Probability a woman is paired with her husband

Question

$2n$ people consisting of $n$ couples are randomly paired together. What is the probability that a particular woman gets paired with her husband?

I know that the answer is $\frac{1}{2n-1}$ because a woman is equally likely to be paired with any of the other $2n-1$ people that are not her, and only one of them is her husband. However, when I first tried to solve this problem, I tried the following:

There are $\frac{(2n)!}{2^n}$ possible ways of making the $n$ couples. Likewise, there should be $\frac{(2n-2)!}{2^{n-1}}$ ways of coupling all the $2n-2$ people that are not the particular woman and her husband. Therefore, the probability of the woman getting paired with her husband should be $\frac{1}{n(2n-1)}$.

My question is: why am I getting this extra $n$ term in the denominator?

score 4 · Accepted Answer · answered Nov 14 '16 at 01:50

4

Here is the problem: say for example $n=3$. Put the people in some predetermined order (say, alphabetical order), and label them with numbers for each couple. So for example $$123321$$ means the first person goes with the last, the second with the second last and the third with the third last. However, your $(2n)!/2^n$ ways of forming couples also include, for example, $$321123\ ,$$ which is actually the same arrangement of couples. The correct number of ways of forming couples is $$\frac{(2n)!}{2^n\,n!}$$ and then your probability calculation is $$\frac{(2n-2)!}{2^{n-1}\,(n-1)!}\bigg/\frac{(2n)!}{2^n\,n!} =\frac1{2n-1}\ .$$

answered Nov 14 '16 at 01:50

David

82,662

And as a further reference, the answers to this question give a couple of explanations of the expression for the number of ways to form the couples. – Brian M. Scott Nov 14 '16 at 01:59
So I am overcounting because I take into account the order of the pairs? – GuPe Nov 14 '16 at 02:43
That's correct. – David Nov 14 '16 at 03:59

Probability a woman is paired with her husband

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