Probability of Getting a number on throwing a dice.












0














The question is that what is the probability of getting the number 6 when throwing a 6 sided die.



Since 6 will either come or it won't the probability of this event should be 0.5.



How come the answer is 1/6 then?



Seriously how do you counter this argument?






probability probability-theory







share|cite|improve this question




















  • 3




    "If I buy a lottery ticket either I win or I lose, so I must have a $50%$ chance of winning!"
    – lulu
    Sep 25 '18 at 12:08












  • Yeah, what's the problem with that? I know that the question is silly.
    – harshit54
    Sep 25 '18 at 12:09






  • 1




    Just because there are only two cases does not mean they are equally likely.
    – lulu
    Sep 25 '18 at 12:16










  • @lulu Yup that looks correct. Thanks
    – harshit54
    Sep 25 '18 at 12:16
















0














The question is that what is the probability of getting the number 6 when throwing a 6 sided die.



Since 6 will either come or it won't the probability of this event should be 0.5.



How come the answer is 1/6 then?



Seriously how do you counter this argument?






probability probability-theory







share|cite|improve this question




















  • 3




    "If I buy a lottery ticket either I win or I lose, so I must have a $50%$ chance of winning!"
    – lulu
    Sep 25 '18 at 12:08












  • Yeah, what's the problem with that? I know that the question is silly.
    – harshit54
    Sep 25 '18 at 12:09






  • 1




    Just because there are only two cases does not mean they are equally likely.
    – lulu
    Sep 25 '18 at 12:16










  • @lulu Yup that looks correct. Thanks
    – harshit54
    Sep 25 '18 at 12:16














0












0








0







The question is that what is the probability of getting the number 6 when throwing a 6 sided die.



Since 6 will either come or it won't the probability of this event should be 0.5.



How come the answer is 1/6 then?



Seriously how do you counter this argument?






probability probability-theory







share|cite|improve this question















The question is that what is the probability of getting the number 6 when throwing a 6 sided die.



Since 6 will either come or it won't the probability of this event should be 0.5.



How come the answer is 1/6 then?



Seriously how do you counter this argument?






probability probability-theory




probability probability-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 18:43

























asked Sep 25 '18 at 12:01









harshit54

346113




346113








  • 3




    "If I buy a lottery ticket either I win or I lose, so I must have a $50%$ chance of winning!"
    – lulu
    Sep 25 '18 at 12:08












  • Yeah, what's the problem with that? I know that the question is silly.
    – harshit54
    Sep 25 '18 at 12:09






  • 1




    Just because there are only two cases does not mean they are equally likely.
    – lulu
    Sep 25 '18 at 12:16










  • @lulu Yup that looks correct. Thanks
    – harshit54
    Sep 25 '18 at 12:16














  • 3




    "If I buy a lottery ticket either I win or I lose, so I must have a $50%$ chance of winning!"
    – lulu
    Sep 25 '18 at 12:08












  • Yeah, what's the problem with that? I know that the question is silly.
    – harshit54
    Sep 25 '18 at 12:09






  • 1




    Just because there are only two cases does not mean they are equally likely.
    – lulu
    Sep 25 '18 at 12:16










  • @lulu Yup that looks correct. Thanks
    – harshit54
    Sep 25 '18 at 12:16








3




3




"If I buy a lottery ticket either I win or I lose, so I must have a $50%$ chance of winning!"
– lulu
Sep 25 '18 at 12:08






"If I buy a lottery ticket either I win or I lose, so I must have a $50%$ chance of winning!"
– lulu
Sep 25 '18 at 12:08














Yeah, what's the problem with that? I know that the question is silly.
– harshit54
Sep 25 '18 at 12:09




Yeah, what's the problem with that? I know that the question is silly.
– harshit54
Sep 25 '18 at 12:09




1




1




Just because there are only two cases does not mean they are equally likely.
– lulu
Sep 25 '18 at 12:16




Just because there are only two cases does not mean they are equally likely.
– lulu
Sep 25 '18 at 12:16












@lulu Yup that looks correct. Thanks
– harshit54
Sep 25 '18 at 12:16




@lulu Yup that looks correct. Thanks
– harshit54
Sep 25 '18 at 12:16










3 Answers
3






active

oldest

votes


















4














To try to answer your question, and your comment 'what is the problem with my Sample Space of 6 and not 6?':



You can have a sample space of '6 and not 6', but these two outcomes are not equally likely. This is an important point: Not all sample spaces have equally likely outcomes.



