Birthday Paradox: why permutations and not combinations?












1












$begingroup$


The Birthday Problem: given $n$ people (typically $n<365$), what is the probability that some pair of them share a birthday (omitting Feb 29th, for simplicity)?



The solution: First, find the probability that all $n$ people have different birthdays. Here is where I am confused. The solutions I have seen all say this probability is:
$$
frac{_{365}P_n}{365^n}
$$



Why isn't this $_{365}C_n$,instead?






probability combinatorics birthday







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    How many ways are there of assigning $n$ of the 365 possible birthdays to $n$ people?
    $endgroup$
    – MJD
    Nov 13 '14 at 19:07
















1












$begingroup$


The Birthday Problem: given $n$ people (typically $n<365$), what is the probability that some pair of them share a birthday (omitting Feb 29th, for simplicity)?



The solution: First, find the probability that all $n$ people have different birthdays. Here is where I am confused. The solutions I have seen all say this probability is:
$$
frac{_{365}P_n}{365^n}
$$



Why isn't this $_{365}C_n$,instead?






probability combinatorics birthday







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    How many ways are there of assigning $n$ of the 365 possible birthdays to $n$ people?
    $endgroup$
    – MJD
    Nov 13 '14 at 19:07














1












1








1


1



$begingroup$


The Birthday Problem: given $n$ people (typically $n<365$), what is the probability that some pair of them share a birthday (omitting Feb 29th, for simplicity)?



The solution: First, find the probability that all $n$ people have different birthdays. Here is where I am confused. The solutions I have seen all say this probability is:
$$
frac{_{365}P_n}{365^n}
$$



Why isn't this $_{365}C_n$,instead?






probability combinatorics birthday







share|cite|improve this question











$endgroup$




The Birthday Problem: given $n$ people (typically $n<365$), what is the probability that some pair of them share a birthday (omitting Feb 29th, for simplicity)?



The solution: First, find the probability that all $n$ people have different birthdays. Here is where I am confused. The solutions I have seen all say this probability is:
$$
frac{_{365}P_n}{365^n}
$$



Why isn't this $_{365}C_n$,instead?






probability combinatorics birthday




probability combinatorics birthday






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 17 '17 at 21:42









Henry

101k481168




101k481168










asked Nov 13 '14 at 19:02









David SteinbergDavid Steinberg

815921




815921








  • 1




    $begingroup$
    How many ways are there of assigning $n$ of the 365 possible birthdays to $n$ people?
    $endgroup$
    – MJD
    Nov 13 '14 at 19:07














  • 1




    $begingroup$
    How many ways are there of assigning $n$ of the 365 possible birthdays to $n$ people?
    $endgroup$
    – MJD
    Nov 13 '14 at 19:07








1




1




$begingroup$
How many ways are there of assigning $n$ of the 365 possible birthdays to $n$ people?
$endgroup$
– MJD
Nov 13 '14 at 19:07




$begingroup$
How many ways are there of assigning $n$ of the 365 possible birthdays to $n$ people?
$endgroup$
– MJD
Nov 13 '14 at 19:07










5 Answers
5






active

oldest

votes


















4












$begingroup$

If you did combinations, you would basically choose the birthdays but not assign them to the $n$ people. However the denominator presumes the birthdays have been assigned (365 choices for Alice, 365 choices for Bob, etc.)






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    what if we will use also compinations ( with repetition ) for the denominator?
    $endgroup$
    – karhas
    Aug 25 '16 at 23:25



















1












$begingroup$

The probability with $n$ people is, by considering the first, then the second, then the third, ..., $$frac{365}{365}times frac{364}{365}times frac{363}{365}times cdots timesfrac{365-n+1}{365}.$$



Now simplify.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Regarding paw88789's response, it is mathematically correct, but omits the "why" of the question, which is why permutations should be used instead of combinations to determine the number of possibilities of the group's birthdays.



