Probability/Combinatorics problem regarding seating at multiple tables with men and women












1












$begingroup$


I recently stumbled across a question which is typically posed in interviews:




If there are $20$ people with $17$ men and $3$ women and they seat themselves randomly at $4$ Tables (A,B,C,D) with $5$ people each. With all arrangements being equally likely, what is the probability that no woman sits at table A ?




I was trying to work out this so I figured you would have to look at the number of combinations in choosing people at each table so $20choose 5$, then $15choose 5$....so on.



But, someone said the answer is just $frac{17choose 5}{20 choose 5}$, which comes out to ~$0.399$, because the number of combinations of men is $17 choose 5$ and the total combinations is $20 choose 5$.



If anyone can provide guidance on how to start this problem that would be great. I am a newbie to this forum so please bear with me. Thank you.






probability combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    The probability that no women sit at table $A$ is equal to the probability that only men sit at table $A$.
    $endgroup$
    – N. F. Taussig
    Dec 16 '18 at 1:03
















1












$begingroup$


I recently stumbled across a question which is typically posed in interviews:




If there are $20$ people with $17$ men and $3$ women and they seat themselves randomly at $4$ Tables (A,B,C,D) with $5$ people each. With all arrangements being equally likely, what is the probability that no woman sits at table A ?




I was trying to work out this so I figured you would have to look at the number of combinations in choosing people at each table so $20choose 5$, then $15choose 5$....so on.



But, someone said the answer is just $frac{17choose 5}{20 choose 5}$, which comes out to ~$0.399$, because the number of combinations of men is $17 choose 5$ and the total combinations is $20 choose 5$.



If anyone can provide guidance on how to start this problem that would be great. I am a newbie to this forum so please bear with me. Thank you.






probability combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    The probability that no women sit at table $A$ is equal to the probability that only men sit at table $A$.
    $endgroup$
    – N. F. Taussig
    Dec 16 '18 at 1:03














1












1








1





$begingroup$


I recently stumbled across a question which is typically posed in interviews:




If there are $20$ people with $17$ men and $3$ women and they seat themselves randomly at $4$ Tables (A,B,C,D) with $5$ people each. With all arrangements being equally likely, what is the probability that no woman sits at table A ?




I was trying to work out this so I figured you would have to look at the number of combinations in choosing people at each table so $20choose 5$, then $15choose 5$....so on.



But, someone said the answer is just $frac{17choose 5}{20 choose 5}$, which comes out to ~$0.399$, because the number of combinations of men is $17 choose 5$ and the total combinations is $20 choose 5$.



If anyone can provide guidance on how to start this problem that would be great. I am a newbie to this forum so please bear with me. Thank you.






probability combinatorics







share|cite|improve this question











$endgroup$




I recently stumbled across a question which is typically posed in interviews:




If there are $20$ people with $17$ men and $3$ women and they seat themselves randomly at $4$ Tables (A,B,C,D) with $5$ people each. With all arrangements being equally likely, what is the probability that no woman sits at table A ?




I was trying to work out this so I figured you would have to look at the number of combinations in choosing people at each table so $20choose 5$, then $15choose 5$....so on.



But, someone said the answer is just $frac{17choose 5}{20 choose 5}$, which comes out to ~$0.399$, because the number of combinations of men is $17 choose 5$ and the total combinations is $20 choose 5$.



If anyone can provide guidance on how to start this problem that would be great. I am a newbie to this forum so please bear with me. Thank you.






probability combinatorics




probability combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 3:03









hardmath

28.9k95297




28.9k95297










asked Dec 16 '18 at 0:41









bkrishbkrish

63




63












  • $begingroup$
    The probability that no women sit at table $A$ is equal to the probability that only men sit at table $A$.
    $endgroup$
    – N. F. Taussig
    Dec 16 '18 at 1:03


















  • $begingroup$
    The probability that no women sit at table $A$ is equal to the probability that only men sit at table $A$.
    $endgroup$
    – N. F. Taussig
    Dec 16 '18 at 1:03
















$begingroup$
The probability that no women sit at table $A$ is equal to the probability that only men sit at table $A$.
$endgroup$
– N. F. Taussig
Dec 16 '18 at 1:03




$begingroup$
The probability that no women sit at table $A$ is equal to the probability that only men sit at table $A$.
$endgroup$
– N. F. Taussig
Dec 16 '18 at 1:03










2 Answers
2






active

oldest

votes


















1












$begingroup$

Formulate the question this way: there are 20 people in a room, 17M and 3W. there's a table with 5 seats, what is the probability that only men sit in this table?



