Coin-flipping game

Question

I'm 13 years old and turning 14 years soon! I Coded this simple text game in around 30 minutes, can someone rate it? and give some feedback

import random
import time
current_coin = 500
def play():
    computer2 = random.choice(['r','p','s'])
    computer1 = random.choice(['r','p','s'])
    if computer2 == computer1:
        return 'tie'
    if is_win(computer2, computer1):
        return 'computer2 wins!!'
    if is_lose(computer2, computer1):
        return 'computer1 wins!!'
    else:
        return 'it looks like you made a typo! try again'

def is_win(computer2, computer1):
    if (computer2 == "r" and computer1 == "s") or (computer2 == "s" and computer1 == "p") or (computer2 == "p" and computer1 == 'r'):
        return True

def is_lose(computer2, computer1):
    if (computer2 == "s" and computer1 == "r") or (computer2 == "p" and computer1 == "s") or (computer2 == "r" and computer1 == 'p'):
        return True

while True:
    print('welcome to rock paper scissors betting simulator!\n')
    time.sleep(1)
    print('1 = computer1\n')
    time.sleep(1)
    print('2 = computer2\n')
    time.sleep(1)
    beton = str(input('please type 1 or 2 to bet!    \n'))
    coinbet = str(input('you currently have ' + str(int(current_coin)) + ' coins! how much would you like to bet!!\n'))
    if int(str(coinbet)) > int(current_coin):
        print('you dont have that much?\n')
        coinbet = str(input('you currently have ' + str(int(current_coin)) + ' coins! how much would you like to bet!!\n'))
    else:
        print('you betted ' + str(int(coinbet)) + ' !')
    result = play()
    time.sleep(0.1)
    if result == 'tie':
        print('welp! they tied! you have ' + str(int(current_coin)) + ' left!!\n')
    if result == 'computer2 wins!!':
        print('welp, since computer2 won. . .\n')
        if beton == '1':
            print('you lost! you now have ' + str(int(current_coin) - int(coinbet)) + '!!')
            current_coin = current_coin - int(coinbet)
        if beton == '2':
            print('you won! you now have ' + str(int(current_coin) + int(coinbet)) + '!!')
            current_coin = current_coin + int(coinbet)
    if result == 'computer1 wins!!':
        print('welp, since computer1 won. . .\n')
        if beton == '2':
            print('you lost! you now have ' + str(int(current_coin) - int(coinbet)) + '!!')
            current_coin = current_coin - int(coinbet)
        if beton == '1':
            print('you won! you now have ' + str(int(current_coin) + int(coinbet)) + '!!')
            current_coin = current_coin + int(coinbet)      
    if current_coin <= 0:
        print('you have no coins left! what a loser!')
        break

Answer 1

Good start! I like that is_win() and is_lose() are separate functions. They could be improved by returning the boolean directly. In general, if you have code that looks like if condition: return True else: Return False, it can be written as return condition directly.

Also, is_lose() can be written as is_win() with the parameters swapped

def is_win(computer2, computer1):
    return (
        (computer2 == "r" and computer1 == "s") or 
        (computer2 == "s" and computer1 == "p") or 
        (computer2 == "p" and computer1 == 'r')
    )

def is_lose(computer2, computer1):
    return is_win(computer1,computer2)

This breaks if the user enters an invalid amount the second time.

beton = str(input('please type 1 or 2 to bet!    \n'))
coinbet = str(input('you currently have ' + str(int(current_coin)) + ' coins! how much would you like to bet!!\n'))
if int(str(coinbet)) > int(current_coin):
    print('you dont have that much?\n')
    coinbet = str(input('you currently have ' + str(int(current_coin)) + ' coins! how much would you like to bet!!\n'))
else:
    print('you betted ' + str(int(coinbet)) + ' !')

This would not be an issue if it was written like so:

coin_bet = None
while coin_bet == None:
    coin_bet_attempt = int(input('you currently have ' + str(int(current_coin)) + ' coins! how much would you like to bet!!\n'))
    if coin_bet_attempt > current_coin:
        print("You don't have that much?")
    else:
        coin_bet = coin_bet_attempt
print('you betted ' + str(coin_bet) + ' !')

Better yet, one could write a function that ONLY exits when a valid input occurs.

def get_coin_bet_from_user(current_coin):
    while True:
        bet = int(input('you currently have ' + str(int(current_coin)) + ' coins! how much would you like to bet!!\n'))
        if bet > current_coin:
            print("You don't have that much?")
        else:
            return bet

# called like so
coinbet = get_coin_bet_from_user(current_coin)

Answer 2

Your

def is_win(computer2, computer1):
    if (computer2 == "r" and computer1 == "s") or (computer2 == "s" and computer1 == "p") or (computer2 == "p" and computer1 == 'r'):
        return True

could be shorter (and always return a bool, instead of True or None):

def is_win(computer2, computer1):
    return (computer2, computer1) in [('r', 's'), ('s', 'p'), ('p', 'r')]

or

def is_win(computer2, computer1):
    return computer2 + computer1 in ['rs', 'sp', 'pr']

or

def is_win(computer2, computer1):
    return computer2 + computer1 in 'rspr'

Not sure why you have this:

    else:
        return 'it looks like you made a typo! try again'

How is that ever supposed to be reached? Both computer2 and computer1 are random.choice(['r','p','s']), which doesn't make typos.