Detecting if a random integer is odd or even using an if statement

-3

I am trying to make a dice game and I want to make it so it will detect if the number on the dice is odd or even. This is what I tried and it didn't work.

from random import randrange



import random



import time



Even = [2, 4, 6]



Odd = [1, 3, 5]



x = random.randint(1, 6)



print("Rolling Dice...")



print("Your number is...." + str(x))



if str(x) == Even:



  print("It is an Even number!")



if str(x) == Odd:



  print("It is an Odd number!")

I need to know how to do the if statments and to make it detect if it's even or odd.

edited Nov 20 '18 at 17:32

jonrsharpe

78k11106215

asked Nov 20 '18 at 17:25

RichardT

1

Possible duplicate of stackoverflow.com/questions/21837208/…

– sla3k
Nov 20 '18 at 17:33

2

Just use a modulo 2 if something % 2 == 0 # this is even and check for the remainder. Odd numbers will have the remainder 1 and even numbers zero.

– Dennis Kuypers
Nov 20 '18 at 17:35

add a comment |

-3

I am trying to make a dice game and I want to make it so it will detect if the number on the dice is odd or even. This is what I tried and it didn't work.

from random import randrange



import random



import time



Even = [2, 4, 6]



Odd = [1, 3, 5]



x = random.randint(1, 6)



print("Rolling Dice...")



print("Your number is...." + str(x))



if str(x) == Even:



  print("It is an Even number!")



if str(x) == Odd:



  print("It is an Odd number!")

I need to know how to do the if statments and to make it detect if it's even or odd.

edited Nov 20 '18 at 17:32

jonrsharpe

78k11106215

asked Nov 20 '18 at 17:25

RichardT

1

Possible duplicate of stackoverflow.com/questions/21837208/…

– sla3k
Nov 20 '18 at 17:33

2

Just use a modulo 2 if something % 2 == 0 # this is even and check for the remainder. Odd numbers will have the remainder 1 and even numbers zero.

– Dennis Kuypers
Nov 20 '18 at 17:35

add a comment |

-3

I am trying to make a dice game and I want to make it so it will detect if the number on the dice is odd or even. This is what I tried and it didn't work.

from random import randrange



import random



import time



Even = [2, 4, 6]



Odd = [1, 3, 5]



x = random.randint(1, 6)



print("Rolling Dice...")



print("Your number is...." + str(x))



if str(x) == Even:



  print("It is an Even number!")



if str(x) == Odd:



  print("It is an Odd number!")

I need to know how to do the if statments and to make it detect if it's even or odd.

edited Nov 20 '18 at 17:32

jonrsharpe

78k11106215

asked Nov 20 '18 at 17:25

RichardT

I am trying to make a dice game and I want to make it so it will detect if the number on the dice is odd or even. This is what I tried and it didn't work.

from random import randrange



import random



import time



Even = [2, 4, 6]



Odd = [1, 3, 5]



x = random.randint(1, 6)



print("Rolling Dice...")



print("Your number is...." + str(x))



if str(x) == Even:



  print("It is an Even number!")



if str(x) == Odd:



  print("It is an Odd number!")

I need to know how to do the if statments and to make it detect if it's even or odd.

python python-3.x

edited Nov 20 '18 at 17:32

jonrsharpe

78k11106215

asked Nov 20 '18 at 17:25

RichardT

edited Nov 20 '18 at 17:32

jonrsharpe

78k11106215

asked Nov 20 '18 at 17:25

RichardT

edited Nov 20 '18 at 17:32

jonrsharpe

78k11106215

edited Nov 20 '18 at 17:32

jonrsharpe

78k11106215

edited Nov 20 '18 at 17:32

jonrsharpe

78k11106215

asked Nov 20 '18 at 17:25

RichardT

asked Nov 20 '18 at 17:25

RichardT

asked Nov 20 '18 at 17:25

RichardT

1

Possible duplicate of stackoverflow.com/questions/21837208/…

– sla3k
Nov 20 '18 at 17:33

2

Just use a modulo 2 if something % 2 == 0 # this is even and check for the remainder. Odd numbers will have the remainder 1 and even numbers zero.

– Dennis Kuypers
Nov 20 '18 at 17:35

add a comment |

1

Possible duplicate of stackoverflow.com/questions/21837208/…

– sla3k
Nov 20 '18 at 17:33

2

Just use a modulo 2 if something % 2 == 0 # this is even and check for the remainder. Odd numbers will have the remainder 1 and even numbers zero.

– Dennis Kuypers
Nov 20 '18 at 17:35

Possible duplicate of stackoverflow.com/questions/21837208/…

– sla3k
Nov 20 '18 at 17:33

Just use a modulo 2 if something % 2 == 0 # this is even and check for the remainder. Odd numbers will have the remainder 1 and even numbers zero.

