A probability question for matching socks in N bins












1














I thought of this problem while working on a separate problem trying to link records from two datasets. Suppose there are N bins, X red socks, and Y blue socks. Suppose we uniformly randomly throw the red socks and blue socks into the N bins. Two socks are considered a pair if they are in the same bin, one of them is red and one of them is blue. For example if there are 5 red socks and 8 blue socks in a bin, then that bin has 5 pairs of socks. What's the mean and variance of the number of pairs of socks we will find?



I ran a few simulations with N = 10 and 50, a fixed number of 100 blue socks, and varying the number of red socks from 10 to 200.



Graph with simulations



Thanks in advance for the help.






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    If a bin has $k$ red socks and $k$ blue socks, does that count as $k$ pairs or $k^2$? Your wording seems to allow a particular sock to be a member of more than one "pair."
    – angryavian
    Jan 4 at 21:09










  • Hi, that would count as k pairs. Thanks.
    – user45264
    Jan 4 at 21:12
















1














I thought of this problem while working on a separate problem trying to link records from two datasets. Suppose there are N bins, X red socks, and Y blue socks. Suppose we uniformly randomly throw the red socks and blue socks into the N bins. Two socks are considered a pair if they are in the same bin, one of them is red and one of them is blue. For example if there are 5 red socks and 8 blue socks in a bin, then that bin has 5 pairs of socks. What's the mean and variance of the number of pairs of socks we will find?



I ran a few simulations with N = 10 and 50, a fixed number of 100 blue socks, and varying the number of red socks from 10 to 200.



Graph with simulations



Thanks in advance for the help.






probability







share|cite|improve this question









New contributor




user45264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    If a bin has $k$ red socks and $k$ blue socks, does that count as $k$ pairs or $k^2$? Your wording seems to allow a particular sock to be a member of more than one "pair."
    – angryavian
    Jan 4 at 21:09










  • Hi, that would count as k pairs. Thanks.
    – user45264
    Jan 4 at 21:12














1












1








1







I thought of this problem while working on a separate problem trying to link records from two datasets. Suppose there are N bins, X red socks, and Y blue socks. Suppose we uniformly randomly throw the red socks and blue socks into the N bins. Two socks are considered a pair if they are in the same bin, one of them is red and one of them is blue. For example if there are 5 red socks and 8 blue socks in a bin, then that bin has 5 pairs of socks. What's the mean and variance of the number of pairs of socks we will find?



I ran a few simulations with N = 10 and 50, a fixed number of 100 blue socks, and varying the number of red socks from 10 to 200.



Graph with simulations



Thanks in advance for the help.






probability







share|cite|improve this question









New contributor




user45264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I thought of this problem while working on a separate problem trying to link records from two datasets. Suppose there are N bins, X red socks, and Y blue socks. Suppose we uniformly randomly throw the red socks and blue socks into the N bins. Two socks are considered a pair if they are in the same bin, one of them is red and one of them is blue. For example if there are 5 red socks and 8 blue socks in a bin, then that bin has 5 pairs of socks. What's the mean and variance of the number of pairs of socks we will find?



I ran a few simulations with N = 10 and 50, a fixed number of 100 blue socks, and varying the number of red socks from 10 to 200.



Graph with simulations



Thanks in advance for the help.






probability




probability






share|cite|improve this question









New contributor




user45264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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user45264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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edited Jan 4 at 21:13







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asked Jan 4 at 21:06









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user45264 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    If a bin has $k$ red socks and $k$ blue socks, does that count as $k$ pairs or $k^2$? Your wording seems to allow a particular sock to be a member of more than one "pair."
    – angryavian
    Jan 4 at 21:09










  • Hi, that would count as k pairs. Thanks.
    – user45264
    Jan 4 at 21:12














  • 1




    If a bin has $k$ red socks and $k$ blue socks, does that count as $k$ pairs or $k^2$? Your wording seems to allow a particular sock to be a member of more than one "pair."
    – angryavian
    Jan 4 at 21:09










  • Hi, that would count as k pairs. Thanks.
    – user45264
    Jan 4 at 21:12








1




1




If a bin has $k$ red socks and $k$ blue socks, does that count as $k$ pairs or $k^2$? Your wording seems to allow a particular sock to be a member of more than one "pair."
– angryavian
Jan 4 at 21:09




If a bin has $k$ red socks and $k$ blue socks, does that count as $k$ pairs or $k^2$? Your wording seems to allow a particular sock to be a member of more than one "pair."
– angryavian
Jan 4 at 21:09












Hi, that would count as k pairs. Thanks.
– user45264
Jan 4 at 21:12




Hi, that would count as k pairs. Thanks.
– user45264
Jan 4 at 21:12










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If $Z_i$ is the number of pairs in the $i$th bin then your random variable of interest is $sum_{i=1}^N Z_i$.



Then $$Eleft[sum_{i=1}^N Z_iright] = sum_{i=1}^N E [Z_i] = N E[Z_1]$$ since the $Z_i$ have the same distribution (one bin is not different from the other when throwing the socks).



