Average number of cards of a particular suit in a hand I am creating a card game AI for an iPhone game and I'm stumped trying to come up with this statistic.
The deck has 4 suits. The cards are numbered 1-14 in each suit and there is one Joker.  So, the deck has 57 cards.  One player will be dealt 18 cards and this person will also get to decide which suit the Joker is (so there are technically 15 cards of that suit) and calls that suit trump.
What is the average number of trump cards that this person will have? Can you help me understand how I would figure this out?  I took a few stats classes in college, and it's all gone!
 A: Assuming that the 18-card player always chooses her longest suit to be trumps, then the calculations are possible but lengthy.
There are 1250 different possible patterns of suits for 18 cards. For example one might be: 2 cards from suit A, 7 cards from suit B, 0 cards from suit C, and 9 cards from suit D, so the longest suit has 9 cards.  The number of ways of producing this pattern is 
$${14 \choose 2}{14 \choose 7}{14 \choose 0}{14 \choose 9} = 91 \times 3432 \times 1 \times 2002 = 625248624$$ 
(or $18!$ times as many if order matters, but we will ignore that here).  Add together all the other ways of having 9 as the number of the longest and you get 3,570,677,566,456 ways.  Do the same for the other possible maxima (which you should be able to see are from 5 to 14) and add up all the possible ways to get ${56 \choose 18}$, the number of ways of choosing 18 cards from 56.  Divide this into the number of ways for each possible maximum length and you get the following probabilities (rounded at the last decimal place)
Max Probability
5   0.168508745
6   0.461233813
7   0.267882939
8   0.083137716
9   0.016816801
10  0.002225768
11  0.000184999
12  0.000008993
13  0.000000224
14  0.000000002

This is enough to work out the expected number in the longest suit, namely about $6.326$.
But that was for 18 cards.  Add the joker and the expected number in the longest suit becomes about $7.326$.  
