Why does my approach to the birthday problem(with 3 people) produce a wrong result?












1












$begingroup$


I'm trying to calculate the probability that at least two people out of three have the same birthday (simplified Birthday Problem). I know that calculating it using the complementary event is easiest, but I want to calculate it directly. So here's an approach that works:



$P(A) = 3cdotfrac{1}{356}cdotfrac{364}{356} + frac{1}{365^2}$
where I account for the three ways 2 people can have the same birthday and the other a different birthday and plus for the event that all three have the same birthday which has a probability of $frac{1}{365^2}$.



But here's a calculation that doesn't work: Let's first define the events:



$E_{1}:$ person 1 and person 2 have the same birthday.



$E_{2}:$ person 1 and person 3 have the same birthday.



$E_{3}:$ person 2 and person 3 have the same birthday.



Now we can calculate the probability of at least two people having the same birthday as:
$$P(E_{1}cup E_{2}cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}cap E_{2}) - P(E_{1}cap E_{3}) - P(E_{2}cap E_{3}) - P(E_{1}cap E_{2}cap E_{3}) $$



with $P(E_{i}) = frac{1}{365}$ , $P(E_{i}cap E_{j}) = frac{1}{365^2}$ and $P(E_{1}cap E_{2} cap E_{3}) = frac{1}{365^2}$



but this produces the wrong answer 0.008189153 instead of the correct one: 0.008204165.



Why does the second calculation produce a wrong answer? What am I not taking into consideration?






probability birthday







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  • 1




    $begingroup$
    Why did you subtract $P(E_1 cap E_2 cap E_3)$? You should have added it.
    $endgroup$
    – fleablood
    Dec 21 '18 at 17:30
















1












$begingroup$


I'm trying to calculate the probability that at least two people out of three have the same birthday (simplified Birthday Problem). I know that calculating it using the complementary event is easiest, but I want to calculate it directly. So here's an approach that works:



$P(A) = 3cdotfrac{1}{356}cdotfrac{364}{356} + frac{1}{365^2}$
where I account for the three ways 2 people can have the same birthday and the other a different birthday and plus for the event that all three have the same birthday which has a probability of $frac{1}{365^2}$.



But here's a calculation that doesn't work: Let's first define the events:



$E_{1}:$ person 1 and person 2 have the same birthday.



$E_{2}:$ person 1 and person 3 have the same birthday.



$E_{3}:$ person 2 and person 3 have the same birthday.



Now we can calculate the probability of at least two people having the same birthday as:
$$P(E_{1}cup E_{2}cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}cap E_{2}) - P(E_{1}cap E_{3}) - P(E_{2}cap E_{3}) - P(E_{1}cap E_{2}cap E_{3}) $$



with $P(E_{i}) = frac{1}{365}$ , $P(E_{i}cap E_{j}) = frac{1}{365^2}$ and $P(E_{1}cap E_{2} cap E_{3}) = frac{1}{365^2}$



but this produces the wrong answer 0.008189153 instead of the correct one: 0.008204165.



Why does the second calculation produce a wrong answer? What am I not taking into consideration?






probability birthday







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Why did you subtract $P(E_1 cap E_2 cap E_3)$? You should have added it.
    $endgroup$
    – fleablood
    Dec 21 '18 at 17:30














1












1








1


1



$begingroup$


I'm trying to calculate the probability that at least two people out of three have the same birthday (simplified Birthday Problem). I know that calculating it using the complementary event is easiest, but I want to calculate it directly. So here's an approach that works:



$P(A) = 3cdotfrac{1}{356}cdotfrac{364}{356} + frac{1}{365^2}$
where I account for the three ways 2 people can have the same birthday and the other a different birthday and plus for the event that all three have the same birthday which has a probability of $frac{1}{365^2}$.



But here's a calculation that doesn't work: Let's first define the events:



$E_{1}:$ person 1 and person 2 have the same birthday.



$E_{2}:$ person 1 and person 3 have the same birthday.



$E_{3}:$ person 2 and person 3 have the same birthday.



Now we can calculate the probability of at least two people having the same birthday as:
$$P(E_{1}cup E_{2}cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}cap E_{2}) - P(E_{1}cap E_{3}) - P(E_{2}cap E_{3}) - P(E_{1}cap E_{2}cap E_{3}) $$



with $P(E_{i}) = frac{1}{365}$ , $P(E_{i}cap E_{j}) = frac{1}{365^2}$ and $P(E_{1}cap E_{2} cap E_{3}) = frac{1}{365^2}$



but this produces the wrong answer 0.008189153 instead of the correct one: 0.008204165.



Why does the second calculation produce a wrong answer? What am I not taking into consideration?






probability birthday







share|cite|improve this question









$endgroup$




I'm trying to calculate the probability that at least two people out of three have the same birthday (simplified Birthday Problem). I know that calculating it using the complementary event is easiest, but I want to calculate it directly. So here's an approach that works:



$P(A) = 3cdotfrac{1}{356}cdotfrac{364}{356} + frac{1}{365^2}$
where I account for the three ways 2 people can have the same birthday and the other a different birthday and plus for the event that all three have the same birthday which has a probability of $frac{1}{365^2}$.



