Player $1 bet: has 10 ways to lose $1 and 1 way to win $9. The total money wagered is $11. The sum of outcomes (10*-1 + 1*9) is -1.
-1/11 ~ -9%
The casino, on the other hand, is betting $9. 1 way to lose $9 and 10 ways to win $1. Total money wagered is $99. The sum of outcomes (-9*1 + 1*10) is +1.
+1/99 ~ +1%.ThatDonGuyThanks for this post from:
’House Edge’ is actually a bit of a misnomer.
The casino isn’t betting $1 against your $1. The casino is betting $9 against your $1. You just have a 9% disadvantage.
The casino has a 1% or so advantage on their $9 bet. You have a 9% (or so) disadvantage on your $1 bet.
This. +1. Whatever the term for ’exactly’ is this week.
The amount of the house edge is the same on both sides of the place bet because you are putting up 9x what the casino is.
The percentage, on the other hand, is different because each side is betting a different amount.ewoks4lifeThank you both for the explanation. It’s much clearer now. I looked over the code 10x and knew it was making the bet and payouts in the exact opposite way so figured it must have just been my lack of understanding the math.
in any case that has to be the most unheard of bet in a casino - to take the large end of the bet. I can’t think of anything like that for the moment
I take it back, the most unheard of thing is to intentionally allow a bet with a player edge
When it comes table games I think Craps has the largest lay bet available at 19 to 41 when laying the 4 or 10 (when factoring in the vig). I can’t think of anything with a smaller payout compared to the original bet.
You can lay much higher odds on sports. The infamous No Safety SB bet comes to mind.
This is just a thought experiment for me, not a serious idea that I’m going to try to sell to casinos or anything. They wouldn’t want this bet. Obviously if a casino did implement this it would not be at 1 to 9. It would be at 1 to 10 or 1 to 11. Much like how 99% of sports betters don’t lay -900 on No Safety in the SB, 99% of players would never lay 1 to 9 on a bet at a Craps table even if they had an advantage so no players would ever lay 1 to 10 or 1 to 11 on a bet where the casino had an advantage so it would just never be implemented.
I’m just messing around trying to intersect my interest in (basic) programming with my interest in gambling.
In this chapter we will develop two programs involving loops that are slightly more involved than the simple examples from the last chapter.1.1. Example: the game of craps¶
For the first example, we’ll write a program that allows a user to play the game of “craps”. It is played by rolling a pair of dice. Normally, the player (and spectators) bet on the outcome. We assume that a player starts with a fixed amount of money, called the “bankroll.”
Craps is played like this. Each round of the game has two “phases”:
In the first phase, you roll the dice once. If the result is 7 or 11, you win immediately and the round ends. If the result is 2, 3, or 12, you lose immediately, and the round ends. In all other cases the number you roll (which has to be a 4, 5, 6, 8, 9, or 10) is called the “point”, and you go on to the second phase.
In the second phase you keep rolling the dice until you roll a 7 or you roll the value of the “point” from the previous phase. If you roll a 7 first, you lose, and if you roll the point first, you win.
The payoff for the bets is always one-to-one: if you bet $10 and win, the bank gives you $10.
We can simulate rolling the dice using the randint function. To make it interactive, we’ll use an input statement to pause until the player presses the “Enter” key. Then we just generate two random values 1 through 6 and add them together.import random# Returns the value of a simulated roll of two dicedef roll_dice(): input(’Press ENTER to roll the dice..’) a = random.randint(1, 6) b = random.randint(1, 6) return a + bprint(roll_dice())
We’ll start by writing a function to play a single round. We can assume that the amount of the bet is a parameter to the function, and the return value of the function is the amount won, where a negative number means a loss. The logic for the first phase just needs a conditional statement.import randomdef play_one_round(bet): # Returns the value of a simulated roll of two dice def roll_dice(): input(’Press ENTER to roll the dice..’) a = random.randint(1, 6) b = random.randint(1, 6) return a + b roll = roll_dice() # First phase: 7 or 11 wins, 2, 3, or 12 loses if roll 7 or roll 11: print(’You win!’) return bet elif roll 2 or roll 3 or roll 12: print (’Sorry, you lose.’) return -bet else: point = roll print() print (’The point is now’, point, ’.’)play_one_round(100000000000)
For the second phase, the player has to keep rolling the dice until she rolls a 7 or rolls the point. There is no way to predict how long that might take, right? So we’ll need a while-loop.Blackjack In Python
Step 1: What is the repeated action?
Step 2: How do we start out?Rolling the dice the first time.
Step 3: How do we know whether to keep doing it?