    Combinations certainly give the number of possible birthday sets, which seems a reasonable way to solve the problem. However, the birthday problem is for a real group of people, and such groups allow for repetition of birthdays. Once repetition is allowed, the number of ways the group can have birthdays is 365^n, for an n-person group. Combinations with repetition are not calculated by nCr [C(n,r)], nor are they related to permutations by a factor of r!, as in nCr = nPr * r!. Although we could calculate the combinations with repetition, it still misses the correct number of possible ways for the group's set of birthdays because a group with a certain combination of birthdays can be put together in possibly more ways than one, and this affects the probability of the result. The basis for the probability of no repetition of birthdays is the number of possible ways for no repetition of birthdays divided by the number of total possible ways, 365^n, for birthdays.



    Only permutation gives the correct number of possible ways a group can have birthdays without repetition.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      You could, in principle, use combinations to do this problem, that is, take the people to be indistinguishable from each other. The difficulty is that the corresponding sample space consists of ordered partitions of $n$ into 365 non-negative pieces, and those partitions are NOT EQUALLY LIKELY (and are hard to count). So it will NOT be simply (size of event) / (size of sample space).



      In general, situations that involve independent identical trials (like people's birthdays, rolling dice, flipping coins, drawing with replacement) are most easily modeled by ordered (i.e. distinguishable) trials, because it is easier to model the sample space with equally likely outcomes that way. For example, the question what is the probability of getting three different numbers when rolling three dice [which IS the birthday problem with 6,3 in place of 365,$n$] is much easier to do if we imagine the dice as distinguishable (i.e. $6times 5
      times 4/6^3$      ) than by enumerating ordered partitions of 3 into 6 pieces and figuring out the relative frequency of each type.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        I agree that the partitions of 365 is tough, but for the dice case, it should be easy enough right: numerator: $frac{6 times 5 times 4}{6}=20$ denominator would be (no repeats + one repeat + all same) = (20+6*5+6) = 56. Which gives prob of no repeats = $frac{20}{56}$ which is vastly different from $frac{6times 5times 4}{6^{3}} = frac{5}{9}$
        $endgroup$
        – tpb261
        Nov 15 '18 at 16:00












      • $begingroup$
        But those 56 items are NOT equally likely -- the multiset [1,1,1] occurs less often than the multiset [1,3,4], for example, and though you can, in theory model the dice rolls by unordered combinations (i.e. multisets) you can not get the probability of events by just dividing the cardinality of the event by the cardinality of the sample space. You must weight the (multiset) outcomes by their frequency which for practical purposes takes you back to viewing them as ordered triples, since that view makes them equally likely.
        $endgroup$
        – Ned
        Nov 15 '18 at 23:47










      • $begingroup$
        @tpb261 see the comment above (sorry I forgot to address you in the comment)
        $endgroup$
        – Ned
        Nov 19 '18 at 2:52










      • $begingroup$
        Thanks!! I did check your comment on phone (from where I don't like to reply), and came to my desktop just now. I missed the point in your answer. My bad.
        $endgroup$
        – tpb261
        Nov 19 '18 at 11:56



















      0












      $begingroup$

      Combinations exclude order, while permutations take order into account.



      So when it comes to assigning ($person_0$,$person_1$, $person_2$) birth dates from ${a,b,c,d}$



      (Arrays used below, index denotes person_number .. so if $a$ has index 0, a has been assigned to $person_0$)



      [a,b,c], [c,b,a], [b,c,a] .. would be considered the same one element.



      So if you are trying to calculate the length of the favorable outcomes space, you'd see why that would be incorrect



      As for the the denominator i.e the length of space of all possible outcomes $365^n$ should not give anyone trouble



      So P = $frac {text{length of favorable outcome space}}{text{length of all possible outcomes space}}$



      So just $C^{365}_n$ would give you a length of set of assignments where order is not important and that is pretty much it. WHen it comes to probability you are missing the deominator



      P.S. $frac{P^{365}_{n}}{365^{n}}$ is the probability of no one sharing the same birthday






      share|cite|improve this answer











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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        If you did combinations, you would basically choose the birthdays but not assign them to the $n$ people. However the denominator presumes the birthdays have been assigned (365 choices for Alice, 365 choices for Bob, etc.)






