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I am trying to solve the following counting problem in Feller's probability book. I checked my answer against that given in the book. My answer to part (a) of the problem is correct (b) but I am difficulty deriving the expression for part (b). Here is my attempt.

Part(a) of the problem was also answered here - Probability/Combinatorics Problem. A closet containing n pairs of shoes.

  1. A closet has $n$ pairs of shoes. If $2r$ shoes are chosen at random (with $2r

Solution.

The sample space S has ${2n \choose 2r}$ sample points.

(a) No complete pair:

Select one of $2n$ shoes. The second shoe can be one of $2n-2$ shoes. The third shoe can be one of $2n-4$ shoes. The $k$'th shoe can be one of $2n-(2k-2)$ shoes.

The $2r$th shoe can be one of $2n-(4r-2)$ shoes. These $2r$ shoes can be permuted in (2r)! ways. As the order does not matter,

$$n(\text{No complete pair})=2^{2r}\frac{n!}{(n-2r)!(2r)!}=2^{2r}{n \choose 2r}$$

$$P(\text{No complete pair})=\frac{2^{2r}{n \choose 2r}}{{2n \choose 2r}}$$

(b) Exactly one complete pair.

Select one of $2n$ shoes. This shoe must have a pair, so that could be selected in only $1$ way. The third shoe can be one of $(2n-2)$ shoes, $k$th shoe is one of 2n-(2k-4). The $2r$th shoe is one of $2n-(4r-4)$. Also, since order is not important, we are overcounting by a factor of $(2r)!$.

$$\begin{alignedat}{1}n(\text{Exactly one pair}) & =(2n)\cdot(2n-2)\cdot(2n-4)\cdot(2n-(4r-4))/(2r)!\\ & =2^{2r-1}\frac{n!}{(n-2r+1)!(2r)!}\\ & =2^{2r-2}\frac{(n-1)!}{(n-2r+1)!(2r-1)!}\cdot\frac{n}{r} \end{alignedat}$$

The correct solution to part (b) of this problem in the book is:

$$\frac{n{{n-1}\choose{2r-2}}2^{2r-2}}{{2n}\choose{2r}}$$

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(a) $\ddot\smile\checkmark$, you are correct, because you are selecting $2r$ distinct pairs and taking one shoe from of each of them.

$$P(\text{no completes}) = \dfrac{~2^{2r}\dbinom n {2r}~}{\dbinom{2n}{2r}}$$

(b) $\ddot{\frown}\times$ The book is correct, because you are selecting one complete pair from $n$, and from the $n-1$ pairs selecting $2r-2$ distinct pairs and taking just one shoe from each of those.   There is no order to these selections.

$$P(\text{exactly one complete}) = \dfrac{2^{2r-2}\dbinom{n}{1}\dbinom{n-1}{2r-2}}{\dbinom{2n}{2r}}$$

[PS: Your logic of selecting one of 2n shoes, then its pair, over counts cases, so is wrong right from the start.]

(c) $\ddot\sim?$ Now can you apply the same principle to find the probability of selecting exactly two complete pairs?

  • 1
    now why didn't that occur to me! First, I would select 2 complete pairs in ${n\choose2}$ ways. Then, I select $2r-4$ distinct pairs from $n-2$ pairs and take one shoe from each of these, ${{2}\choose{1}}^{2r-4}$. The probability that two complete pairs are selected is: $$P(\text{exactly two complete})=\frac{2^{2r-4}{n\choose2}{{n-2}\choose{2r-4}}}{{2n}\choose{2r}}$$ – 2017-01-19