What is the expected number of suits in a hand of 4 cards? To find the expected number of suits the formula is
$E(Num Suits) = 1*P(1 Suit) + 2*P(2 Suit) + 3*P(3 Suit) + 4*P(4 Suit)$
For the probability of getting 4 suits I got
${13 \choose 1}^4 {4 \choose 4}/{52 \choose 4}$
For the probability of getting just 1 suit I got
${4 \choose 1} {13 \choose 4} / {52 \choose 4}$
I'm having difficulty finding the probabilities for the 2 and 3 suits. I know for the two suits that we have the possibility of getting AAAB or AABB. 
Thank you!
 A: This is a nice exercise, because the method described by @grand_chat is so much easier than the direct method.  Using OP's notation, the probability of having exactly one suit is
$$
P(AAAA)=\frac{4\cdot{{13}\choose{4}}}{{52}\choose{4}}\approx 0.0106;
$$
the probability of having two suits is
$$
P(AAAB)+P(AABB)=\frac{4\cdot {{13}\choose{3}}\cdot 3 \cdot {{13}\choose{1}} + {{4}\choose{2}}\cdot{{13}\choose{2}}^2} {{52}\choose{4}}\approx 0.2996;
$$
the probability of having three suits is
$$
P(AABC)=\frac{4\cdot{{13}\choose{2}}\cdot{{3}\choose{2}}\cdot{{13}\choose{1}}^2}{{52}\choose{4}}\approx 0.5843;
$$
and the probability of having four suits is
$$
P(ABCD)=\frac{{{13}\choose{1}}^4}{{52}\choose{4}}\approx 0.1055.
$$
These sum to $1$, as they should.  The expected number of suits is therefore
$$
(1\cdot 0.0106) + (2\cdot 0.2996) + (3\cdot 0.5843) + (4\cdot 0.1055) = 2.7847.
$$
Alternately, we calculate the probability of having any of a particular suit, which is
$$
P(\#A>0)=1-P(\#A=0)=1-\frac{{39}\choose{4}}{{52}\choose{4}}\approx 0.6962,
$$
and multiply that by four, obtaining the same result ($2.7847$) much more simply.
A: Hint: The number of suits is the sum of four indicators: $N = X_1+X_2+X_3+X_4$ where $X_k$ equals 1 if suit $k$ is present, and equals zero otherwise, with $k$ ranging over the four possible suits.
Now the expectation you seek is $E(N) = E(X_1)+E(X_2)+E(X_3)+E(X_4)=4E(X_1)$ by symmetry. But $E(X_1)=P(\text{suit #1 is present})$, which you can calculate by asking "What's the probability that suit #1 is absent from my hand of four cards?"
A: Interestingly, E(N) can be estimated by a polynomial: $4-4\bigg(1-\frac{N}{52}\bigg)^{13}$
E(4) = 2.5870, which is 7% off the correct answer, but not bad over all values of N = 1...52
Has anyone ever seen this before? Is this a known result?