– Dennis Kuypers
Nov 20 '18 at 17:35

add a comment |

2 Answers
2

active

oldest

votes

For checking number being odd or even, you may not need the Even, Odd lists, you might check out the following code:

from random import randrange



import random



import time



x = random.randint(1, 6)



print("Rolling Dice...")



time.sleep(2)



print("Your number is....{}".format(x))



if x % 2 == 0:



  print("It is an Even number!")



else:



  print("It is an Odd number!")

edited Nov 20 '18 at 17:49

answered Nov 20 '18 at 17:40

Preetkaran Singh

114111

This could be further simplified with the second condition being else: instead of another if, but it's definitely more pythonic than what's already there

– G. Anderson
Nov 20 '18 at 17:45

1

Yes ofcourse, @G.Anderson, I have edited the answer.

– Preetkaran Singh
Nov 20 '18 at 17:49

add a comment |

if x % 2 == 0: print("It is an Even number!") else: print("It is an Odd number!")

answered Nov 20 '18 at 17:58

Dharmesh

8501212

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

For checking number being odd or even, you may not need the Even, Odd lists, you might check out the following code:

from random import randrange



import random



import time



x = random.randint(1, 6)



print("Rolling Dice...")



time.sleep(2)



print("Your number is....{}".format(x))



if x % 2 == 0:



  print("It is an Even number!")



else:



  print("It is an Odd number!")

edited Nov 20 '18 at 17:49

answered Nov 20 '18 at 17:40

Preetkaran Singh

114111

This could be further simplified with the second condition being else: instead of another if, but it's definitely more pythonic than what's already there

– G. Anderson
Nov 20 '18 at 17:45

1

Yes ofcourse, @G.Anderson, I have edited the answer.

– Preetkaran Singh
Nov 20 '18 at 17:49

add a comment |

For checking number being odd or even, you may not need the Even, Odd lists, you might check out the following code:

from random import randrange



import random



import time



x = random.randint(1, 6)



print("Rolling Dice...")



time.sleep(2)



print("Your number is....{}".format(x))



if x % 2 == 0:



  print("It is an Even number!")



else:



  print("It is an Odd number!")

edited Nov 20 '18 at 17:49

answered Nov 20 '18 at 17:40

Preetkaran Singh

114111

This could be further simplified with the second condition being else: instead of another if, but it's definitely more pythonic than what's already there

– G. Anderson
Nov 20 '18 at 17:45

1

Yes ofcourse, @G.Anderson, I have edited the answer.

– Preetkaran Singh
Nov 20 '18 at 17:49

add a comment |

For checking number being odd or even, you may not need the Even, Odd lists, you might check out the following code:

from random import randrange



import random



import time



x = random.randint(1, 6)



print("Rolling Dice...")



time.sleep(2)



print("Your number is....{}".format(x))



if x % 2 == 0:



  print("It is an Even number!")



else:



  print("It is an Odd number!")

edited Nov 20 '18 at 17:49

answered Nov 20 '18 at 17:40

Preetkaran Singh

114111

For checking number being odd or even, you may not need the Even, Odd lists, you might check out the following code:

from random import randrange



import random



import time



x = random.randint(1, 6)



print("Rolling Dice...")



time.sleep(2)



print("Your number is....{}".format(x))



if x % 2 == 0:



  print("It is an Even number!")



else:



  print("It is an Odd number!")

edited Nov 20 '18 at 17:49

answered Nov 20 '18 at 17:40

Preetkaran Singh

114111

edited Nov 20 '18 at 17:49

answered Nov 20 '18 at 17:40

Preetkaran Singh

114111

answered Nov 20 '18 at 17:40

Preetkaran Singh

114111

answered Nov 20 '18 at 17:40

Preetkaran Singh

114111

This could be further simplified with the second condition being else: instead of another if, but it's definitely more pythonic than what's already there

– G. Anderson
Nov 20 '18 at 17:45

1

Yes ofcourse, @G.Anderson, I have edited the answer.

– Preetkaran Singh
Nov 20 '18 at 17:49

add a comment |

This could be further simplified with the second condition being else: instead of another if, but it's definitely more pythonic than what's already there

– G. Anderson
Nov 20 '18 at 17:45

1

Yes ofcourse, @G.Anderson, I have edited the answer.

– Preetkaran Singh
Nov 20 '18 at 17:49

This could be further simplified with the second condition being else: instead of another if, but it's definitely more pythonic than what's already there

– G. Anderson
Nov 20 '18 at 17:45

Yes ofcourse, @G.Anderson, I have edited the answer.

– Preetkaran Singh
Nov 20 '18 at 17:49

add a comment |

if x % 2 == 0: print("It is an Even number!") else: print("It is an Odd number!")

answered Nov 20 '18 at 17:58

Dharmesh

8501212

add a comment |

if x % 2 == 0: print("It is an Even number!") else: print("It is an Odd number!")

answered Nov 20 '18 at 17:58

Dharmesh

8501212

add a comment |

if x % 2 == 0: print("It is an Even number!") else: print("It is an Odd number!")

answered Nov 20 '18 at 17:58

Dharmesh

8501212

if x % 2 == 0: print("It is an Even number!") else: print("It is an Odd number!")

answered Nov 20 '18 at 17:58

Dharmesh

8501212

answered Nov 20 '18 at 17:58

Dharmesh

8501212

answered Nov 20 '18 at 17:58

Dharmesh

8501212

answered Nov 20 '18 at 17:58

Dharmesh

8501212

add a comment |

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