The probability a given day has no birthdays is $\left(1-\frac{1}{365}\right)^{10}$ so the expected number of days with no birthday is $365$ times this, about $355.1224$.
The probability a given day has exactly one birthday is $10 \times \frac{1}{365} \times \left(1-\frac{1}{365}\right)^9$ so the expected number of days with exactly one birthday is $365$ times this, about $9.7561$.
Subtract these two figures from $365$ and you get a result of about $0.1215$ for the expected number of days with two or more birthdays.
Let's rephrase the question.
Find the probability that any $2$ people have their birthday in the same day of the year in a group $4n$ people, of which $n$ born were in the leap year and $3n$ people were born in the regular year.
The complement of the event "any 2 people have their birthday" is "all people have birthday on different days of the year".
Consider two mutually exclusive possibilites: $\mathcal{A}_1$ - one of these birthdays falls on "Feb. 29", and $\mathcal{A}_2$ -- none of these birthdays fall on the leap day.
The number of configurations of the second kind is
$$
N\left(\mathcal{A}_2\right) = \prod_{k=1}^{4 n} \left( 365 - (k-1) \right)
$$
and the number of configurations of the first kind is
$$
N\left(\mathcal{A}_1\right) = n \prod_{k=1}^{4 n-1} \left( 365 - (k-1) \right)
$$
The total configuration size is $N_T=366^n 365^{3n}$, hence the probability of having no birthday duplicates is
$$
p_c = \frac{ N\left(\mathcal{A}_1\right) + N\left(\mathcal{A}_2\right)}{N_T}
$$
Here is a confirmation in Mathematica:
(* computed probability expression *)
p[n_Integer] = (Product[365 - (k - 1), {k, 1, 4 n}] +
n Product[365 - (k - 1), {k, 1, 4 n - 1}])/(366^n 365^(3 n));
(* simulation result *)
In[45]:= With[{n=7,reps=10^8},
Sum[Boole[DuplicateFreeQ[Join[
RandomChoice[Range[1,366],n],
RandomChoice[Range[1,365],3n]]]],{k,1,reps}]/reps]//N
Out[45]= 0.34599694
In[53]:= N[p[7]]
Out[53]= 0.3460041650022472
(* verify that computed value belongs to the sampling
confidence interval *)
In[55]:= With[{pSample = Out[45], pExpected = p[7]},
IntervalMemberQ[
Interval[pSample + {-1, 1} Sqrt[(pSample (1 - pSample))/10^8]],
pExpected]]
Out[55]= True
Best Answer
By linearity of expectation, this is just $365$ times the probability that exactly $k$ people are born on a given day, which is $\binom nk(1/365)^k(364/365)^{n-k}=\binom nk364^{n-k}/365^n$, so the expected number of such days is $\binom nk364^{n-k}/365^{n-1}$.