Drunk men finding their tents

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Problem:



There are four couples camping at the lakeside. They start drinking and as the women get bored and tired, they all go to sleep into their tents. The men continue and get very drunk. In the morning they all go randomly into a tent (but each to a separate one). What is the probability that



  1. P (all men go to their own tents)

  2. P (3 go their own tent, and 1 to a foreign one)

  3. P (2 go their own tent, and 2 to a foreign one)

  4. P (1 go their own tent, and 3 to a foreign one)

  5. P (all men are mistaken)

My attempt:



  1. P (all men go to their own tents) = 1/4 * 1/3 * 1/2 * 1 = 1/24 = 0.04167

because the P (1 man finds his tent) = 1/4, then the P the next one gets it right is 1/3 because he can go to one less place, etc.



  1. P (3 go their own tent, and 1 to a foreign one) = 0 because impossible


  2. P (2 go their own tent, and 2 to a foreign one) = (4! / 2!*2!) / 4! = 6/24 = 0.25


because we need to select two men who go to their right place, two that do not, and (in the denominator:) they can altogether be placed 4! different ways.



  1. P (1 go their own tent, and 3 to a foreign one) = (4! / 1!*3!) / 4! = 4/24 = 0.16667

because the same logic as no.3.



  1. P (all men are mistaken) = 3/4 * 2/3 * 1/2 * 1 = 6/24 = O.25

because the P that the first man goes to a wrong place is 3/4, the probability that the second one goes to a wrong place is one less: 2/3, etc.



Question:



But there is something wrong with my attempt as the probabilities do not add up to 100%. Where do I go wrong?



  1. 0.04167

  2. 0.00000

  3. 0.25000

  4. 0.16667


  5. 0.25000



    0.70834







probability







share|cite|improve this question

























    up vote
    2
    down vote

    favorite












    Problem:



    There are four couples camping at the lakeside. They start drinking and as the women get bored and tired, they all go to sleep into their tents. The men continue and get very drunk. In the morning they all go randomly into a tent (but each to a separate one). What is the probability that



    1. P (all men go to their own tents)

    2. P (3 go their own tent, and 1 to a foreign one)

    3. P (2 go their own tent, and 2 to a foreign one)

    4. P (1 go their own tent, and 3 to a foreign one)

    5. P (all men are mistaken)

    My attempt:



    1. P (all men go to their own tents) = 1/4 * 1/3 * 1/2 * 1 = 1/24 = 0.04167

    because the P (1 man finds his tent) = 1/4, then the P the next one gets it right is 1/3 because he can go to one less place, etc.



    1. P (3 go their own tent, and 1 to a foreign one) = 0 because impossible


    2. P (2 go their own tent, and 2 to a foreign one) = (4! / 2!*2!) / 4! = 6/24 = 0.25


    because we need to select two men who go to their right place, two that do not, and (in the denominator:) they can altogether be placed 4! different ways.



    1. P (1 go their own tent, and 3 to a foreign one) = (4! / 1!*3!) / 4! = 4/24 = 0.16667

    because the same logic as no.3.



    1. P (all men are mistaken) = 3/4 * 2/3 * 1/2 * 1 = 6/24 = O.25

    because the P that the first man goes to a wrong place is 3/4, the probability that the second one goes to a wrong place is one less: 2/3, etc.



    Question:



    But there is something wrong with my attempt as the probabilities do not add up to 100%. Where do I go wrong?



    1. 0.04167

    2. 0.00000

    3. 0.25000

    4. 0.16667


    5. 0.25000



      0.70834







    probability







    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Problem:



      There are four couples camping at the lakeside. They start drinking and as the women get bored and tired, they all go to sleep into their tents. The men continue and get very drunk. In the morning they all go randomly into a tent (but each to a separate one). What is the probability that



      1. P (all men go to their own tents)

      2. P (3 go their own tent, and 1 to a foreign one)

      3. P (2 go their own tent, and 2 to a foreign one)

      4. P (1 go their own tent, and 3 to a foreign one)

      5. P (all men are mistaken)

      My attempt:



      1. P (all men go to their own tents) = 1/4 * 1/3 * 1/2 * 1 = 1/24 = 0.04167

      because the P (1 man finds his tent) = 1/4, then the P the next one gets it right is 1/3 because he can go to one less place, etc.



