If I have 10 different pairs of socks and have washed 10 socks, what are the chances that none will match?

Question

I have 10 pairs of different types of socks. I randomly (let's just assume it was true randomness) washed 10 individual socks. It turns out none of them match! What are the chances of this?

I've done 20 choose 10 and got 184,756. So are the chances 1/184,756 that 10 randomly selected individual socks don't match out of 10 pairs of socks? This doesn't seem to make sense to me since those are the odds of any 10 socks randomly being chosen.

Answer 1

We have $20$ socks in $10$ pairs as follows - call them $1a,1b,2a,2b,...,10a,10b$ (we assume each sock is distinct). There are $2^{10}$ ways of choosing $10$ non-matching socks (we have to choose exactly one sock from $\{1a,1b\}$, exactly one from $\{2a,2b\}$, etc). There are $(20\; choose \;10)$ ways of choosing any $10$ socks from these $20$.

So the chance is $2^{10}/(20\; choose \; 10) \approx 0.554\%$.

Answer 2

Let $n=10$ just to keep notations clean and easier to generalize.

$C^{n}_{2n}=\frac{(2n)!}{n!n!}$ is the total number of possible choices (unsorted). Now the question is, how many among them make none match? Since you have only 2 choices in each pair, so your total number of non-matching choices is $2^n$.

So the probability is $\frac{2^nn!^2}{(2n)!}=\frac{1024}{184756}\approx0.55\%$.

Answer 3

Chances of this won't be $\frac{1}{20 \choose 10}$ because when you choose one sock from all 10 pairs, there are still 2 socks in each pair to be chosen from.

The actual chances of this happening will be: $ \frac{2^{10}}{20 \choose 10} $ since from each pair you have 2 socks to choose. This value comes out to be: $\frac{1,024}{184,756} = \frac{256}{46,189}$

Answer 4

You need to choose one from each pair to wash and get rid of. That would be 2^10 ways to choose for each pair of socks which to get rid of, then just divide by what you did! You weren't that far off!

Answer 5

You have 20$\cdot$18$\cdot$16$\cdot$14 ..... ways of avoiding pairs,

against 20$\cdot$19$\cdot$18$\cdot$17 ...... total ways

Thus Pr = $\frac{20\cdot 18\cdot 16\cdot 14\cdot 12 \cdot 10\cdot 8\cdot 6 \cdot 4 \cdot 2}{20\cdot19\cdot18\cdot17\cdot16\cdot15\cdot14\cdot13\cdot12\cdot11}$

This can be simply written as $\frac{20!!}{^{20}P_{10}}$ = $\frac{256}{46189}$, where the numerator is called a double factorial