Question

Practice: • function • loops • if condition • random Roll a dice many times, you will see that each face value will count of roughly 1/6 • write a function • roll a dice once, return value will be 1-6, • use your functin I call your function 6000 times • count how many times of face 1 • count how many times of face 2 • print out the numbers of total face 1, 2, ..., 6 you counted • I wrote a flip coin code for your reference, you can make the rolling dice similarily. • function return 1/0 • call function 10000 times and print out

Answer 1

1 import random 3- def roll_dice(): x = random.random() if x< 1/6: return 1 if x < 2/6: return 2 if x < 3/6: return 3 if x < 4/6: return 4 if x < 5/6: return 5 return 6 ones = 0 18 twos = 0 threes = 0 20 fours = 0 21 fives = 0 sixes = 0 WNnmmmmmmmmm amun 24 - for i in range (6000): 25 roll = roll_dice 26 - if roll == 1: ones += 1 elif roll == 2: twos += 1 elif roll == 3: threes += 1 elif roll == 4: fours += 1 elif roll == 5: fives += 1 elif roll == 6: sixes += 1 39 40 41 42 43 44 print("1s =", ones) print("2s =", twos) print("3s =", threes) print("4s +", fours) print("5s =", fives) print("6s =", sixes)

CODE

import random

def roll_dice():
x = random.random()
if x < 1/6:
return 1
if x < 2/6:
return 2
if x < 3/6:
return 3
if x < 4/6:
return 4
if x < 5/6:
return 5
return 6

ones = 0
twos = 0
threes = 0
fours = 0
fives = 0
sixes = 0

for i in range(6000):
roll = roll_dice()
if roll == 1:
ones += 1
elif roll == 2:
twos += 1
elif roll == 3:
threes += 1
elif roll == 4:
fours += 1
elif roll == 5:
fives += 1
elif roll == 6:
sixes += 1

print("1s =", ones)
print("2s =", twos)
print("3s =", threes)
print("4s =", fours)
print("5s =", fives)
print("6s =", sixes)

SAMPLE OUTPUT

1s = 1013
2s = 998
3s = 1028
4s = 1007
5s = 992
6s = 962

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