        share|cite|improve this answer









        $endgroup$









        • 1




          $begingroup$
          what if we will use also compinations ( with repetition ) for the denominator?
          $endgroup$
          – karhas
          Aug 25 '16 at 23:25
















        4












        $begingroup$

        If you did combinations, you would basically choose the birthdays but not assign them to the $n$ people. However the denominator presumes the birthdays have been assigned (365 choices for Alice, 365 choices for Bob, etc.)






        share|cite|improve this answer









        $endgroup$









        • 1




          $begingroup$
          what if we will use also compinations ( with repetition ) for the denominator?
          $endgroup$
          – karhas
          Aug 25 '16 at 23:25














        4












        4








        4





        $begingroup$

        If you did combinations, you would basically choose the birthdays but not assign them to the $n$ people. However the denominator presumes the birthdays have been assigned (365 choices for Alice, 365 choices for Bob, etc.)






        share|cite|improve this answer









        $endgroup$



        If you did combinations, you would basically choose the birthdays but not assign them to the $n$ people. However the denominator presumes the birthdays have been assigned (365 choices for Alice, 365 choices for Bob, etc.)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 13 '14 at 19:13









        paw88789paw88789

        29.4k12349




        29.4k12349








        • 1




          $begingroup$
          what if we will use also compinations ( with repetition ) for the denominator?
          $endgroup$
          – karhas
          Aug 25 '16 at 23:25














        • 1




          $begingroup$
          what if we will use also compinations ( with repetition ) for the denominator?
          $endgroup$
          – karhas
          Aug 25 '16 at 23:25








        1




        1




        $begingroup$
        what if we will use also compinations ( with repetition ) for the denominator?
        $endgroup$
        – karhas
        Aug 25 '16 at 23:25




        $begingroup$
        what if we will use also compinations ( with repetition ) for the denominator?
        $endgroup$
        – karhas
        Aug 25 '16 at 23:25











        1












        $begingroup$

        The probability with $n$ people is, by considering the first, then the second, then the third, ..., $$frac{365}{365}times frac{364}{365}times frac{363}{365}times cdots timesfrac{365-n+1}{365}.$$



        Now simplify.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          The probability with $n$ people is, by considering the first, then the second, then the third, ..., $$frac{365}{365}times frac{364}{365}times frac{363}{365}times cdots timesfrac{365-n+1}{365}.$$



          Now simplify.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            The probability with $n$ people is, by considering the first, then the second, then the third, ..., $$frac{365}{365}times frac{364}{365}times frac{363}{365}times cdots timesfrac{365-n+1}{365}.$$



            Now simplify.






            share|cite|improve this answer









            $endgroup$



            The probability with $n$ people is, by considering the first, then the second, then the third, ..., $$frac{365}{365}times frac{364}{365}times frac{363}{365}times cdots timesfrac{365-n+1}{365}.$$



            Now simplify.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 13 '14 at 19:14









            HenryHenry

            101k481168




            101k481168























                0












                $begingroup$

                Regarding paw88789's response, it is mathematically correct, but omits the "why" of the question, which is why permutations should be used instead of combinations to determine the number of possibilities of the group's birthdays.



                Combinations certainly give the number of possible birthday sets, which seems a reasonable way to solve the problem. However, the birthday problem is for a real group of people, and such groups allow for repetition of birthdays. Once repetition is allowed, the number of ways the group can have birthdays is 365^n, for an n-person group. Combinations with repetition are not calculated by nCr [C(n,r)], nor are they related to permutations by a factor of r!, as in nCr = nPr * r!. Although we could calculate the combinations with repetition, it still misses the correct number of possible ways for the group's set of birthdays because a group with a certain combination of birthdays can be put together in possibly more ways than one, and this affects the probability of the result. The basis for the probability of no repetition of birthdays is the number of possible ways for no repetition of birthdays divided by the number of total possible ways, 365^n, for birthdays.