Part 1: the universe:



How many combination of 5 can you form from 20? Answer $dbinom{20}{5}$



Part 2: your request:



You request to know how many M only can be seated in this table? Answer $dbinom{17}{5}$




Part 3: Result

so the answer is:

$frac{17choose 5}{20choose 5}$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    I think that the someone is correct. His or her reasoning goes along the lines: let's define only men sitting at table A a success. The probability that there will be only be men at table A is then the number of ways we can choose $5$ men from $20$ as a proportion of the number of ways we can choose $5$ people from $20$:



    $$frac{17choose 5}{20choose 5}$$



    From this point, we don't care who sits at the other tables, so the other tables cannot influence whether the experiment is a success.



    Your previous reasoning can be applied correctly, but with unnecessary detail. The reasoning would be along the lines: what is the probability that only men sit at table A, any combination of men and women sit at table B, any combination at C and any combination at D? The solution to this would be the above result times $1^3$, which is, of course, equivalent.



    Edit



    Just to address the iterative solution described in the comments below:



    I see a couple of issues with the iterative solution as you've expressed it.




    1. The probability that the second person is male, for example, depends on whether the first person selected was male. It seems that you are assuming that the previous guest was male in each iteration, which is not necessarily the case.


    2. If we are randomly assigning tables as the guests arrive, then the probability that someone is assigned to table A is a fixed $frac 1 4$ until one of the tables is filled. (It doesn't follow the sequence $frac{5}{20}$, $frac{4}{19}$, $dots$)



    In coming up with your iterative solution it seems as though you are (correctly) interpreting the given information as follows:




    20 (17 male + 3 female) guests arrive in a random order (such that all permutations are equally likely) and are randomly assigned an available table from A, B, C and D (such that each available table is equally likely to be assigned) as they arrive.




    To see how your iterative solution for this problem fails, think about how we would solve the following (different from the original) problem:




    What is the probability that the first five guests are male and are assigned to table A?




    Then the solution would be:



    $$left(frac 1 4right)^5left(frac{17}{20}right)left(frac{16}{19}right)left(frac{15}{18}right)left(frac{14}{17}right)left(frac{13}{16}right)$$



    which seems to be where you were heading with your calculations.



    Note that this is very different from the problem:




    What is the probability that the five guests assigned to table A once all the guests have arrived are all male?




    The answer to the first problem is the probability of only one of the (many) potential outcomes that leads to table A being filled with men only. (Another outcome, for example, is that the three women arrive first but are assigned to different tables.)



    Note that the iterative solution, while it would work (for example in computer simulations), would require us to keep track of a huge tree in order for us to identify all the successes.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I see thanks for the clarification. But if I think about this iteratively, If I were someone observing or"randomly" picking, the first person to sit at a table has a (17/20) chance of being a male. What is the probability that person sits at table A? (5/20). The second person has a (16/19) chance of being male. The probability that person also sits at table A is now (4/19), since one seat is taken. So this becomes a productive multiplication of (17/20) * (5/20) * (16/19) * (4/19).. so on so forth which is much lower than the result. Where am I going wrong here ?
      $endgroup$
      – bkrish
      Dec 16 '18 at 5:53












    • $begingroup$
      Are you not assuming that table A always gets filled first in your solution ?
      $endgroup$
      – bkrish
      Dec 16 '18 at 5:55










    • $begingroup$
      Answer edited to address this.
      $endgroup$
      – Richard Ambler
      Dec 16 '18 at 6:53










    • $begingroup$
      Oh wow ok so I see where my reasoning went off. So you are essentially saying, that the probability the first 5 people are male and assigned to any random seat is equivalent to the first answer to the problem (i.e) (17C5)/(20C5) which is what the iterative solution converges to without the (1/4) factor. Is that correct ?
      $endgroup$
      – bkrish
      Dec 16 '18 at 7:40










    • $begingroup$
      Yep. Just to clarify: the iterative interpretation is also valid (and would give the same answer to the original problem if we were to track all the "success" branches). The only thing you did wrong before (besides the table assignment probabilities) was only track one of the many "success" branches.
      $endgroup$
      – Richard Ambler
      Dec 16 '18 at 8:04











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Formulate the question this way: there are 20 people in a room, 17M and 3W. there's a table with 5 seats, what is the probability that only men sit in this table?