When probabilities are assigned to outcomes in a discrete sample space, two rules must be obeyed (to suit the definition of a probability distribution): Every individual outcome probability must be a real number between $0$ and $1$ inclusive; and the sum of the outcome probabilities must equal $1$. Anything that fits these two conditions is a probability distribution on the sample space. However, to be most useful, we select a probability distribution that will best reflect reality. Assigning 6 and not 6 each a probability of $frac12$ will likely not reflect reality very well at all (in the sense that it is highly unlikely that in the long term half your rolls will come up $6$--assuming that your die is fair).






share|cite|improve this answer





















  • Yup. Thanks for the answer.
    – harshit54
    Sep 25 '18 at 12:24






  • 1




    +1 It's actually this answer that makes me understand the problem of the OP.
    – drhab
    Sep 25 '18 at 12:25



















0














On a six sided dice there are 6 possible outcomes, rolling a 1, 2, 3, 4, 5, or 6. Rolling a 6 is only one of these possible outcomes and so the probability of it happening is $frac 16$






share|cite|improve this answer





















  • Yeah I know that. But what's the problem with my argument??
    – harshit54
    Sep 25 '18 at 12:09






  • 1




    I've just explained that, there are 6 outcomes and you want the probability of getting one of them. If you wanted the probability of getting an even number,the answer would be $frac 36 = frac 12$ because there are 3 even numbers you can roll from a possible 6
    – lioness99a
    Sep 25 '18 at 12:11












  • But what is the problem with my Sample Space of 6 and not 6?
    – harshit54
    Sep 25 '18 at 12:12










  • I don't understand what you mean by that
    – lioness99a
    Sep 25 '18 at 12:13



















0














probability is the ratio of possible outcome to possible event, in this question possible events are six, and they are mutually exclusive, and you have to count them all, when I am saying one is not possible, that is happening in 5 separate events, not one, the logic is (6)U(5)U(4)U(3)U(2), they can not happen in same time so you have to consider OR here.






share|cite|improve this answer





















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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    To try to answer your question, and your comment 'what is the problem with my Sample Space of 6 and not 6?':



    You can have a sample space of '6 and not 6', but these two outcomes are not equally likely. This is an important point: Not all sample spaces have equally likely outcomes.



    When probabilities are assigned to outcomes in a discrete sample space, two rules must be obeyed (to suit the definition of a probability distribution): Every individual outcome probability must be a real number between $0$ and $1$ inclusive; and the sum of the outcome probabilities must equal $1$. Anything that fits these two conditions is a probability distribution on the sample space. However, to be most useful, we select a probability distribution that will best reflect reality. Assigning 6 and not 6 each a probability of $frac12$ will likely not reflect reality very well at all (in the sense that it is highly unlikely that in the long term half your rolls will come up $6$--assuming that your die is fair).






    share|cite|improve this answer





















    • Yup. Thanks for the answer.
      – harshit54
      Sep 25 '18 at 12:24






    • 1




      +1 It's actually this answer that makes me understand the problem of the OP.
      – drhab
      Sep 25 '18 at 12:25
















    4














    To try to answer your question, and your comment 'what is the problem with my Sample Space of 6 and not 6?':



    You can have a sample space of '6 and not 6', but these two outcomes are not equally likely. This is an important point: Not all sample spaces have equally likely outcomes.



    When probabilities are assigned to outcomes in a discrete sample space, two rules must be obeyed (to suit the definition of a probability distribution): Every individual outcome probability must be a real number between $0$ and $1$ inclusive; and the sum of the outcome probabilities must equal $1$. Anything that fits these two conditions is a probability distribution on the sample space. However, to be most useful, we select a probability distribution that will best reflect reality. Assigning 6 and not 6 each a probability of $frac12$ will likely not reflect reality very well at all (in the sense that it is highly unlikely that in the long term half your rolls will come up $6$--assuming that your die is fair).






    share|cite|improve this answer





















    • Yup. Thanks for the answer.
      – harshit54
      Sep 25 '18 at 12:24






    • 1




      +1 It's actually this answer that makes me understand the problem of the OP.
      – drhab
      Sep 25 '18 at 12:25














    4












    4








    4






    To try to answer your question, and your comment 'what is the problem with my Sample Space of 6 and not 6?':



    You can have a sample space of '6 and not 6', but these two outcomes are not equally likely. This is an important point: Not all sample spaces have equally likely outcomes.