      1. P (3 go their own tent, and 1 to a foreign one) = 0 because impossible


      2. P (2 go their own tent, and 2 to a foreign one) = (4! / 2!*2!) / 4! = 6/24 = 0.25


      because we need to select two men who go to their right place, two that do not, and (in the denominator:) they can altogether be placed 4! different ways.



      1. P (1 go their own tent, and 3 to a foreign one) = (4! / 1!*3!) / 4! = 4/24 = 0.16667

      because the same logic as no.3.



      1. P (all men are mistaken) = 3/4 * 2/3 * 1/2 * 1 = 6/24 = O.25

      because the P that the first man goes to a wrong place is 3/4, the probability that the second one goes to a wrong place is one less: 2/3, etc.



      Question:



      But there is something wrong with my attempt as the probabilities do not add up to 100%. Where do I go wrong?



      1. 0.04167

      2. 0.00000

      3. 0.25000

      4. 0.16667


      5. 0.25000



        0.70834







      probability







      share|cite|improve this question













      Problem:



      There are four couples camping at the lakeside. They start drinking and as the women get bored and tired, they all go to sleep into their tents. The men continue and get very drunk. In the morning they all go randomly into a tent (but each to a separate one). What is the probability that



      1. P (all men go to their own tents)

      2. P (3 go their own tent, and 1 to a foreign one)

      3. P (2 go their own tent, and 2 to a foreign one)

      4. P (1 go their own tent, and 3 to a foreign one)

      5. P (all men are mistaken)

      My attempt:



      1. P (all men go to their own tents) = 1/4 * 1/3 * 1/2 * 1 = 1/24 = 0.04167

      because the P (1 man finds his tent) = 1/4, then the P the next one gets it right is 1/3 because he can go to one less place, etc.



      1. P (3 go their own tent, and 1 to a foreign one) = 0 because impossible


      2. P (2 go their own tent, and 2 to a foreign one) = (4! / 2!*2!) / 4! = 6/24 = 0.25


      because we need to select two men who go to their right place, two that do not, and (in the denominator:) they can altogether be placed 4! different ways.



      1. P (1 go their own tent, and 3 to a foreign one) = (4! / 1!*3!) / 4! = 4/24 = 0.16667

      because the same logic as no.3.



      1. P (all men are mistaken) = 3/4 * 2/3 * 1/2 * 1 = 6/24 = O.25

      because the P that the first man goes to a wrong place is 3/4, the probability that the second one goes to a wrong place is one less: 2/3, etc.



      Question:



      But there is something wrong with my attempt as the probabilities do not add up to 100%. Where do I go wrong?



      1. 0.04167

      2. 0.00000

      3. 0.25000

      4. 0.16667


      5. 0.25000



        0.70834







      probability




      probability






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      share|cite|improve this question











      share|cite|improve this question




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      asked 2 hours ago









      malasi

      354




      354




















          2 Answers
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          The number of ways for $k$ men to get to the correct tent is
          $$
          overbrace binom4k ^substacktextways to choose\textthe correct tentsoverbrace mathcalD_4-k vphantombinom4k^substacktextways to choose\textthe wrong tents
          $$
          where $mathcalD_k$ is the number of derangements of $k$ items. That is,
          $$
          beginarrayc
          k&binom4kmathcalD_4-k&P(k)\hline
          0&9&frac38\
          1&8&frac13\
          2&6&frac14\
          3&0&0\
          4&1&frac124
          endarray
          $$






          share|cite|improve this answer



























            up vote
            0
            down vote













            Robjohn gives the right answer, here are the flaws in your reasoning :



            1. There are 4 ways of choosing which guy goes to the right tent, but then there are 2 ways (not only 1) for the three other guys to be all in a wrong tent.