                Only permutation gives the correct number of possible ways a group can have birthdays without repetition.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Regarding paw88789's response, it is mathematically correct, but omits the "why" of the question, which is why permutations should be used instead of combinations to determine the number of possibilities of the group's birthdays.



                  Combinations certainly give the number of possible birthday sets, which seems a reasonable way to solve the problem. However, the birthday problem is for a real group of people, and such groups allow for repetition of birthdays. Once repetition is allowed, the number of ways the group can have birthdays is 365^n, for an n-person group. Combinations with repetition are not calculated by nCr [C(n,r)], nor are they related to permutations by a factor of r!, as in nCr = nPr * r!. Although we could calculate the combinations with repetition, it still misses the correct number of possible ways for the group's set of birthdays because a group with a certain combination of birthdays can be put together in possibly more ways than one, and this affects the probability of the result. The basis for the probability of no repetition of birthdays is the number of possible ways for no repetition of birthdays divided by the number of total possible ways, 365^n, for birthdays.



                  Only permutation gives the correct number of possible ways a group can have birthdays without repetition.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Regarding paw88789's response, it is mathematically correct, but omits the "why" of the question, which is why permutations should be used instead of combinations to determine the number of possibilities of the group's birthdays.



                    Combinations certainly give the number of possible birthday sets, which seems a reasonable way to solve the problem. However, the birthday problem is for a real group of people, and such groups allow for repetition of birthdays. Once repetition is allowed, the number of ways the group can have birthdays is 365^n, for an n-person group. Combinations with repetition are not calculated by nCr [C(n,r)], nor are they related to permutations by a factor of r!, as in nCr = nPr * r!. Although we could calculate the combinations with repetition, it still misses the correct number of possible ways for the group's set of birthdays because a group with a certain combination of birthdays can be put together in possibly more ways than one, and this affects the probability of the result. The basis for the probability of no repetition of birthdays is the number of possible ways for no repetition of birthdays divided by the number of total possible ways, 365^n, for birthdays.



                    Only permutation gives the correct number of possible ways a group can have birthdays without repetition.






                    share|cite|improve this answer









                    $endgroup$



                    Regarding paw88789's response, it is mathematically correct, but omits the "why" of the question, which is why permutations should be used instead of combinations to determine the number of possibilities of the group's birthdays.



                    Combinations certainly give the number of possible birthday sets, which seems a reasonable way to solve the problem. However, the birthday problem is for a real group of people, and such groups allow for repetition of birthdays. Once repetition is allowed, the number of ways the group can have birthdays is 365^n, for an n-person group. Combinations with repetition are not calculated by nCr [C(n,r)], nor are they related to permutations by a factor of r!, as in nCr = nPr * r!. Although we could calculate the combinations with repetition, it still misses the correct number of possible ways for the group's set of birthdays because a group with a certain combination of birthdays can be put together in possibly more ways than one, and this affects the probability of the result. The basis for the probability of no repetition of birthdays is the number of possible ways for no repetition of birthdays divided by the number of total possible ways, 365^n, for birthdays.



                    Only permutation gives the correct number of possible ways a group can have birthdays without repetition.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 21 '18 at 13:39









                    John KribsJohn Kribs

                    1




                    1























                        0












                        $begingroup$

                        You could, in principle, use combinations to do this problem, that is, take the people to be indistinguishable from each other. The difficulty is that the corresponding sample space consists of ordered partitions of $n$ into 365 non-negative pieces, and those partitions are NOT EQUALLY LIKELY (and are hard to count). So it will NOT be simply (size of event) / (size of sample space).



                        In general, situations that involve independent identical trials (like people's birthdays, rolling dice, flipping coins, drawing with replacement) are most easily modeled by ordered (i.e. distinguishable) trials, because it is easier to model the sample space with equally likely outcomes that way. For example, the question what is the probability of getting three different numbers when rolling three dice [which IS the birthday problem with 6,3 in place of 365,$n$] is much easier to do if we imagine the dice as distinguishable (i.e. $6times 5
                        times 4/6^3$                        ) than by enumerating ordered partitions of 3 into 6 pieces and figuring out the relative frequency of each type.