    When probabilities are assigned to outcomes in a discrete sample space, two rules must be obeyed (to suit the definition of a probability distribution): Every individual outcome probability must be a real number between $0$ and $1$ inclusive; and the sum of the outcome probabilities must equal $1$. Anything that fits these two conditions is a probability distribution on the sample space. However, to be most useful, we select a probability distribution that will best reflect reality. Assigning 6 and not 6 each a probability of $frac12$ will likely not reflect reality very well at all (in the sense that it is highly unlikely that in the long term half your rolls will come up $6$--assuming that your die is fair).






    share|cite|improve this answer












    To try to answer your question, and your comment 'what is the problem with my Sample Space of 6 and not 6?':



    You can have a sample space of '6 and not 6', but these two outcomes are not equally likely. This is an important point: Not all sample spaces have equally likely outcomes.



    When probabilities are assigned to outcomes in a discrete sample space, two rules must be obeyed (to suit the definition of a probability distribution): Every individual outcome probability must be a real number between $0$ and $1$ inclusive; and the sum of the outcome probabilities must equal $1$. Anything that fits these two conditions is a probability distribution on the sample space. However, to be most useful, we select a probability distribution that will best reflect reality. Assigning 6 and not 6 each a probability of $frac12$ will likely not reflect reality very well at all (in the sense that it is highly unlikely that in the long term half your rolls will come up $6$--assuming that your die is fair).







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Sep 25 '18 at 12:22









    paw88789

    29k12349




    29k12349












    • Yup. Thanks for the answer.
      – harshit54
      Sep 25 '18 at 12:24






    • 1




      +1 It's actually this answer that makes me understand the problem of the OP.
      – drhab
      Sep 25 '18 at 12:25


















    • Yup. Thanks for the answer.
      – harshit54
      Sep 25 '18 at 12:24






    • 1




      +1 It's actually this answer that makes me understand the problem of the OP.
      – drhab
      Sep 25 '18 at 12:25
















    Yup. Thanks for the answer.
    – harshit54
    Sep 25 '18 at 12:24




    Yup. Thanks for the answer.
    – harshit54
    Sep 25 '18 at 12:24




    1




    1




    +1 It's actually this answer that makes me understand the problem of the OP.
    – drhab
    Sep 25 '18 at 12:25




    +1 It's actually this answer that makes me understand the problem of the OP.
    – drhab
    Sep 25 '18 at 12:25











    0














    On a six sided dice there are 6 possible outcomes, rolling a 1, 2, 3, 4, 5, or 6. Rolling a 6 is only one of these possible outcomes and so the probability of it happening is $frac 16$






    share|cite|improve this answer





















    • Yeah I know that. But what's the problem with my argument??
      – harshit54
      Sep 25 '18 at 12:09






    • 1




      I've just explained that, there are 6 outcomes and you want the probability of getting one of them. If you wanted the probability of getting an even number,the answer would be $frac 36 = frac 12$ because there are 3 even numbers you can roll from a possible 6
      – lioness99a
      Sep 25 '18 at 12:11












    • But what is the problem with my Sample Space of 6 and not 6?
      – harshit54
      Sep 25 '18 at 12:12










    • I don't understand what you mean by that
      – lioness99a
      Sep 25 '18 at 12:13
















    0














    On a six sided dice there are 6 possible outcomes, rolling a 1, 2, 3, 4, 5, or 6. Rolling a 6 is only one of these possible outcomes and so the probability of it happening is $frac 16$






    share|cite|improve this answer





















    • Yeah I know that. But what's the problem with my argument??
      – harshit54
      Sep 25 '18 at 12:09






    • 1




      I've just explained that, there are 6 outcomes and you want the probability of getting one of them. If you wanted the probability of getting an even number,the answer would be $frac 36 = frac 12$ because there are 3 even numbers you can roll from a possible 6
      – lioness99a
      Sep 25 '18 at 12:11












    • But what is the problem with my Sample Space of 6 and not 6?
      – harshit54
      Sep 25 '18 at 12:12










    • I don't understand what you mean by that
      – lioness99a
      Sep 25 '18 at 12:13














