Mixing
A probability question for matching socks in N bins
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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
asked Jan 4 at 21:06
user45264user45264
62
$endgroup$
add a comment |
$begingroup$
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
asked Jan 4 at 21:06
user45264user45264
62
$endgroup$
1
$begingroup$
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."
$endgroup$
– angryavian
Jan 4 at 21:09
$begingroup$
Hi, that would count as k pairs. Thanks.
$endgroup$
– user45264
Jan 4 at 21:12
add a comment |
$begingroup$
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
asked Jan 4 at 21:06
user45264user45264
62
$endgroup$
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
asked Jan 4 at 21:06
user45264user45264
62
asked Jan 4 at 21:06
user45264user45264
62
edited Jan 4 at 21:13
user45264
asked Jan 4 at 21:06
user45264user45264
62
asked Jan 4 at 21:06
user45264user45264
62
asked Jan 4 at 21:06
user45264user45264
62
62
1
$begingroup$
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."
$endgroup$
– angryavian
Jan 4 at 21:09
$begingroup$
Hi, that would count as k pairs. Thanks.
$endgroup$
– user45264
Jan 4 at 21:12
add a comment |
1
$begingroup$
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."
$endgroup$
– angryavian
Jan 4 at 21:09
$begingroup$
Hi, that would count as k pairs. Thanks.
$endgroup$
– user45264
Jan 4 at 21:12
1
1
$begingroup$
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."
$endgroup$
– angryavian
Jan 4 at 21:09
$begingroup$
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."
$endgroup$
– angryavian
Jan 4 at 21:09
$begingroup$
Hi, that would count as k pairs. Thanks.
$endgroup$
– user45264
Jan 4 at 21:12
$begingroup$
Hi, that would count as k pairs. Thanks.
$endgroup$
– user45264
Jan 4 at 21:12
add a comment |
1 Answer
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oldest
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$begingroup$
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.
answered Jan 4 at 21:58
angryavianangryavian
40.3k23280
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1 Answer
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$begingroup$
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.
answered Jan 4 at 21:58
angryavianangryavian
40.3k23280
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 4 at 21:58
angryavianangryavian
40.3k23280
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 4 at 21:58
angryavianangryavian
40.3k23280
$endgroup$
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.
answered Jan 4 at 21:58
angryavianangryavian
40.3k23280
edited Jan 4 at 22:04
answered Jan 4 at 21:58
angryavianangryavian
40.3k23280
answered Jan 4 at 21:58
angryavianangryavian
40.3k23280
answered Jan 4 at 21:58
angryavianangryavian
40.3k23280
40.3k23280
add a comment |
add a comment |
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$begingroup$
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."
$endgroup$
– angryavian
Jan 4 at 21:09
$begingroup$
Hi, that would count as k pairs. Thanks.
$endgroup$
– user45264
Jan 4 at 21:12