            2. When the first guy has chosen a wrong tent with probability 3/4, the probability for the second guy to be wrong too is not always 2/3. It is 1 in one case out of three: when his own tent is already occupied by the first guy.






            share|cite|improve this answer




















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              2 Answers
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              2 Answers
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              active

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              active

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              up vote
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              The number of ways for $k$ men to get to the correct tent is
              $$
              overbrace binom4k ^substacktextways to choose\textthe correct tentsoverbrace mathcalD_4-k vphantombinom4k^substacktextways to choose\textthe wrong tents
              $$
              where $mathcalD_k$ is the number of derangements of $k$ items. That is,
              $$
              beginarrayc
              k&binom4kmathcalD_4-k&P(k)\hline
              0&9&frac38\
              1&8&frac13\
              2&6&frac14\
              3&0&0\
              4&1&frac124
              endarray
              $$






              share|cite|improve this answer
























                up vote
                3
                down vote













                The number of ways for $k$ men to get to the correct tent is
                $$
                overbrace binom4k ^substacktextways to choose\textthe correct tentsoverbrace mathcalD_4-k vphantombinom4k^substacktextways to choose\textthe wrong tents
                $$
                where $mathcalD_k$ is the number of derangements of $k$ items. That is,
                $$
                beginarrayc
                k&binom4kmathcalD_4-k&P(k)\hline
                0&9&frac38\
                1&8&frac13\
                2&6&frac14\
                3&0&0\
                4&1&frac124
                endarray
                $$






                share|cite|improve this answer






















                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  The number of ways for $k$ men to get to the correct tent is
                  $$
                  overbrace binom4k ^substacktextways to choose\textthe correct tentsoverbrace mathcalD_4-k vphantombinom4k^substacktextways to choose\textthe wrong tents
                  $$
                  where $mathcalD_k$ is the number of derangements of $k$ items. That is,
                  $$
                  beginarrayc
                  k&binom4kmathcalD_4-k&P(k)\hline
                  0&9&frac38\
                  1&8&frac13\
                  2&6&frac14\
                  3&0&0\
                  4&1&frac124
                  endarray
                  $$






                  share|cite|improve this answer












                  The number of ways for $k$ men to get to the correct tent is
                  $$
                  overbrace binom4k ^substacktextways to choose\textthe correct tentsoverbrace mathcalD_4-k vphantombinom4k^substacktextways to choose\textthe wrong tents
                  $$
                  where $mathcalD_k$ is the number of derangements of $k$ items. That is,
                  $$
                  beginarrayc
                  k&binom4kmathcalD_4-k&P(k)\hline
                  0&9&frac38\
                  1&8&frac13\
                  2&6&frac14\
                  3&0&0\
                  4&1&frac124
                  endarray
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  robjohn♦

                  259k26299614




                  259k26299614




















                      up vote
                      0
                      down vote













                      Robjohn gives the right answer, here are the flaws in your reasoning :



                      1. There are 4 ways of choosing which guy goes to the right tent, but then there are 2 ways (not only 1) for the three other guys to be all in a wrong tent.


                      2. When the first guy has chosen a wrong tent with probability 3/4, the probability for the second guy to be wrong too is not always 2/3. It is 1 in one case out of three: when his own tent is already occupied by the first guy.






                      share|cite|improve this answer
























                        up vote
                        0
                        down vote













                        Robjohn gives the right answer, here are the flaws in your reasoning :



                        1. There are 4 ways of choosing which guy goes to the right tent, but then there are 2 ways (not only 1) for the three other guys to be all in a wrong tent.


                        2. When the first guy has chosen a wrong tent with probability 3/4, the probability for the second guy to be wrong too is not always 2/3. It is 1 in one case out of three: when his own tent is already occupied by the first guy.






                        share|cite|improve this answer






















                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          Robjohn gives the right answer, here are the flaws in your reasoning :



                          1. There are 4 ways of choosing which guy goes to the right tent, but then there are 2 ways (not only 1) for the three other guys to be all in a wrong tent.


                          2. When the first guy has chosen a wrong tent with probability 3/4, the probability for the second guy to be wrong too is not always 2/3. It is 1 in one case out of three: when his own tent is already occupied by the first guy.






                          share|cite|improve this answer












                          Robjohn gives the right answer, here are the flaws in your reasoning :



                          1. There are 4 ways of choosing which guy goes to the right tent, but then there are 2 ways (not only 1) for the three other guys to be all in a wrong tent.


                          2. When the first guy has chosen a wrong tent with probability 3/4, the probability for the second guy to be wrong too is not always 2/3. It is 1 in one case out of three: when his own tent is already occupied by the first guy.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 11 mins ago









                          Evargalo

                          2,30218




                          2,30218



























                               

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