Matching Socks

Great solution!

Btw, anyone who isn’t so good at math, the exclamation points show factorials, meaning x+(x-1) (x-2) ...*1

And, the * is a multiplication sign.

Nice approach! @Zee Ell

Yep... Damn... i did nt multiply the 5 in the corner... and i did try out of whimse.. at the last trial.

This seems wrong to me... in the denominator, you’re assuming identical socks are interchangeable. In the numerator, you’re not. Can you find my error?

I’m getting: Six pairs in any order = 6! Four of each color can be in any order = 4!4!4! Total orders = 12!

So it should be 6!4!4!4!/12!

Why do we need the 2!2!2! in your formula? The order of identical socks is already accounted for in the 4! terms

For anyone not satisfied with this formula, there's another way to get it: by induction. Let there be $a$ black pairs, $b$ blue pairs and $c$ white pairs and let's show that the probability is $\boxed{P_{a,b,c}=\dfrac{(a+b+c)!·(2a)!·(2b)!·(2c)!}{(2(a+b+c))!·a!·b!·c!}}$ .

Let's call $f(a, b, c):=\dfrac{(a+b+c)!·(2a)!·(2b)!·(2c)!}{(2(a+b+c))!·a!·b!·c!}$ .

• The formula is correct with only one color: $P_{a,0,0}=1=f(a, 0, 0)$ .

Using $x!!=x·(x-2)·(x-4)·\cdots·1$ for $x$ odd, so $7!!=7·5·3·1=105$ , it is left to the reader to show that $(2k-1)!!=\dfrac{(2k)!}{2^k·k!}$ and therefore that:

$f(a, b, c)=\boxed{\ \dfrac{(2a-1)!!·(2b-1)!!·(2c-1)!!}{(2(a+b+c)-1)!!}\ }$

Note: This formula is indirectly used in other answers: $f(2,2,2)=\dfrac{(3·1)·(3·1)·(3·1)}{11·9·7·5·3·1}=\dfrac{3}{11}·\dfrac{1}{9}·\dfrac{3}{7}·\dfrac{1}{5}$

• Now, suppose $P_{a,b,c}=f(a, b, c)$ , let's show that $P_{a+1,b,c}=f(a+1,b,c)$ :

As $a+1>0$ , there's a place with a black sock, next to a pairing sock which has probability $\dfrac{2(a+1)-1}{2((a+1)+b+c)-1}$ to be also black. So we have $P_{a+1,b,c}=\dfrac{2(a+1)-1}{2((a+1)+b+c)-1}·P_{a,b,c}$ $\overset{\text{by ind.}}{=}\dfrac{2(a+1)-1}{2((a+1)+b+c)-1}·\dfrac{(2a-1)!!·(2b-1)!!·(2c-1)!!}{(2(a+b+c)-1)!!}$

$=\dfrac{2a+1}{2(a+b+c)+1}·\dfrac{(2a-1)!!·(2b-1)!!·(2c-1)!!}{(2(a+b+c)-1)!!}\ =\ \dfrac{(2a+1)!!·(2b-1)!!·(2c-1)!!}{(2(a+b+c)+1)!!}$

$=\ \dfrac{(2(a+1)-1)!!·(2b-1)!!·(2c-1)!!}{(2((a+1)+b+c)-1)!!}\ =\ f(a+1, b, c)$ .

Note: $P_{a,b+1,c}=f(a, b+1, c)$ and $P_{a,b,c+1}=f(a, b, c+1)$ can be done the same way, but it's obvious because $P_{a,b,c}=P_{b,a,c}=P_{c,b,a}$ .

• That shows the formula holds for any $a$ , $b$ and $c$ .

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If you're not satisfied with picking a place where there's a black sock, there are two other ways:

1) Have to cases, one where the first sock is black, and one where the first is, WLOG, blue. Show that $P_{a+1,b,c}=f(a+1, b, c)$ in both cases. Only difference is that induction must be made on $n=a+b+c$ and beginning with showing that $P_{a,b,c}=f(a, b, c)$ for $a+b+c\leq1$ .

2) Show that the probability doesn't change with the color of the first sock (i.e. knowing which color is the first sock doesn't increase or decrease your chances of success) and then just assume the first one is black to do the maths. To show that, see that you can alter the row of socks without changing the probability that socks are paired. If some black sock is on an even place, exchange it with its matching sock -- the one just before. The pair remains two blacks, or of different colors. Now, you have at least one black sock on an odd place. Exchange it and its following neighbor with the first and second one. This doesn't change the probability either.