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How do you feel about guys wearing girly socks?
Posted on Tuesday, April 2, 2013 by Abi Samsuri
Q. Thigh high socks, striped socks, fuzzy socks, toe socks, thigh high toe socks hehe what do you think??
A. I think it's adorable, and sexy if you do it with a confident and carefree attitude.
My S.O. borrows my socks all the time. lol.
My S.O. borrows my socks all the time. lol.
What is the probability that all pairs taken will consist of one red and one blue sock?
Q. A drawer contains n red and n blue socks. Socks are taken at random by pairs without replacement. What is the probability that all pairs taken will consist of one red and one blue sock?
A. Here is a concise answer, followed by my original working out of the problem:
Will draw sock #1, sock #2, sock #3, sock #4, ..., sock #(2n1), sock #(2n)
Total number of orders is (2n)!
For each pair (12), (34), ... (2n1), (2n),
one of them has to be red.
So for each pair, there are two choices for the red sock.
Then the others are blue.
Now we have assigned every sock position in the order a color in one of 2^n ways.
Then we can assign the individual socks, first the reds in n! ways, then the blues in n! ways.
So total number of mismatchedpairarrangements is
2^n n! n! and probability that it happens at random is
2^ n! n! / (2n)!

And for the benefit of those who would like to know that concise solutions don't just pop out of the air, here is how I first approached the problem.

First sock is some color.
Second sock must be other color: (n)/(2n1)
Third sock is some color:
Fourth sock must be other color: (n1)/(2n3)
and so on
Last but one sock is some color.
Last sock must be other color: 1/1
multiplying all that together: (n) (n1) ... (1) / (2n1) (2n3) ... (3) (1)
Numerator is n!
Denominator is (2n1) ! / (2n 2) (2n 4) (2n6) ... 2
Subdenominator is 2(n1) 2 (n2) 2(n3) ... 2
= 2^(n1) (n1)!
 Final formulas 
So we have
2^(n1) n! (n1)! / (2n 1)!
Multiplying by 2n/2n gives
2^(n) (n!)^2 / (2n)!

The 2^n comes from the fact that you can pull either the red or the blue first, so each mismatched pair can come in 2 orders. ivan t overlooked that. His answer has the number of arrangments where the first sock is always the same color.
Example:
3 pairs
Formula given:
2^2 3! 2! / 5! = 4 * 6 * 2 / 120 = 48/120 = 2/5
or
2^3 3! 3! / 6! = 288 / 720 = 2/5
Pick a sock, let's say it's red.
Next must be blue: 3/5
Pick a sock, let's say it's blue.
Next must be red: 2/3
Now just 1 red, 1 blue left.
3/5 * 2/3 = 2/5
Total arrangments: 6! = 720
Alternating arrangements:
6 * 3 * 4 * 2 * 2 * 1 = 288
288 / 720 = 2/5
Will draw sock #1, sock #2, sock #3, sock #4, ..., sock #(2n1), sock #(2n)
Total number of orders is (2n)!
For each pair (12), (34), ... (2n1), (2n),
one of them has to be red.
So for each pair, there are two choices for the red sock.
Then the others are blue.
Now we have assigned every sock position in the order a color in one of 2^n ways.
Then we can assign the individual socks, first the reds in n! ways, then the blues in n! ways.
So total number of mismatchedpairarrangements is
2^n n! n! and probability that it happens at random is
2^ n! n! / (2n)!

And for the benefit of those who would like to know that concise solutions don't just pop out of the air, here is how I first approached the problem.

First sock is some color.
Second sock must be other color: (n)/(2n1)
Third sock is some color:
Fourth sock must be other color: (n1)/(2n3)
and so on
Last but one sock is some color.
Last sock must be other color: 1/1
multiplying all that together: (n) (n1) ... (1) / (2n1) (2n3) ... (3) (1)
Numerator is n!
Denominator is (2n1) ! / (2n 2) (2n 4) (2n6) ... 2
Subdenominator is 2(n1) 2 (n2) 2(n3) ... 2
= 2^(n1) (n1)!
 Final formulas 
So we have
2^(n1) n! (n1)! / (2n 1)!
Multiplying by 2n/2n gives
2^(n) (n!)^2 / (2n)!

The 2^n comes from the fact that you can pull either the red or the blue first, so each mismatched pair can come in 2 orders. ivan t overlooked that. His answer has the number of arrangments where the first sock is always the same color.
Example:
3 pairs
Formula given:
2^2 3! 2! / 5! = 4 * 6 * 2 / 120 = 48/120 = 2/5
or
2^3 3! 3! / 6! = 288 / 720 = 2/5
Pick a sock, let's say it's red.
Next must be blue: 3/5
Pick a sock, let's say it's blue.
Next must be red: 2/3
Now just 1 red, 1 blue left.
3/5 * 2/3 = 2/5
Total arrangments: 6! = 720
Alternating arrangements:
6 * 3 * 4 * 2 * 2 * 1 = 288
288 / 720 = 2/5
What kind of socks are best to wear with skate shoes?
Q. High socks, ankle socks, or noshow socks. I am a male.
A. High socks or no show socks :p
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