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Python

Posted by abrar hamzah
abrar hamzah
199 Points

Coin flip

So i want to make a coin flip program that if a user input number of flips, it will print the answer as many times the input is. import random

coin =random.randint(1, 2)

count=int(input("number of flips: "))

for goes in range(count): if coin == 1: print("The coin came up Heads")

else: print("The coin came up Tails")

the problem is, if the input is 6, the result is 6 of either heads or tails. How to make 6 of random results?

2 Answers

Alexander Davison
Alexander Davison
65,469 Points

You need to make a new random number for each flip, not set the random number in the beginning and use that same number for every flip.

Your fixed program might look like this:

import random

count = int(input("Enter the number of flips: "))

for goes in range(count):
    coin = random.randint(1, 2)
    if coin == 1:
        print("The coin came up as heads")
    else:
        print("The coin came up as tails")

Also, your program will cause a ValueError if the user enters something like "a". Python causes an error because "a" isn't a valid number, and can't be converted to a number. So, to mkae sure the user's input is safe, you can do something like this:

import random

while True:
    try:
        count = int(input("Enter the number of flips: "))
    except ValueError:
        print("That's not a valid number!")
    else:
        break

for goes in range(count):
    coin = random.randint(1, 2)
    if coin == 1:
        print("The coin came up as heads")
    else:
        print("The coin came up as tails")

This program will ask for input until the input is a valid number.

I hope this helps :grin:

Happy C0D1NG! :tada:

:dizzy: ~Alex :dizzy:

Chris Sehnert
Chris Sehnert
30,857 Points
Chris Sehnert
Chris Sehnert
30,857 Points

Hello Abrar...

when you call... coin =random.randint(1, 2) that represents a coin flip..... at this point your code only performs 1 flip.....and then asks "count" times..... "What was the result?"