CODEWITHSUNDEEP
pythonloops
user4378959
user4378959

Reputation: 19

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

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

Upvotes: 0

Views: 61

Answers (2)

niaei
niaei

Reputation: 2399

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')

Upvotes: 1

Neox
Neox

Reputation: 279

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.

Upvotes: 0

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