Count the number of time a specific number occurs in a for loop-CodePudding

So i am trying to make a dice game that rolls the dice 10 times and is you get 2 1's you get extra points

This is my for loop

import random

rolls = 10
total = 0     

for i in range(1, rolls   1):
    n = random.randint(1, 6)
    total  = n
    print(f'Roll {i}: [{n}]')

How to make it tell me if there are 2 1's in the any of the rolls

CodePudding user response：

Two dices means two randomized input. So you'll need do two randints. Then you can check if both are 1.

import random

rolls = 10
total = 0     

for i in range(1, rolls   1):
    n1 = random.randint(1, 6)
    n2 = random.randint(1, 6)
    if n1 == 1 and n2 == 1:
        total  = 1

print(f'rolled {n} times. Got both 1s {total} times')

CodePudding user response：

You can keep track of how many 1's with a variable, let's call in numberOfOnes and increment it every time we roll a 1.

import random

rolls = 10
total = 0
numberOfOnes = 0    

for i in range(1, rolls   1):
    n = random.randint(1, 6)
    total  = n
    if n == 1:
        numberOfOnes  = 1
    print(f'Roll {i}: [{n}]. {numberOfOnes} 1\'s rolled so far!')

#here we'll check if we rolled 1's at least twice
if numberOfOnes >= 2:
    print("You got an extra point!")

If you want to handle two dices (rolled both 10 times), you can check @MSH's answer.