                        share|cite|improve this answer











                        $endgroup$













                        • $begingroup$
                          I agree that the partitions of 365 is tough, but for the dice case, it should be easy enough right: numerator: $frac{6 times 5 times 4}{6}=20$ denominator would be (no repeats + one repeat + all same) = (20+6*5+6) = 56. Which gives prob of no repeats = $frac{20}{56}$ which is vastly different from $frac{6times 5times 4}{6^{3}} = frac{5}{9}$
                          $endgroup$
                          – tpb261
                          Nov 15 '18 at 16:00












                        • $begingroup$
                          But those 56 items are NOT equally likely -- the multiset [1,1,1] occurs less often than the multiset [1,3,4], for example, and though you can, in theory model the dice rolls by unordered combinations (i.e. multisets) you can not get the probability of events by just dividing the cardinality of the event by the cardinality of the sample space. You must weight the (multiset) outcomes by their frequency which for practical purposes takes you back to viewing them as ordered triples, since that view makes them equally likely.
                          $endgroup$
                          – Ned
                          Nov 15 '18 at 23:47










                        • $begingroup$
                          @tpb261 see the comment above (sorry I forgot to address you in the comment)
                          $endgroup$
                          – Ned
                          Nov 19 '18 at 2:52










                        • $begingroup$
                          Thanks!! I did check your comment on phone (from where I don't like to reply), and came to my desktop just now. I missed the point in your answer. My bad.
                          $endgroup$
                          – tpb261
                          Nov 19 '18 at 11:56
















                        0












                        $begingroup$

                        You could, in principle, use combinations to do this problem, that is, take the people to be indistinguishable from each other. The difficulty is that the corresponding sample space consists of ordered partitions of $n$ into 365 non-negative pieces, and those partitions are NOT EQUALLY LIKELY (and are hard to count). So it will NOT be simply (size of event) / (size of sample space).



                        In general, situations that involve independent identical trials (like people's birthdays, rolling dice, flipping coins, drawing with replacement) are most easily modeled by ordered (i.e. distinguishable) trials, because it is easier to model the sample space with equally likely outcomes that way. For example, the question what is the probability of getting three different numbers when rolling three dice [which IS the birthday problem with 6,3 in place of 365,$n$] is much easier to do if we imagine the dice as distinguishable (i.e. $6times 5
                        times 4/6^3$                        ) than by enumerating ordered partitions of 3 into 6 pieces and figuring out the relative frequency of each type.






                        share|cite|improve this answer











                        $endgroup$













                        • $begingroup$
                          I agree that the partitions of 365 is tough, but for the dice case, it should be easy enough right: numerator: $frac{6 times 5 times 4}{6}=20$ denominator would be (no repeats + one repeat + all same) = (20+6*5+6) = 56. Which gives prob of no repeats = $frac{20}{56}$ which is vastly different from $frac{6times 5times 4}{6^{3}} = frac{5}{9}$
                          $endgroup$
                          – tpb261
                          Nov 15 '18 at 16:00












                        • $begingroup$
                          But those 56 items are NOT equally likely -- the multiset [1,1,1] occurs less often than the multiset [1,3,4], for example, and though you can, in theory model the dice rolls by unordered combinations (i.e. multisets) you can not get the probability of events by just dividing the cardinality of the event by the cardinality of the sample space. You must weight the (multiset) outcomes by their frequency which for practical purposes takes you back to viewing them as ordered triples, since that view makes them equally likely.
                          $endgroup$
                          – Ned
                          Nov 15 '18 at 23:47










                        • $begingroup$
                          @tpb261 see the comment above (sorry I forgot to address you in the comment)
                          $endgroup$
                          – Ned
                          Nov 19 '18 at 2:52










                        • $begingroup$
                          Thanks!! I did check your comment on phone (from where I don't like to reply), and came to my desktop just now. I missed the point in your answer. My bad.
                          $endgroup$
                          – tpb261
                          Nov 19 '18 at 11:56














                        0












                        0








                        0





                        $begingroup$

                        You could, in principle, use combinations to do this problem, that is, take the people to be indistinguishable from each other. The difficulty is that the corresponding sample space consists of ordered partitions of $n$ into 365 non-negative pieces, and those partitions are NOT EQUALLY LIKELY (and are hard to count). So it will NOT be simply (size of event) / (size of sample space).



