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At a dance there are $n=3$ married couples: Ann and Andy, Betty and Boris, and Danielle and Dan. The wives select a husband at random with whom to share a dance. What is the probability that each of the three men dances with a woman other than his spouse.

In each pair, I denoted wives with a lowercase letter and husbands with an uppercase. The pairs are aA, bB, and dD. I counted $9$ ways in which women could select men. aA, aB, aD bB, bA, bD dD, dA, dB

So there are $6$ ways to select other spouses. The probability is $\frac{6}{9}$, which is $0.67$, but the answer is $0.33$. What's wrong? Maybe I misunderstood the question, because English is my second language. Please help.

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  • $\begingroup$ Which triples of assignments are simultaneously compatible? For instance $\{aA, aB, \langle\text{anything}\rangle\}$ is not a possibility. The question is not about choices taken independently, it's about simultaneous triples of choices that are compatible. (An example of a compatible triple is $\{aB, bD, dA\}$. What fraction of compatible triples have no wife-husband pairing at all?) $\endgroup$ Mar 9, 2014 at 2:30

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Answer: The total number of ways you can pair them is ofcourse 9. A can choose to dance with b or c. Suppose he chooses to dance with b. B will have a choice between a and c. She has to select a because C cannot dance with c. Similarly, If A chooses to dance with c, C will have a choice to dance with b, because B cannot dance with b. Thus either way, there are only three possible choices and thus 3/9 = 0.333

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  • $\begingroup$ You yourself just counted 6 ways they can divide into pairs. Didn't you? $\endgroup$ Mar 10, 2014 at 20:50
  • $\begingroup$ Oh I got you so you mean that the probability that the first man will dance with the woman other than his wife is 2/3, for the second one it is 1/2 and for the third it is 1. Multiplying them we get 1/3. Ok. thanks! $\endgroup$ Mar 10, 2014 at 20:59

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