    Part 1: the universe:



    How many combination of 5 can you form from 20? Answer $dbinom{20}{5}$



    Part 2: your request:



    You request to know how many M only can be seated in this table? Answer $dbinom{17}{5}$




    Part 3: Result

    so the answer is:

    $frac{17choose 5}{20choose 5}$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Formulate the question this way: there are 20 people in a room, 17M and 3W. there's a table with 5 seats, what is the probability that only men sit in this table?



      Part 1: the universe:



      How many combination of 5 can you form from 20? Answer $dbinom{20}{5}$



      Part 2: your request:



      You request to know how many M only can be seated in this table? Answer $dbinom{17}{5}$




      Part 3: Result

      so the answer is:

      $frac{17choose 5}{20choose 5}$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Formulate the question this way: there are 20 people in a room, 17M and 3W. there's a table with 5 seats, what is the probability that only men sit in this table?



        Part 1: the universe:



        How many combination of 5 can you form from 20? Answer $dbinom{20}{5}$



        Part 2: your request:



        You request to know how many M only can be seated in this table? Answer $dbinom{17}{5}$




        Part 3: Result

        so the answer is:

        $frac{17choose 5}{20choose 5}$






        share|cite|improve this answer











        $endgroup$



        Formulate the question this way: there are 20 people in a room, 17M and 3W. there's a table with 5 seats, what is the probability that only men sit in this table?



        Part 1: the universe:



        How many combination of 5 can you form from 20? Answer $dbinom{20}{5}$



        Part 2: your request:



        You request to know how many M only can be seated in this table? Answer $dbinom{17}{5}$




        Part 3: Result

        so the answer is:

        $frac{17choose 5}{20choose 5}$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 16 '18 at 2:59

























        answered Dec 16 '18 at 2:53









        adhgadhg

        340413




        340413























            1












            $begingroup$

            I think that the someone is correct. His or her reasoning goes along the lines: let's define only men sitting at table A a success. The probability that there will be only be men at table A is then the number of ways we can choose $5$ men from $20$ as a proportion of the number of ways we can choose $5$ people from $20$:



            $$frac{17choose 5}{20choose 5}$$



            From this point, we don't care who sits at the other tables, so the other tables cannot influence whether the experiment is a success.



            Your previous reasoning can be applied correctly, but with unnecessary detail. The reasoning would be along the lines: what is the probability that only men sit at table A, any combination of men and women sit at table B, any combination at C and any combination at D? The solution to this would be the above result times $1^3$, which is, of course, equivalent.



            Edit



            Just to address the iterative solution described in the comments below:



            I see a couple of issues with the iterative solution as you've expressed it.




            1. The probability that the second person is male, for example, depends on whether the first person selected was male. It seems that you are assuming that the previous guest was male in each iteration, which is not necessarily the case.


            2. If we are randomly assigning tables as the guests arrive, then the probability that someone is assigned to table A is a fixed $frac 1 4$ until one of the tables is filled. (It doesn't follow the sequence $frac{5}{20}$, $frac{4}{19}$, $dots$)



            In coming up with your iterative solution it seems as though you are (correctly) interpreting the given information as follows:




            20 (17 male + 3 female) guests arrive in a random order (such that all permutations are equally likely) and are randomly assigned an available table from A, B, C and D (such that each available table is equally likely to be assigned) as they arrive.




            To see how your iterative solution for this problem fails, think about how we would solve the following (different from the original) problem:




            What is the probability that the first five guests are male and are assigned to table A?




            Then the solution would be:



            $$left(frac 1 4right)^5left(frac{17}{20}right)left(frac{16}{19}right)left(frac{15}{18}right)left(frac{14}{17}right)left(frac{13}{16}right)$$



            which seems to be where you were heading with your calculations.



            Note that this is very different from the problem:




            What is the probability that the five guests assigned to table A once all the guests have arrived are all male?




