You are in a magical forest where fairies have enchanted the leaves of the trees with different colours. There are 6 pairs of enchanted leaves (12 leaves in total) in the forest, each labelled with a distinct number from 1 to 6. When you pick leaves blindly from the trees and pair them randomly, what is the probability that the paired leaves have colours differing by at most 1?
What I've tried:
We know, as far as I am able to understand the question, there are 12 leaves with 12 different colours. 6 pairs are formed and each pair (i.e each individual leaf) is given a number between 1 and 6. Now we need to create 6 new pairs from our 12 leaves such that each pair has a difference between the numbers on the two leaves of atmost 1.
Let $F(n)$ denote the number of pairings such that the given condition is satisfied for $n$ pairs i.e $2n$ leaves. then, I get the recurrence $$F(n)=F(n-1)+2F(n-2)$$ The $F(n-1)$ term comes from pairing both leaves with number n together and the $2F(n-2)$ term comes from pairing each $n$ leaf with an $n-1$ leaf (which there are 2 ways to do, since both $n$ leaves and $n-1$ leaves all have different colors) Combining this with the base conditions $F(1)=1, F(2)=3$ I get $$F(6)=43$$ The only problem I am having is figuring out the denominator. I doubt it would just be the number of pairings, $\frac{12!}{2^6 6!}$. I also feel that there is something problematic with my recurrence somehow. Any help is appreciated!
To a certain extent, this response is off-point, because I am not addressing where the original poster went wrong. The reason that I am ducking the issue is that I don't understand the OP's reasoning behind the idea that
$$F(n)=F(n-1)+2F(n-2).$$
$\underline{\text{My Approach}}$
This is an unusual combinatorics/probability problem that is more easily solved by avoiding combinatorics, and focusing solely on probability of events.
I will first solve the case of $~n = 6,~$ and then solve the more general case.
$~1/3~$ of the time, the first leaf chosen will be numbered either 1 or 6. When that happens, there will then be $~3~$ satisfying leaves from the remaining 11 leaves.
$~2/3~$ of the time, the first leaf chosen will not be numbered either 1 or 6. When that happens, there will then be $~5~$ satisfying leaves from the remaining 11 leaves.
Therefore the desired computation (when $~n=6~$) is
$$\left[ ~\frac{1}{3} \times \frac{3}{11} ~\right] + \left[ ~\frac{2}{3} \times \frac{5}{11} ~\right]. \tag1 $$
The computation in (1) above easily generalizes to the case of $~2n~$ leaves.
$~2/n~$ of the time, the first leaf chosen will be numbered either 1 or n. When that happens, there will then be $~3~$ satisfying leaves from the remaining $~(2n-1)~$ leaves.
$~(n-2)/n~$ of the time, the first leaf chosen will not be numbered either 1 or n. When that happens, there will then be $~5~$ satisfying leaves from the remaining $~(2n-1)~$ leaves.
Therefore the desired computation is
$$\left[ ~\frac{2}{n} \times \frac{3}{2n-1} ~\right] + \left[ ~\frac{n-2}{n} \times \frac{5}{2n-1} ~\right]. \tag2 $$