What is the expected number of suits in a hand of 5 cards? Suppose we draw 5 cards out of a deck of 52.What is the expected number of
different suits in our hand? For example, if we draw K♠ 3♠ 10♥ 8♥ 6♣, there
are three different suits in our hand.
Answer: 3.114
Here’s what I’ve tried. 
N = number of suits
E[X] = (1*P(N=1) + 2*P(N=2) + 3*P(N=3) + 4*P(N=4))/52C5
Now to calculate the probability of N suits is where I get a problem.
P(N=1) = (4C1 *13C5)/52C5 


*

*My Reasoning: (Pick a suit * choose 5 cards from it)


P(N=2) = (4C2 *13C1*13C1*24C3)/52C5


*

*My Reasoning: (Pick two suits * pick 1 card from each * choose 3 from the remaining cards in those two suits)


P(N=3) = (4C3 *13C1*13C1*13C1*36C2)/52C5


*

*My Reasoning: (Pick three suits * pick 1 card from each * choose 2 from the remaining cards in those three suits)


P(N=4) = (4C4 *13C1^4*48C1)/52C5


*

*My Reasoning: (Pick four suits * pick 1 card from each * choose 1 from the remaining cards in those four suits)


This leaves me with: 1(5148)+2(2052336)+3(5536440)+4(1370928)/2598960
Which equals 10.002
They’re aren’t even 10 suits in a deck so I’ve done something very wrong. 
 A: Your counting is a bit off.


*

*There are $4\binom{13}5$ hands with just one suit, so you’re fine there.  

*A hand with exactly two suits can have have one card in one suit and four in the other, or it can have two cards in one suit and three in the other. There are $4$ ways to choose the suit with fewer cards and then $3$ ways to choose the other suit, so the first of these possibilities gives rise to $4\binom{13}1\cdot3\binom{13}4$ hands, and the second gives rise to $4\binom{13}2\cdot3\binom{13}3$. The total number of two-suited hands is therefore $4\binom{13}1\cdot3\binom{13}4+4\binom{13}2\cdot3\binom{13}3$. Notice that you can’t just pick two suits: it matters which one is which.  

*A hand with exactly three suits can have a suit breakdown of $3$-$1$-$1$ or of $2$-$2$-$1$; as in the previous case, you have to count the possibilities separately. I’ll do the second and leave the first to you. There are $\binom42$ ways to choose the two suits that will have two cards each, and $\binom{13}2^2$ ways to choose those four cards. (This time it doesn’t matter which suit is which, since they supply the same number of cards.) There are $2$ ways to choose the remaining suit and $\binom{13}1$ ways to pick a card from it, so this case supplies $\binom42\binom{13}2^2\cdot2\binom{13}1$ hands.  

*A hand with all four suits must have a $2$-$1$-$1$-$1$ suit breakdown. There are $4$ ways to choose the suit that supplies two cards and $\binom{13}2$ ways to pick those two cards. The remaining three cards can be chosen in $\binom{13}1^3=13^3$ ways, for a total of $\binom{13}2\cdot13^3$ four-suited hands.

A: As another approach: Let $X_i$ be the indicator variable for the $i^{th}$ suit (so $1$ if the suit is present, $0$ if it is not). The Probability that a fixed suit is missing is $$q=\frac {39\times 38\times37\times 36\times 35}{52\times 51\times 50\times 49\times 48}=\frac {2109}{9520}\sim .22153$$
Thus $E[X_i]=1-q=\frac {7411}{9520}\sim .7785$
By the Linearity of  Expectation the answer you want is $$4\times E[X_i]=\frac {7411}{2380}\sim \fbox {3.1139}$$