            The answer to the first problem is the probability of only one of the (many) potential outcomes that leads to table A being filled with men only. (Another outcome, for example, is that the three women arrive first but are assigned to different tables.)



            Note that the iterative solution, while it would work (for example in computer simulations), would require us to keep track of a huge tree in order for us to identify all the successes.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I see thanks for the clarification. But if I think about this iteratively, If I were someone observing or"randomly" picking, the first person to sit at a table has a (17/20) chance of being a male. What is the probability that person sits at table A? (5/20). The second person has a (16/19) chance of being male. The probability that person also sits at table A is now (4/19), since one seat is taken. So this becomes a productive multiplication of (17/20) * (5/20) * (16/19) * (4/19).. so on so forth which is much lower than the result. Where am I going wrong here ?
              $endgroup$
              – bkrish
              Dec 16 '18 at 5:53












            • $begingroup$
              Are you not assuming that table A always gets filled first in your solution ?
              $endgroup$
              – bkrish
              Dec 16 '18 at 5:55










            • $begingroup$
              Answer edited to address this.
              $endgroup$
              – Richard Ambler
              Dec 16 '18 at 6:53










            • $begingroup$
              Oh wow ok so I see where my reasoning went off. So you are essentially saying, that the probability the first 5 people are male and assigned to any random seat is equivalent to the first answer to the problem (i.e) (17C5)/(20C5) which is what the iterative solution converges to without the (1/4) factor. Is that correct ?
              $endgroup$
              – bkrish
              Dec 16 '18 at 7:40










            • $begingroup$
              Yep. Just to clarify: the iterative interpretation is also valid (and would give the same answer to the original problem if we were to track all the "success" branches). The only thing you did wrong before (besides the table assignment probabilities) was only track one of the many "success" branches.
              $endgroup$
              – Richard Ambler
              Dec 16 '18 at 8:04
















            1












            $begingroup$

            I think that the someone is correct. His or her reasoning goes along the lines: let's define only men sitting at table A a success. The probability that there will be only be men at table A is then the number of ways we can choose $5$ men from $20$ as a proportion of the number of ways we can choose $5$ people from $20$:



            $$frac{17choose 5}{20choose 5}$$



            From this point, we don't care who sits at the other tables, so the other tables cannot influence whether the experiment is a success.



            Your previous reasoning can be applied correctly, but with unnecessary detail. The reasoning would be along the lines: what is the probability that only men sit at table A, any combination of men and women sit at table B, any combination at C and any combination at D? The solution to this would be the above result times $1^3$, which is, of course, equivalent.



            Edit



            Just to address the iterative solution described in the comments below:



            I see a couple of issues with the iterative solution as you've expressed it.




            1. The probability that the second person is male, for example, depends on whether the first person selected was male. It seems that you are assuming that the previous guest was male in each iteration, which is not necessarily the case.


            2. If we are randomly assigning tables as the guests arrive, then the probability that someone is assigned to table A is a fixed $frac 1 4$ until one of the tables is filled. (It doesn't follow the sequence $frac{5}{20}$, $frac{4}{19}$, $dots$)



            In coming up with your iterative solution it seems as though you are (correctly) interpreting the given information as follows:




            20 (17 male + 3 female) guests arrive in a random order (such that all permutations are equally likely) and are randomly assigned an available table from A, B, C and D (such that each available table is equally likely to be assigned) as they arrive.




            To see how your iterative solution for this problem fails, think about how we would solve the following (different from the original) problem:




            What is the probability that the first five guests are male and are assigned to table A?




            Then the solution would be:



            $$left(frac 1 4right)^5left(frac{17}{20}right)left(frac{16}{19}right)left(frac{15}{18}right)left(frac{14}{17}right)left(frac{13}{16}right)$$



            which seems to be where you were heading with your calculations.



            Note that this is very different from the problem:




            What is the probability that the five guests assigned to table A once all the guests have arrived are all male?




            The answer to the first problem is the probability of only one of the (many) potential outcomes that leads to table A being filled with men only. (Another outcome, for example, is that the three women arrive first but are assigned to different tables.)