When the loop finishes, we know that the value of ?roll? Poker table casino near me. is either equal to 7 or equal to the point. We just have to check which one, in order to determine whether the player won or lost.
Here is the complete function:import random# Returns the value of a simulated roll of two dicedef roll_dice(): input(’Press ENTER to roll the dice..’) a = random.randint(1, 6) b = random.randint(1, 6) return a + b# Plays one round of craps using the given amount as the bet. Returns the# amount won as a positive number or the amount lost as a negative number.def play_one_round(bet): roll = roll_dice() # First phase: 7 or 11 wins, 2, 3, or 12 loses if roll 7 or roll 11: print (’You win!’) return bet elif roll 2 or roll 3 or roll 12: print (’Sorry, you lose.’) return -bet else: point = roll print() print (’The point is now’, point, ’.’) # Second phase roll = roll_dice() print (’You rolled ’, roll) while roll != 7 and roll != point: roll = roll_dice() print (’You rolled ’, roll) # after loop, roll is 7 or is equal to point if roll 7: print (’Sorry, you lose.’) return -bet else: print (’You win!’) return betplay_one_round(100000000000)
To make this into a complete application, we need a user interface. The interface should allow the user to play again or quit, it should allow her to choose how much to bet, and should keep track of how much she’s won. This could be similar to the user interface for the number-guessing game in the last chapter. This part is left as an exercise.1.2. Example: a loan table¶
In this section, the problem we are solving is based on the following scenario. Suppose you buy a car and you are making monthly payments. Two years later, you discover you need to trade it in for a minivan. How much do you still owe on your car?
The answer to a question like this is found by creating something called an “amortization table” for the loan. This is just a table that shows for each payment, how much went to pay interest, how much went towards paying off what you owe, called the principal, and how much you still owe, called the balance.
As always, we’ll start with a small concrete example. Suppose you borrow $100 at an interest rate of 10% per month, and you make monthly payments of $30. A month goes by, and you make your first payment. After that, how much do you owe?
The interest you owed for the month is 10% of 100 dollars, or 10 dollars.
The amount paid on your loan is 30 - 10 , or 20 dollars.
The balance you owe is now 100 - 20, or 80 dollars.
We can summarize all this by writing one line in a table:
Then, you make your second payment.
Next month you make another payment.
The process is similar for the next payment.
But now something different happens.
The interest you owe is 10% of 7.18, or 72 cents.The total amount you still owe is the 7.18 balance + 72 cents interest, which is a total of 7.90.Another 30 dollar payment would be too much! So you make a final payment of 7.90, and then the balance is zero
When will online gambling be legal in pa. We’ll put the code into a function. The function needs three pieces of information: the amount of the loan, the payment amount, and the interest rate, so our function will have three parameters.
defprint_loan_table(amount,payment,annual_rate):
Let?s analyze what we have done and try to write the code.
Step 1: what is the repeated action?
Look at the calculations we did over and over again:
Introducing some appropriate variable names, we can write these steps by following what we did for the examples. For instance, in the first step, we had a balance of 100 dollars. We found the interest by multiplying the balance by the monthly rate of 10 percent. We found the amount paid on the principal by subracting the interest from the payment amount. We found the new balance by subtracting the principal from the balance.
To print a line of the table, we?ll use a format string with 12 spaces per column. Using a format specifier of ?percent 12 point 2 f? will round everything to two decimal places.print(’%12.2f%12.2f%12.2f’%(interest,principal,balance))
That takes care of the loop body.
Step 2: how do we start out?
Initially the balance is the original amount borrowed. The monthly rate is the annual rate, divided by 12. At this point we have this much written. The statements representing the “repeated action” become the loop body.Craps Game In Python
Step 3: How do we know whether to keep doing it?
Looking back at our example, we knew it was the end when the amount we owed was only 7.90, which was less than the payment. So the loop should look something like this:Craps In Python
How did we know the amount owed? It was the balance, plus the interest for the month, that is:
balance+balance*monthly_rate
So we could write the condition asCraps Game In Python
while(balance+balance*monthly_rate)<=payment:Craps In Python
That takes care of the loop, but we’re not quite done. We still have to write the very last line of the table showing the final payment.
The complete function is shown below.# Prints out an amortization tabledef print_loan_table(amount, payment, annual_rate): balance = amount monthly_rate = annual_rate / 12.0 while payment <= balance + balance * monthly_rate: interest = balance * monthly_rate principal = payment - interest balance = balance - principal print (’%12.2f%12.2f%12.2f’ % (interest, principal, balance)) # print last line interest = balance * monthly_rate print (’%12.2f%12.2f%12.2f’ % (interest, balance, 0))print_loan_table(100, 20, 1.2)
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