                        In general, situations that involve independent identical trials (like people's birthdays, rolling dice, flipping coins, drawing with replacement) are most easily modeled by ordered (i.e. distinguishable) trials, because it is easier to model the sample space with equally likely outcomes that way. For example, the question what is the probability of getting three different numbers when rolling three dice [which IS the birthday problem with 6,3 in place of 365,$n$] is much easier to do if we imagine the dice as distinguishable (i.e. $6times 5
                        times 4/6^3$                        ) than by enumerating ordered partitions of 3 into 6 pieces and figuring out the relative frequency of each type.






                        share|cite|improve this answer











                        $endgroup$



                        You could, in principle, use combinations to do this problem, that is, take the people to be indistinguishable from each other. The difficulty is that the corresponding sample space consists of ordered partitions of $n$ into 365 non-negative pieces, and those partitions are NOT EQUALLY LIKELY (and are hard to count). So it will NOT be simply (size of event) / (size of sample space).



                        In general, situations that involve independent identical trials (like people's birthdays, rolling dice, flipping coins, drawing with replacement) are most easily modeled by ordered (i.e. distinguishable) trials, because it is easier to model the sample space with equally likely outcomes that way. For example, the question what is the probability of getting three different numbers when rolling three dice [which IS the birthday problem with 6,3 in place of 365,$n$] is much easier to do if we imagine the dice as distinguishable (i.e. $6times 5
                        times 4/6^3$                        ) than by enumerating ordered partitions of 3 into 6 pieces and figuring out the relative frequency of each type.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Nov 15 '18 at 16:40









                        Namaste

                        1




                        1










                        answered Mar 21 '18 at 14:31









                        NedNed

                        2,048910




                        2,048910












                        • $begingroup$
                          I agree that the partitions of 365 is tough, but for the dice case, it should be easy enough right: numerator: $frac{6 times 5 times 4}{6}=20$ denominator would be (no repeats + one repeat + all same) = (20+6*5+6) = 56. Which gives prob of no repeats = $frac{20}{56}$ which is vastly different from $frac{6times 5times 4}{6^{3}} = frac{5}{9}$
                          $endgroup$
                          – tpb261
                          Nov 15 '18 at 16:00












                        • $begingroup$
                          But those 56 items are NOT equally likely -- the multiset [1,1,1] occurs less often than the multiset [1,3,4], for example, and though you can, in theory model the dice rolls by unordered combinations (i.e. multisets) you can not get the probability of events by just dividing the cardinality of the event by the cardinality of the sample space. You must weight the (multiset) outcomes by their frequency which for practical purposes takes you back to viewing them as ordered triples, since that view makes them equally likely.
                          $endgroup$
                          – Ned
                          Nov 15 '18 at 23:47










                        • $begingroup$
                          @tpb261 see the comment above (sorry I forgot to address you in the comment)
                          $endgroup$
                          – Ned
                          Nov 19 '18 at 2:52










                        • $begingroup$
                          Thanks!! I did check your comment on phone (from where I don't like to reply), and came to my desktop just now. I missed the point in your answer. My bad.
                          $endgroup$
                          – tpb261
                          Nov 19 '18 at 11:56


















                        • $begingroup$
                          I agree that the partitions of 365 is tough, but for the dice case, it should be easy enough right: numerator: $frac{6 times 5 times 4}{6}=20$ denominator would be (no repeats + one repeat + all same) = (20+6*5+6) = 56. Which gives prob of no repeats = $frac{20}{56}$ which is vastly different from $frac{6times 5times 4}{6^{3}} = frac{5}{9}$
                          $endgroup$
                          – tpb261
                          Nov 15 '18 at 16:00