            Note that the iterative solution, while it would work (for example in computer simulations), would require us to keep track of a huge tree in order for us to identify all the successes.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I see thanks for the clarification. But if I think about this iteratively, If I were someone observing or"randomly" picking, the first person to sit at a table has a (17/20) chance of being a male. What is the probability that person sits at table A? (5/20). The second person has a (16/19) chance of being male. The probability that person also sits at table A is now (4/19), since one seat is taken. So this becomes a productive multiplication of (17/20) * (5/20) * (16/19) * (4/19).. so on so forth which is much lower than the result. Where am I going wrong here ?
              $endgroup$
              – bkrish
              Dec 16 '18 at 5:53












            • $begingroup$
              Are you not assuming that table A always gets filled first in your solution ?
              $endgroup$
              – bkrish
              Dec 16 '18 at 5:55










            • $begingroup$
              Answer edited to address this.
              $endgroup$
              – Richard Ambler
              Dec 16 '18 at 6:53










            • $begingroup$
              Oh wow ok so I see where my reasoning went off. So you are essentially saying, that the probability the first 5 people are male and assigned to any random seat is equivalent to the first answer to the problem (i.e) (17C5)/(20C5) which is what the iterative solution converges to without the (1/4) factor. Is that correct ?
              $endgroup$
              – bkrish
              Dec 16 '18 at 7:40










            • $begingroup$
              Yep. Just to clarify: the iterative interpretation is also valid (and would give the same answer to the original problem if we were to track all the "success" branches). The only thing you did wrong before (besides the table assignment probabilities) was only track one of the many "success" branches.
              $endgroup$
              – Richard Ambler
              Dec 16 '18 at 8:04














            1












            1








            1





            $begingroup$

            I think that the someone is correct. His or her reasoning goes along the lines: let's define only men sitting at table A a success. The probability that there will be only be men at table A is then the number of ways we can choose $5$ men from $20$ as a proportion of the number of ways we can choose $5$ people from $20$:



            $$frac{17choose 5}{20choose 5}$$



            From this point, we don't care who sits at the other tables, so the other tables cannot influence whether the experiment is a success.



            Your previous reasoning can be applied correctly, but with unnecessary detail. The reasoning would be along the lines: what is the probability that only men sit at table A, any combination of men and women sit at table B, any combination at C and any combination at D? The solution to this would be the above result times $1^3$, which is, of course, equivalent.



            Edit



            Just to address the iterative solution described in the comments below:



            I see a couple of issues with the iterative solution as you've expressed it.




            1. The probability that the second person is male, for example, depends on whether the first person selected was male. It seems that you are assuming that the previous guest was male in each iteration, which is not necessarily the case.


            2. If we are randomly assigning tables as the guests arrive, then the probability that someone is assigned to table A is a fixed $frac 1 4$ until one of the tables is filled. (It doesn't follow the sequence $frac{5}{20}$, $frac{4}{19}$, $dots$)



            In coming up with your iterative solution it seems as though you are (correctly) interpreting the given information as follows:




            20 (17 male + 3 female) guests arrive in a random order (such that all permutations are equally likely) and are randomly assigned an available table from A, B, C and D (such that each available table is equally likely to be assigned) as they arrive.




            To see how your iterative solution for this problem fails, think about how we would solve the following (different from the original) problem:




            What is the probability that the first five guests are male and are assigned to table A?




            Then the solution would be:



            $$left(frac 1 4right)^5left(frac{17}{20}right)left(frac{16}{19}right)left(frac{15}{18}right)left(frac{14}{17}right)left(frac{13}{16}right)$$



            which seems to be where you were heading with your calculations.



            Note that this is very different from the problem:




            What is the probability that the five guests assigned to table A once all the guests have arrived are all male?




            The answer to the first problem is the probability of only one of the (many) potential outcomes that leads to table A being filled with men only. (Another outcome, for example, is that the three women arrive first but are assigned to different tables.)



            Note that the iterative solution, while it would work (for example in computer simulations), would require us to keep track of a huge tree in order for us to identify all the successes.






            share|cite|improve this answer











            $endgroup$



            I think that the someone is correct. His or her reasoning goes along the lines: let's define only men sitting at table A a success. The probability that there will be only be men at table A is then the number of ways we can choose $5$ men from $20$ as a proportion of the number of ways we can choose $5$ people from $20$:



            $$frac{17choose 5}{20choose 5}$$



            From this point, we don't care who sits at the other tables, so the other tables cannot influence whether the experiment is a success.



