Average number of pairs of distinct adjacent objects? 

(1) Given 8 red balls and 7 blue balls arranged randomly in a sequential fashion, what is the average number of pairs of distinct adjacent balls ?




For example the sequence $\mathbf{RBR}RRRRR\mathbf{RB}BBBBB$ would contribute 3 pairs, $\mathbf{RB}$ in slot $(1,2)$ , $\mathbf{BR}$ in slot $(2,3)$ and $\mathbf{RB}$ in slot $(9,10)$



2) Same question but with 6 red balls, 5 blue balls, 4 white balls




*

*My attempt for (1) is to consider the first 2 slots, and state that these will form a pair if it is either BR or RB, so that the probability of the first 2 slots being a pair is:
$$ \frac {7} {7+8} * \frac {8} {7+7} + \frac {8} {7+8} * \frac {7} {7+7} $$


but then for slots (2,3) , given that there is no replacement and overlap with slot (2) for slots (1,2) , i am not sure ...


*

*I guess if I can manage (1) then (2) is similar ?


Use three  generating functions  ($A, B$ and  $C$) in  three variables
($z$, $w$, and $v$) to enumerate strings ending in one of three colors
with the  variable $u$ marking  pairs of distinct adjacent  colors. We
have $a$  instances of the first color,  $b$ of the second  and $c$ of
the third.
We obtain
$$A - z = Az + Buz + Cuz,
\\ B - w = Auw + Bw + Cuw,
\\ C - v = Auv + Buv + Cv.$$
Putting $Q = A + B + C$ we seek
$$P= \left. \frac{d}{du} Q \right|_{u=1}.$$
Differentiate to obtain
$$A' = A'z + Bz + B'uz + Cz + C'uz.
\\ B' = Aw + A'uw + B'w + Cw + C'uw,
\\ C' = Av + A'uv + Bv + B'uv + C'v.$$
Add these and put $u=1$ to get
$$P = P (z+w+v) + 
\left.Q (z+w+v)\right|_{u=1} 
- \left.(Az + Bw + Cv)\right|_{u=1}.$$
We have by inspection that
$$\left.Q \right|_{u=1}  = \frac{z+w+v}{1-z-w-v}$$
and
$$\left.(Az + Bw + Cv)\right|_{u=1}
=  \frac{z^2+w^2+v^2}{1-z-w-v}.$$
This yields
$$P(1-z-w-v) = \frac{(z+w+v)^2}{1-z-w-v}
- \frac{z^2+w^2+v^2}{1-z-w-v}$$
or
$$P = \frac{2zw+2zv+2wv}{(1-z-w-v)^2}.$$
We  now skip  ahead and  show how  to solve  the general  case  of $m$
colors. We get for the generating function
$$P = \frac{2\sum_{1\le p \lt q\le m} w_p w_q}
{\left(1-\sum_{p=1}^m w_p\right)^2}.$$
Extracting  coefficients  on  $[w_1^{d_1} w_2^{d_2}\cdots  w_m^{d_m}]$
where  $d  =  \sum_{p=1}^m  d_p$   we  use  the  Newton  binomial  and
obtain
$$2 \sum_{1\le p \lt q\le m} 
(d-1) {d-2\choose d_1, d_2, \ldots d_p-1, \ldots d_q-1, \ldots d_m}
\\ = {d\choose d_1, d_2, \ldots d_p, \ldots d_q, \ldots d_m}
\\  \times 2 \sum_{1\le p \lt q\le m} (d-1)
\frac{1}{d(d-1)} d_p d_q.$$
Divide by the multinomial coefficient to obtain the expectation
$$\bbox[5px,border:2px solid #00A000]{
\frac{2}{d_1+d_2+\cdots+d_m} 
\sum_{1\le p \lt q\le m} d_p d_q.}$$
The original problem by the OP then produces
$$\bbox[5px,border:2px solid #00A000]{
\frac{2ab+2ac+2bc}{a+b+c}.}$$
In particular we get for $(6,5,4)$ the exact value and the numerics
$$\bbox[5px,border:2px solid #00A000]{
\frac{148}{15} \approx 9.866666667.}$$
There  is a  Maple script  for this  which goes  as  follows (warning:
enumeration -- use on small values):

with(combinat);
ENUM :=
proc(L)
option remember;
local m, d, all, perm, pos, res, src, flips;

    m := nops(L); d := add(p, p in L);

    all := 0; res := 0;

    src := [seq(seq(p, q=1..L[p]), p=1..m)];

    for perm in permute(src) do
        flips := 0;
        for pos to d-1 do
            if perm[pos] <> perm[pos+1] then
                flips := flips + 1;
            fi;
        od;

        res := res + flips;
        all := all + 1;
    od;

    res/all;
end;

X :=
proc(L)
option remember;
    local m, d;

    m := nops(L); d := add(p, p in L);
    2/d*add(add(L[p]*L[q], q=p+1..m), p=1..m);
end;

An elementary argument is sure to appear now that the answer, which is
very simple, has been posted.
 What we have here is essentially the DFA method, a legacy algorithm.