                        • $begingroup$
                          But those 56 items are NOT equally likely -- the multiset [1,1,1] occurs less often than the multiset [1,3,4], for example, and though you can, in theory model the dice rolls by unordered combinations (i.e. multisets) you can not get the probability of events by just dividing the cardinality of the event by the cardinality of the sample space. You must weight the (multiset) outcomes by their frequency which for practical purposes takes you back to viewing them as ordered triples, since that view makes them equally likely.
                          $endgroup$
                          – Ned
                          Nov 15 '18 at 23:47










                        • $begingroup$
                          @tpb261 see the comment above (sorry I forgot to address you in the comment)
                          $endgroup$
                          – Ned
                          Nov 19 '18 at 2:52










                        • $begingroup$
                          Thanks!! I did check your comment on phone (from where I don't like to reply), and came to my desktop just now. I missed the point in your answer. My bad.
                          $endgroup$
                          – tpb261
                          Nov 19 '18 at 11:56
















                        $begingroup$
                        I agree that the partitions of 365 is tough, but for the dice case, it should be easy enough right: numerator: $frac{6 times 5 times 4}{6}=20$ denominator would be (no repeats + one repeat + all same) = (20+6*5+6) = 56. Which gives prob of no repeats = $frac{20}{56}$ which is vastly different from $frac{6times 5times 4}{6^{3}} = frac{5}{9}$
                        $endgroup$
                        – tpb261
                        Nov 15 '18 at 16:00






                        $begingroup$
                        I agree that the partitions of 365 is tough, but for the dice case, it should be easy enough right: numerator: $frac{6 times 5 times 4}{6}=20$ denominator would be (no repeats + one repeat + all same) = (20+6*5+6) = 56. Which gives prob of no repeats = $frac{20}{56}$ which is vastly different from $frac{6times 5times 4}{6^{3}} = frac{5}{9}$
                        $endgroup$
                        – tpb261
                        Nov 15 '18 at 16:00














                        $begingroup$
                        But those 56 items are NOT equally likely -- the multiset [1,1,1] occurs less often than the multiset [1,3,4], for example, and though you can, in theory model the dice rolls by unordered combinations (i.e. multisets) you can not get the probability of events by just dividing the cardinality of the event by the cardinality of the sample space. You must weight the (multiset) outcomes by their frequency which for practical purposes takes you back to viewing them as ordered triples, since that view makes them equally likely.
                        $endgroup$
                        – Ned
                        Nov 15 '18 at 23:47




                        $begingroup$
                        But those 56 items are NOT equally likely -- the multiset [1,1,1] occurs less often than the multiset [1,3,4], for example, and though you can, in theory model the dice rolls by unordered combinations (i.e. multisets) you can not get the probability of events by just dividing the cardinality of the event by the cardinality of the sample space. You must weight the (multiset) outcomes by their frequency which for practical purposes takes you back to viewing them as ordered triples, since that view makes them equally likely.
                        $endgroup$
                        – Ned
                        Nov 15 '18 at 23:47












                        $begingroup$
                        @tpb261 see the comment above (sorry I forgot to address you in the comment)
                        $endgroup$
                        – Ned
                        Nov 19 '18 at 2:52




                        $begingroup$
                        @tpb261 see the comment above (sorry I forgot to address you in the comment)
                        $endgroup$
                        – Ned
                        Nov 19 '18 at 2:52












                        $begingroup$
                        Thanks!! I did check your comment on phone (from where I don't like to reply), and came to my desktop just now. I missed the point in your answer. My bad.
                        $endgroup$
                        – tpb261
                        Nov 19 '18 at 11:56




                        $begingroup$
                        Thanks!! I did check your comment on phone (from where I don't like to reply), and came to my desktop just now. I missed the point in your answer. My bad.
                        $endgroup$
                        – tpb261
                        Nov 19 '18 at 11:56











                        0












                        $begingroup$

                        Combinations exclude order, while permutations take order into account.