            Your previous reasoning can be applied correctly, but with unnecessary detail. The reasoning would be along the lines: what is the probability that only men sit at table A, any combination of men and women sit at table B, any combination at C and any combination at D? The solution to this would be the above result times $1^3$, which is, of course, equivalent.



            Edit



            Just to address the iterative solution described in the comments below:



            I see a couple of issues with the iterative solution as you've expressed it.




            1. The probability that the second person is male, for example, depends on whether the first person selected was male. It seems that you are assuming that the previous guest was male in each iteration, which is not necessarily the case.


            2. If we are randomly assigning tables as the guests arrive, then the probability that someone is assigned to table A is a fixed $frac 1 4$ until one of the tables is filled. (It doesn't follow the sequence $frac{5}{20}$, $frac{4}{19}$, $dots$)



            In coming up with your iterative solution it seems as though you are (correctly) interpreting the given information as follows:




            20 (17 male + 3 female) guests arrive in a random order (such that all permutations are equally likely) and are randomly assigned an available table from A, B, C and D (such that each available table is equally likely to be assigned) as they arrive.




            To see how your iterative solution for this problem fails, think about how we would solve the following (different from the original) problem:




            What is the probability that the first five guests are male and are assigned to table A?




            Then the solution would be:



            $$left(frac 1 4right)^5left(frac{17}{20}right)left(frac{16}{19}right)left(frac{15}{18}right)left(frac{14}{17}right)left(frac{13}{16}right)$$



            which seems to be where you were heading with your calculations.



            Note that this is very different from the problem:




            What is the probability that the five guests assigned to table A once all the guests have arrived are all male?




            The answer to the first problem is the probability of only one of the (many) potential outcomes that leads to table A being filled with men only. (Another outcome, for example, is that the three women arrive first but are assigned to different tables.)



            Note that the iterative solution, while it would work (for example in computer simulations), would require us to keep track of a huge tree in order for us to identify all the successes.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 16 '18 at 6:58

























            answered Dec 16 '18 at 2:58









            Richard AmblerRichard Ambler

            1,298515




            1,298515












            • $begingroup$
              I see thanks for the clarification. But if I think about this iteratively, If I were someone observing or"randomly" picking, the first person to sit at a table has a (17/20) chance of being a male. What is the probability that person sits at table A? (5/20). The second person has a (16/19) chance of being male. The probability that person also sits at table A is now (4/19), since one seat is taken. So this becomes a productive multiplication of (17/20) * (5/20) * (16/19) * (4/19).. so on so forth which is much lower than the result. Where am I going wrong here ?
              $endgroup$
              – bkrish
              Dec 16 '18 at 5:53












            • $begingroup$
              Are you not assuming that table A always gets filled first in your solution ?
              $endgroup$
              – bkrish
              Dec 16 '18 at 5:55










            • $begingroup$
              Answer edited to address this.
              $endgroup$
              – Richard Ambler
              Dec 16 '18 at 6:53










            • $begingroup$
              Oh wow ok so I see where my reasoning went off. So you are essentially saying, that the probability the first 5 people are male and assigned to any random seat is equivalent to the first answer to the problem (i.e) (17C5)/(20C5) which is what the iterative solution converges to without the (1/4) factor. Is that correct ?
              $endgroup$
              – bkrish
              Dec 16 '18 at 7:40










            • $begingroup$
              Yep. Just to clarify: the iterative interpretation is also valid (and would give the same answer to the original problem if we were to track all the "success" branches). The only thing you did wrong before (besides the table assignment probabilities) was only track one of the many "success" branches.
              $endgroup$
              – Richard Ambler
              Dec 16 '18 at 8:04


















            • $begingroup$
              I see thanks for the clarification. But if I think about this iteratively, If I were someone observing or"randomly" picking, the first person to sit at a table has a (17/20) chance of being a male. What is the probability that person sits at table A? (5/20). The second person has a (16/19) chance of being male. The probability that person also sits at table A is now (4/19), since one seat is taken. So this becomes a productive multiplication of (17/20) * (5/20) * (16/19) * (4/19).. so on so forth which is much lower than the result. Where am I going wrong here ?
              $endgroup$
              – bkrish
              Dec 16 '18 at 5:53












