Probability of successful Secret Santa selection












0












$begingroup$


In the following Secret Santa Scenario:



There are 12 people, each have written their name on a scrap of paper and put it into a bowl. The bowl is then passed around and each of the 12 picks out a name and reads it.



What is the probability that no one picks out their own name?



At first I thought it would simply be 11/12 x 10/11 x 9/10...etc. But I thought that wouldn’t work because there are situations where someone’s name has already been drawn and there is 0 chance for them to pick out their name, so I’m stuck.






probability permutations combinations







share|cite|improve this question









$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/Derangement
    $endgroup$
    – Lord Shark the Unknown
    Dec 15 '18 at 7:16
















0












$begingroup$


In the following Secret Santa Scenario:



There are 12 people, each have written their name on a scrap of paper and put it into a bowl. The bowl is then passed around and each of the 12 picks out a name and reads it.



What is the probability that no one picks out their own name?



At first I thought it would simply be 11/12 x 10/11 x 9/10...etc. But I thought that wouldn’t work because there are situations where someone’s name has already been drawn and there is 0 chance for them to pick out their name, so I’m stuck.






probability permutations combinations







share|cite|improve this question









$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/Derangement
    $endgroup$
    – Lord Shark the Unknown
    Dec 15 '18 at 7:16














0












0








0





$begingroup$


In the following Secret Santa Scenario:



There are 12 people, each have written their name on a scrap of paper and put it into a bowl. The bowl is then passed around and each of the 12 picks out a name and reads it.



What is the probability that no one picks out their own name?



At first I thought it would simply be 11/12 x 10/11 x 9/10...etc. But I thought that wouldn’t work because there are situations where someone’s name has already been drawn and there is 0 chance for them to pick out their name, so I’m stuck.






probability permutations combinations







share|cite|improve this question









$endgroup$




In the following Secret Santa Scenario:



There are 12 people, each have written their name on a scrap of paper and put it into a bowl. The bowl is then passed around and each of the 12 picks out a name and reads it.



What is the probability that no one picks out their own name?



At first I thought it would simply be 11/12 x 10/11 x 9/10...etc. But I thought that wouldn’t work because there are situations where someone’s name has already been drawn and there is 0 chance for them to pick out their name, so I’m stuck.






probability permutations combinations




probability permutations combinations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 15 '18 at 7:11









M. WeateM. Weate

445




445












  • $begingroup$
    en.wikipedia.org/wiki/Derangement
    $endgroup$
    – Lord Shark the Unknown
    Dec 15 '18 at 7:16


















  • $begingroup$
    en.wikipedia.org/wiki/Derangement
    $endgroup$
    – Lord Shark the Unknown
    Dec 15 '18 at 7:16
















$begingroup$
en.wikipedia.org/wiki/Derangement
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 7:16




$begingroup$
en.wikipedia.org/wiki/Derangement
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 7:16










2 Answers
2






active

oldest

votes


















1












$begingroup$

There is actually a term for something like this: it's called a derangement. (https://en.wikipedia.org/wiki/Derangement)



Generally, if you have $n$ objects, and want to rearrange them without them ending up in the same spot, the number of such rearrangements is given by



$$text{# of derangements} = n! cdot sum_{k=1}^n frac{(-1)^k}{k!}$$



among other potential relations. This is commonly denoted $!n$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Consider all the possible permutations, and we have to consider how many of those are derangement, that is permutation where no element appear in the original position.



    The answer is



    $$frac{!12}{12!}=frac{lfloor frac{12!}e + frac12 rfloor}{12!}=frac{176214841}{12!}approx 0.3679$$






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      There is actually a term for something like this: it's called a derangement. (https://en.wikipedia.org/wiki/Derangement)



      Generally, if you have $n$ objects, and want to rearrange them without them ending up in the same spot, the number of such rearrangements is given by



      $$text{# of derangements} = n! cdot sum_{k=1}^n frac{(-1)^k}{k!}$$



      among other potential relations. This is commonly denoted $!n$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        There is actually a term for something like this: it's called a derangement. (https://en.wikipedia.org/wiki/Derangement)



        Generally, if you have $n$ objects, and want to rearrange them without them ending up in the same spot, the number of such rearrangements is given by



        $$text{# of derangements} = n! cdot sum_{k=1}^n frac{(-1)^k}{k!}$$



        among other potential relations. This is commonly denoted $!n$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          There is actually a term for something like this: it's called a derangement. (https://en.wikipedia.org/wiki/Derangement)



          Generally, if you have $n$ objects, and want to rearrange them without them ending up in the same spot, the number of such rearrangements is given by



          $$text{# of derangements} = n! cdot sum_{k=1}^n frac{(-1)^k}{k!}$$



          among other potential relations. This is commonly denoted $!n$.






          share|cite|improve this answer









          $endgroup$



          There is actually a term for something like this: it's called a derangement. (https://en.wikipedia.org/wiki/Derangement)



          Generally, if you have $n$ objects, and want to rearrange them without them ending up in the same spot, the number of such rearrangements is given by



          $$text{# of derangements} = n! cdot sum_{k=1}^n frac{(-1)^k}{k!}$$



          among other potential relations. This is commonly denoted $!n$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 7:15









          Eevee TrainerEevee Trainer

          10.4k31742




          10.4k31742























              0












              $begingroup$

              Consider all the possible permutations, and we have to consider how many of those are derangement, that is permutation where no element appear in the original position.



              The answer is



              $$frac{!12}{12!}=frac{lfloor frac{12!}e + frac12 rfloor}{12!}=frac{176214841}{12!}approx 0.3679$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Consider all the possible permutations, and we have to consider how many of those are derangement, that is permutation where no element appear in the original position.



                The answer is



                $$frac{!12}{12!}=frac{lfloor frac{12!}e + frac12 rfloor}{12!}=frac{176214841}{12!}approx 0.3679$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Consider all the possible permutations, and we have to consider how many of those are derangement, that is permutation where no element appear in the original position.



                  The answer is



                  $$frac{!12}{12!}=frac{lfloor frac{12!}e + frac12 rfloor}{12!}=frac{176214841}{12!}approx 0.3679$$






                  share|cite|improve this answer









                  $endgroup$



                  Consider all the possible permutations, and we have to consider how many of those are derangement, that is permutation where no element appear in the original position.



                  The answer is



                  $$frac{!12}{12!}=frac{lfloor frac{12!}e + frac12 rfloor}{12!}=frac{176214841}{12!}approx 0.3679$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 15 '18 at 7:17









                  Siong Thye GohSiong Thye Goh

                  104k1468120




                  104k1468120






























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