                        So when it comes to assigning ($person_0$,$person_1$, $person_2$) birth dates from ${a,b,c,d}$



                        (Arrays used below, index denotes person_number .. so if $a$ has index 0, a has been assigned to $person_0$)



                        [a,b,c], [c,b,a], [b,c,a] .. would be considered the same one element.



                        So if you are trying to calculate the length of the favorable outcomes space, you'd see why that would be incorrect



                        As for the the denominator i.e the length of space of all possible outcomes $365^n$ should not give anyone trouble



                        So P = $frac {text{length of favorable outcome space}}{text{length of all possible outcomes space}}$



                        So just $C^{365}_n$ would give you a length of set of assignments where order is not important and that is pretty much it. WHen it comes to probability you are missing the deominator



                        P.S. $frac{P^{365}_{n}}{365^{n}}$ is the probability of no one sharing the same birthday






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          Combinations exclude order, while permutations take order into account.



                          So when it comes to assigning ($person_0$,$person_1$, $person_2$) birth dates from ${a,b,c,d}$



                          (Arrays used below, index denotes person_number .. so if $a$ has index 0, a has been assigned to $person_0$)



                          [a,b,c], [c,b,a], [b,c,a] .. would be considered the same one element.



                          So if you are trying to calculate the length of the favorable outcomes space, you'd see why that would be incorrect



                          As for the the denominator i.e the length of space of all possible outcomes $365^n$ should not give anyone trouble



                          So P = $frac {text{length of favorable outcome space}}{text{length of all possible outcomes space}}$



                          So just $C^{365}_n$ would give you a length of set of assignments where order is not important and that is pretty much it. WHen it comes to probability you are missing the deominator



                          P.S. $frac{P^{365}_{n}}{365^{n}}$ is the probability of no one sharing the same birthday






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Combinations exclude order, while permutations take order into account.



                            So when it comes to assigning ($person_0$,$person_1$, $person_2$) birth dates from ${a,b,c,d}$



                            (Arrays used below, index denotes person_number .. so if $a$ has index 0, a has been assigned to $person_0$)



                            [a,b,c], [c,b,a], [b,c,a] .. would be considered the same one element.



                            So if you are trying to calculate the length of the favorable outcomes space, you'd see why that would be incorrect



                            As for the the denominator i.e the length of space of all possible outcomes $365^n$ should not give anyone trouble



                            So P = $frac {text{length of favorable outcome space}}{text{length of all possible outcomes space}}$



                            So just $C^{365}_n$ would give you a length of set of assignments where order is not important and that is pretty much it. WHen it comes to probability you are missing the deominator



                            P.S. $frac{P^{365}_{n}}{365^{n}}$ is the probability of no one sharing the same birthday






                            share|cite|improve this answer











                            $endgroup$



                            Combinations exclude order, while permutations take order into account.



                            So when it comes to assigning ($person_0$,$person_1$, $person_2$) birth dates from ${a,b,c,d}$



                            (Arrays used below, index denotes person_number .. so if $a$ has index 0, a has been assigned to $person_0$)



                            [a,b,c], [c,b,a], [b,c,a] .. would be considered the same one element.



                            So if you are trying to calculate the length of the favorable outcomes space, you'd see why that would be incorrect



                            As for the the denominator i.e the length of space of all possible outcomes $365^n$ should not give anyone trouble



                            So P = $frac {text{length of favorable outcome space}}{text{length of all possible outcomes space}}$



                            So just $C^{365}_n$ would give you a length of set of assignments where order is not important and that is pretty much it. WHen it comes to probability you are missing the deominator



                            P.S. $frac{P^{365}_{n}}{365^{n}}$ is the probability of no one sharing the same birthday







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 5 at 4:04

























                            answered Jan 5 at 3:38









                            functor-soupfunctor-soup

                            989




                            989






























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