I have 3 sets of 15 elements, sets A B and C. Each element in set A matches with one and only one element in set B, and one and only one element in set C. For example there are 15 distinct colours in set A, and the same 15 colours in set B and C.
One picks an element from set A, then set B, and finally set C, which forms a group (group number 1). This will then repeat until all the elements have been grouped. I want to calculate the probability of there being at least one matching group of elements 'correct' (all 3 colours are the same) when all 15 groups have been made by picking the elements at random from their respective sets (still only one element from each set).
Sorry about the lackluster delivery of the question, I find it very hard to phrase.
Arrange $15$ elements in a row, say $1,2,3,.......15$, and we need to arrange two more rows of $15$
We shall select sets of $3$ from columns formed along with the other two rows.
If no set of three has to match, one from rows $2$ and $3$ need to be deranged w.r.t. row $1$.
Denoting the number of derangements of $n$ objects by $D_n$, a simple formula is $D_n = \left[\dfrac{n!}{e}\right],$
where $[x]$ is the nearest integer function, e.g. $D_4 = [\frac{4!}{e}] = 9$
So # of ways in which no column is identical $= (D_{15})\cdot15!$ against a total of $(15!)^2$ arrangements,
and using the complement, P(at least one column has the same colors) $= 1 - \dfrac{(D_{15})\cdot15!}{(15!)^2} =1 -\dfrac{(D_{15})}{(15!)} $
Added
Wolframalpha gives the numerical value as $\approx 0.6321$