The event ${Z_1 ge j}$ is equivalent to "there are at least $j$ red socks and at least $j$ blue socks in the first bin." Thus, by the tail sum formula for expectation,
$$E[Z_1] = sum_{j=1}^{min(X,Y)} P(Z_1 ge j) = sum_{j=1}^{min(X,Y)} sum_{m=j}^X binom{X}{m} frac{(N-1)^{X-m}}{N^X} sum_{l = j}^Y binom{Y}{m} frac{(N-1)^{Y-l}}{N^Y}.$$
I could not think of a way to simplify this, nor of a different route that would be simpler. Perhaps someone more savvy could help you further.






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    If $Z_i$ is the number of pairs in the $i$th bin then your random variable of interest is $sum_{i=1}^N Z_i$.



    Then $$Eleft[sum_{i=1}^N Z_iright] = sum_{i=1}^N E [Z_i] = N E[Z_1]$$ since the $Z_i$ have the same distribution (one bin is not different from the other when throwing the socks).



    The event ${Z_1 ge j}$ is equivalent to "there are at least $j$ red socks and at least $j$ blue socks in the first bin." Thus, by the tail sum formula for expectation,
    $$E[Z_1] = sum_{j=1}^{min(X,Y)} P(Z_1 ge j) = sum_{j=1}^{min(X,Y)} sum_{m=j}^X binom{X}{m} frac{(N-1)^{X-m}}{N^X} sum_{l = j}^Y binom{Y}{m} frac{(N-1)^{Y-l}}{N^Y}.$$
    I could not think of a way to simplify this, nor of a different route that would be simpler. Perhaps someone more savvy could help you further.






    share|cite|improve this answer




























      1














      If $Z_i$ is the number of pairs in the $i$th bin then your random variable of interest is $sum_{i=1}^N Z_i$.



      Then $$Eleft[sum_{i=1}^N Z_iright] = sum_{i=1}^N E [Z_i] = N E[Z_1]$$ since the $Z_i$ have the same distribution (one bin is not different from the other when throwing the socks).



      The event ${Z_1 ge j}$ is equivalent to "there are at least $j$ red socks and at least $j$ blue socks in the first bin." Thus, by the tail sum formula for expectation,
      $$E[Z_1] = sum_{j=1}^{min(X,Y)} P(Z_1 ge j) = sum_{j=1}^{min(X,Y)} sum_{m=j}^X binom{X}{m} frac{(N-1)^{X-m}}{N^X} sum_{l = j}^Y binom{Y}{m} frac{(N-1)^{Y-l}}{N^Y}.$$
      I could not think of a way to simplify this, nor of a different route that would be simpler. Perhaps someone more savvy could help you further.






      share|cite|improve this answer


























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        If $Z_i$ is the number of pairs in the $i$th bin then your random variable of interest is $sum_{i=1}^N Z_i$.



        Then $$Eleft[sum_{i=1}^N Z_iright] = sum_{i=1}^N E [Z_i] = N E[Z_1]$$ since the $Z_i$ have the same distribution (one bin is not different from the other when throwing the socks).



        The event ${Z_1 ge j}$ is equivalent to "there are at least $j$ red socks and at least $j$ blue socks in the first bin." Thus, by the tail sum formula for expectation,
        $$E[Z_1] = sum_{j=1}^{min(X,Y)} P(Z_1 ge j) = sum_{j=1}^{min(X,Y)} sum_{m=j}^X binom{X}{m} frac{(N-1)^{X-m}}{N^X} sum_{l = j}^Y binom{Y}{m} frac{(N-1)^{Y-l}}{N^Y}.$$
        I could not think of a way to simplify this, nor of a different route that would be simpler. Perhaps someone more savvy could help you further.






        share|cite|improve this answer














        If $Z_i$ is the number of pairs in the $i$th bin then your random variable of interest is $sum_{i=1}^N Z_i$.



        Then $$Eleft[sum_{i=1}^N Z_iright] = sum_{i=1}^N E [Z_i] = N E[Z_1]$$ since the $Z_i$ have the same distribution (one bin is not different from the other when throwing the socks).



        The event ${Z_1 ge j}$ is equivalent to "there are at least $j$ red socks and at least $j$ blue socks in the first bin." Thus, by the tail sum formula for expectation,
        $$E[Z_1] = sum_{j=1}^{min(X,Y)} P(Z_1 ge j) = sum_{j=1}^{min(X,Y)} sum_{m=j}^X binom{X}{m} frac{(N-1)^{X-m}}{N^X} sum_{l = j}^Y binom{Y}{m} frac{(N-1)^{Y-l}}{N^Y}.$$
        I could not think of a way to simplify this, nor of a different route that would be simpler. Perhaps someone more savvy could help you further.







        share|cite|improve this answer














        share|cite|improve this answer



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        edited Jan 4 at 22:04

























        answered Jan 4 at 21:58









        angryavianangryavian

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