            • $begingroup$
              Are you not assuming that table A always gets filled first in your solution ?
              $endgroup$
              – bkrish
              Dec 16 '18 at 5:55










            • $begingroup$
              Answer edited to address this.
              $endgroup$
              – Richard Ambler
              Dec 16 '18 at 6:53










            • $begingroup$
              Oh wow ok so I see where my reasoning went off. So you are essentially saying, that the probability the first 5 people are male and assigned to any random seat is equivalent to the first answer to the problem (i.e) (17C5)/(20C5) which is what the iterative solution converges to without the (1/4) factor. Is that correct ?
              $endgroup$
              – bkrish
              Dec 16 '18 at 7:40










            • $begingroup$
              Yep. Just to clarify: the iterative interpretation is also valid (and would give the same answer to the original problem if we were to track all the "success" branches). The only thing you did wrong before (besides the table assignment probabilities) was only track one of the many "success" branches.
              $endgroup$
              – Richard Ambler
              Dec 16 '18 at 8:04
















            $begingroup$
            I see thanks for the clarification. But if I think about this iteratively, If I were someone observing or"randomly" picking, the first person to sit at a table has a (17/20) chance of being a male. What is the probability that person sits at table A? (5/20). The second person has a (16/19) chance of being male. The probability that person also sits at table A is now (4/19), since one seat is taken. So this becomes a productive multiplication of (17/20) * (5/20) * (16/19) * (4/19).. so on so forth which is much lower than the result. Where am I going wrong here ?
            $endgroup$
            – bkrish
            Dec 16 '18 at 5:53






            $begingroup$
            I see thanks for the clarification. But if I think about this iteratively, If I were someone observing or"randomly" picking, the first person to sit at a table has a (17/20) chance of being a male. What is the probability that person sits at table A? (5/20). The second person has a (16/19) chance of being male. The probability that person also sits at table A is now (4/19), since one seat is taken. So this becomes a productive multiplication of (17/20) * (5/20) * (16/19) * (4/19).. so on so forth which is much lower than the result. Where am I going wrong here ?
            $endgroup$
            – bkrish
            Dec 16 '18 at 5:53














            $begingroup$
            Are you not assuming that table A always gets filled first in your solution ?
            $endgroup$
            – bkrish
            Dec 16 '18 at 5:55




            $begingroup$
            Are you not assuming that table A always gets filled first in your solution ?
            $endgroup$
            – bkrish
            Dec 16 '18 at 5:55












            $begingroup$
            Answer edited to address this.
            $endgroup$
            – Richard Ambler
            Dec 16 '18 at 6:53




            $begingroup$
            Answer edited to address this.
            $endgroup$
            – Richard Ambler
            Dec 16 '18 at 6:53












            $begingroup$
            Oh wow ok so I see where my reasoning went off. So you are essentially saying, that the probability the first 5 people are male and assigned to any random seat is equivalent to the first answer to the problem (i.e) (17C5)/(20C5) which is what the iterative solution converges to without the (1/4) factor. Is that correct ?
            $endgroup$
            – bkrish
            Dec 16 '18 at 7:40




            $begingroup$
            Oh wow ok so I see where my reasoning went off. So you are essentially saying, that the probability the first 5 people are male and assigned to any random seat is equivalent to the first answer to the problem (i.e) (17C5)/(20C5) which is what the iterative solution converges to without the (1/4) factor. Is that correct ?
            $endgroup$
            – bkrish
            Dec 16 '18 at 7:40












            $begingroup$
            Yep. Just to clarify: the iterative interpretation is also valid (and would give the same answer to the original problem if we were to track all the "success" branches). The only thing you did wrong before (besides the table assignment probabilities) was only track one of the many "success" branches.
            $endgroup$
            – Richard Ambler
            Dec 16 '18 at 8:04




            $begingroup$
            Yep. Just to clarify: the iterative interpretation is also valid (and would give the same answer to the original problem if we were to track all the "success" branches). The only thing you did wrong before (besides the table assignment probabilities) was only track one of the many "success" branches.
            $endgroup$
            – Richard Ambler
            Dec 16 '18 at 8:04


















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