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:

Any input on the matter would be greatly appreciated :)
 A: 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.
A: We have to be careful, since there are two ends to every rope. Suppose that Adam has an end already. Then there are $2n-1$ remaining ends, exactly one of which is the other end of Adam's rope. So the chance they are holding the same rope is $1/(2n-1)$.
A: You can treat the ends are identical or unique/distinguishable. It doesn't matter.
If the ropes are identical then which one the first person picks doesn't matter. Then the second person can pick one end out of the $2n-1$ remaining. So the probability is $\frac{1}{2n-1}$.
If the ends of the rope are distinct then there are $2n$ ends. The first person has $2n$ choices. The second person has $2n-1$ choices. You have $2n$ desirable outcomes. Either the first person is holding rope $n$ by end number one and the second person is holding rope $n$ by end number two OR they are holding the opposite ends. So the probability is $\frac{2n}{2n(2n-1}=\frac{1}{2n-1}$.
A: There are $2n(2n-1)$ possible couples of rope ends (Adam and Ben respective choices). For any rope end chosen by Adam out of the $2n$ available, only one out of the remaining $(2n-1)$ is the right one. So there are actually $2n$ compatible couples (with Adam holding one end, and Ben holding the other).
I'm quite sure the answer is $\frac{2n}{2n(2n-1)}=\frac{1}{2n-1}$.
The comment you made on Adam starting with the other end of the same rope is wrong. Let's say you "tag" every rope with one black end and one white end (end 1 and end 2, as you said).
Then the probability of Adam taking precisely a black end is not $\frac{2}{2n}$. It is actually $\frac{1}{2n}$. So, in making that miscounting, you actually counted already the possibility of Adam taking one rope, no matter which end.
Summarizing:


*

*The $\frac{1}{n}$ in the answer is wrong.

*The double counting is already made implicitly in your answer.


My suggestion is to solve the problem directly using rope ends choices. 