But here's a calculation that doesn't work: Let's first define the events:



$E_{1}:$ person 1 and person 2 have the same birthday.



$E_{2}:$ person 1 and person 3 have the same birthday.



$E_{3}:$ person 2 and person 3 have the same birthday.



Now we can calculate the probability of at least two people having the same birthday as:
$$P(E_{1}cup E_{2}cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}cap E_{2}) - P(E_{1}cap E_{3}) - P(E_{2}cap E_{3}) - P(E_{1}cap E_{2}cap E_{3}) $$



with $P(E_{i}) = frac{1}{365}$ , $P(E_{i}cap E_{j}) = frac{1}{365^2}$ and $P(E_{1}cap E_{2} cap E_{3}) = frac{1}{365^2}$



but this produces the wrong answer 0.008189153 instead of the correct one: 0.008204165.



Why does the second calculation produce a wrong answer? What am I not taking into consideration?






probability birthday




probability birthday






share|cite|improve this question













share|cite|improve this question











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asked Dec 21 '18 at 17:14









user3071028user3071028

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  • 1




    $begingroup$
    Why did you subtract $P(E_1 cap E_2 cap E_3)$? You should have added it.
    $endgroup$
    – fleablood
    Dec 21 '18 at 17:30














  • 1




    $begingroup$
    Why did you subtract $P(E_1 cap E_2 cap E_3)$? You should have added it.
    $endgroup$
    – fleablood
    Dec 21 '18 at 17:30








1




1




$begingroup$
Why did you subtract $P(E_1 cap E_2 cap E_3)$? You should have added it.
$endgroup$
– fleablood
Dec 21 '18 at 17:30




$begingroup$
Why did you subtract $P(E_1 cap E_2 cap E_3)$? You should have added it.
$endgroup$
– fleablood
Dec 21 '18 at 17:30










1 Answer
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$begingroup$

You have done your inclusion-exclusion principle incorrectly. Instead of:



$$P(E_{1}cup E_{2}cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}cap E_{2}) - P(E_{1}cap E_{3}) - P(E_{2}cap E_{3}) - P(E_{1}cap E_{2}cap E_{3}) $$



It should have been:



$$P(E_{1}cup E_{2}cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}cap E_{2}) - P(E_{1}cap E_{3}) - P(E_{2}cap E_{3}) color{red}+ P(E_{1}cap E_{2}cap E_{3}) $$






share|cite|improve this answer











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    $begingroup$

    You have done your inclusion-exclusion principle incorrectly. Instead of:



    $$P(E_{1}cup E_{2}cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}cap E_{2}) - P(E_{1}cap E_{3}) - P(E_{2}cap E_{3}) - P(E_{1}cap E_{2}cap E_{3}) $$



    It should have been:



    $$P(E_{1}cup E_{2}cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}cap E_{2}) - P(E_{1}cap E_{3}) - P(E_{2}cap E_{3}) color{red}+ P(E_{1}cap E_{2}cap E_{3}) $$






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      You have done your inclusion-exclusion principle incorrectly. Instead of:



      $$P(E_{1}cup E_{2}cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}cap E_{2}) - P(E_{1}cap E_{3}) - P(E_{2}cap E_{3}) - P(E_{1}cap E_{2}cap E_{3}) $$



      It should have been:



      $$P(E_{1}cup E_{2}cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}cap E_{2}) - P(E_{1}cap E_{3}) - P(E_{2}cap E_{3}) color{red}+ P(E_{1}cap E_{2}cap E_{3}) $$






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        You have done your inclusion-exclusion principle incorrectly. Instead of:



        $$P(E_{1}cup E_{2}cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}cap E_{2}) - P(E_{1}cap E_{3}) - P(E_{2}cap E_{3}) - P(E_{1}cap E_{2}cap E_{3}) $$



        It should have been:



        $$P(E_{1}cup E_{2}cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}cap E_{2}) - P(E_{1}cap E_{3}) - P(E_{2}cap E_{3}) color{red}+ P(E_{1}cap E_{2}cap E_{3}) $$






        share|cite|improve this answer











        $endgroup$



        You have done your inclusion-exclusion principle incorrectly. Instead of:



        $$P(E_{1}cup E_{2}cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}cap E_{2}) - P(E_{1}cap E_{3}) - P(E_{2}cap E_{3}) - P(E_{1}cap E_{2}cap E_{3}) $$



        It should have been:



        $$P(E_{1}cup E_{2}cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1}cap E_{2}) - P(E_{1}cap E_{3}) - P(E_{2}cap E_{3}) color{red}+ P(E_{1}cap E_{2}cap E_{3}) $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 21 '18 at 17:38









        Bram28

        63.2k44793




        63.2k44793










        answered Dec 21 '18 at 17:19









        DubsDubs

        55926




        55926






























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