Probability of two people holding the same rope The original question: > Adam and Ben find a jumble of n ropes lying on the floor. Each takes hold of one loose end. What is the probability that they are both holding the same rope? * * * There is a fair portion of every grade 12 maths teacher in _Johannesburg_ coming up with a different solution to this seemingly basic probability question. Here is what I've come up with so far: !my solution Any input on the matter would be greatly appreciated :)

There are $n$ ropes and $2n$ ends. Suppose Adam grabs one of those $2n$ ends first; Ben then has to pick from $2n-1$ ends and only one of those ends belongs to the same rope Adam is holding. Therefore the probability they hold the same rope is $$\frac1{2n-1}$$ Alternatively, suppose the ends of the first rope are labelled 1 and 2, the ends of the second rope are labelled 3 and 4 and so on until the last rope's ends are labelled $2n-1$ and $2n$. Then Adam and Ben can select the ends in $2n(2n-1)$ ways, of which $2n$ ways result in them holding the same rope. We get the same probability as above.