    0












    0








    0






    On a six sided dice there are 6 possible outcomes, rolling a 1, 2, 3, 4, 5, or 6. Rolling a 6 is only one of these possible outcomes and so the probability of it happening is $frac 16$






    share|cite|improve this answer












    On a six sided dice there are 6 possible outcomes, rolling a 1, 2, 3, 4, 5, or 6. Rolling a 6 is only one of these possible outcomes and so the probability of it happening is $frac 16$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Sep 25 '18 at 12:09









    lioness99a

    3,6262727




    3,6262727












    • Yeah I know that. But what's the problem with my argument??
      – harshit54
      Sep 25 '18 at 12:09






    • 1




      I've just explained that, there are 6 outcomes and you want the probability of getting one of them. If you wanted the probability of getting an even number,the answer would be $frac 36 = frac 12$ because there are 3 even numbers you can roll from a possible 6
      – lioness99a
      Sep 25 '18 at 12:11












    • But what is the problem with my Sample Space of 6 and not 6?
      – harshit54
      Sep 25 '18 at 12:12










    • I don't understand what you mean by that
      – lioness99a
      Sep 25 '18 at 12:13


















    • Yeah I know that. But what's the problem with my argument??
      – harshit54
      Sep 25 '18 at 12:09






    • 1




      I've just explained that, there are 6 outcomes and you want the probability of getting one of them. If you wanted the probability of getting an even number,the answer would be $frac 36 = frac 12$ because there are 3 even numbers you can roll from a possible 6
      – lioness99a
      Sep 25 '18 at 12:11












    • But what is the problem with my Sample Space of 6 and not 6?
      – harshit54
      Sep 25 '18 at 12:12










    • I don't understand what you mean by that
      – lioness99a
      Sep 25 '18 at 12:13
















    Yeah I know that. But what's the problem with my argument??
    – harshit54
    Sep 25 '18 at 12:09




    Yeah I know that. But what's the problem with my argument??
    – harshit54
    Sep 25 '18 at 12:09




    1




    1




    I've just explained that, there are 6 outcomes and you want the probability of getting one of them. If you wanted the probability of getting an even number,the answer would be $frac 36 = frac 12$ because there are 3 even numbers you can roll from a possible 6
    – lioness99a
    Sep 25 '18 at 12:11






    I've just explained that, there are 6 outcomes and you want the probability of getting one of them. If you wanted the probability of getting an even number,the answer would be $frac 36 = frac 12$ because there are 3 even numbers you can roll from a possible 6
    – lioness99a
    Sep 25 '18 at 12:11














    But what is the problem with my Sample Space of 6 and not 6?
    – harshit54
    Sep 25 '18 at 12:12




    But what is the problem with my Sample Space of 6 and not 6?
    – harshit54
    Sep 25 '18 at 12:12












    I don't understand what you mean by that
    – lioness99a
    Sep 25 '18 at 12:13




    I don't understand what you mean by that
    – lioness99a
    Sep 25 '18 at 12:13











    0














    probability is the ratio of possible outcome to possible event, in this question possible events are six, and they are mutually exclusive, and you have to count them all, when I am saying one is not possible, that is happening in 5 separate events, not one, the logic is (6)U(5)U(4)U(3)U(2), they can not happen in same time so you have to consider OR here.






    share|cite|improve this answer


























      0














      probability is the ratio of possible outcome to possible event, in this question possible events are six, and they are mutually exclusive, and you have to count them all, when I am saying one is not possible, that is happening in 5 separate events, not one, the logic is (6)U(5)U(4)U(3)U(2), they can not happen in same time so you have to consider OR here.






      share|cite|improve this answer
























        0












        0








        0






        probability is the ratio of possible outcome to possible event, in this question possible events are six, and they are mutually exclusive, and you have to count them all, when I am saying one is not possible, that is happening in 5 separate events, not one, the logic is (6)U(5)U(4)U(3)U(2), they can not happen in same time so you have to consider OR here.






        share|cite|improve this answer












        probability is the ratio of possible outcome to possible event, in this question possible events are six, and they are mutually exclusive, and you have to count them all, when I am saying one is not possible, that is happening in 5 separate events, not one, the logic is (6)U(5)U(4)U(3)U(2), they can not happen in same time so you have to consider OR here.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 25 '18 at 12:12









        explorer

        2916




        2916






























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