You want to set up a games stall at the Winter Wonderland. You have a single pack of 52
cards and decide to play card games with your customers. The game is designed as follows. A
customer draws 4 cards at random. For every “Ace” that the customer draws, he/she wins £3
and for every face card (“Jack”, “Queen” or “King”), he/she wins £1.
Let the two discrete random variables U and V, be the number of aces and face cards
obtained, respectively.
(i) Derive the joint probability mass function p(U, V).
(ii) Find the marginal probability mass functions of U and V.
(iii) Are U and V independent? Specify the reason.
(iv) Find E(U), E(V), Var(U), Var(V) and Cov(U, V) and interpret the obtained values.
(v) Find the minimum price for the game, so that you don’t bear a loss if a very large
number of games are played.
(vi) If you set the price for each game by rounding up the value found in (v), find the
percentage of people who will go back home with positive winnings.
That is a "serious" problem
I will use A for ace
F for facecard
and L for non-ace, non-face (L for loser)
I will also assume that none of the 4 cards drawn in a game are returned to the deck, but they would be returned after a game is played.
so the possible combinations of cards drawn is
AAAA
AAAF
AAAL
AAFF
AFFF
FFFF
AAFL
AFFL
AALL
LLFF
LLAF
LLLF
LLLA
LLLL
(hope I got them all)
Let's just take one of these combinations:
AAFL
this will win 3+3+1+0 or £ 7
prob of it happening
= (4/52)(3/51)(12/50)(36/49)
(notice that no matter which one we pick, the denominator would always be the same in the original calculation )
BUT, AAFL can be arranged in 4!/2! ways
so the expected winnings for any arrangement of AAFJ
= 7 * (4/52)(3/51)(12/50)(36/49) * (4!/2!)
= .....
I will leave the fun part up to you
Unless I have completely mis-interpreted the question .....