Question

A standard deck of 52 cards is shuffled and dealt. Let X1 be the number of cards appearing before the first ace, X2 the number of cards between the first and second ace (not counting either ace), X3 the number between the second and third ace, X4 the number between the third and forth ace, and X5 the number after the last ace. It can be shown that each of these random variables Xi had the same distribution, i=1,2,...,5, and you can assume this to be true.

a) Write down a formula for P(Xi=k), 0≤k≤48

b) Show that E(Xi)= 9.6 (Hint:Don't use your answer to part a)

c) Are X1,X2,...,X5 pairwise independent? Prove your answer.

Answer 1

Let X1 be the number of cards appearing before the first ace

P(X1 = 0) = 4/52 = 1/13 (since it's P(first card is Ace))
P(X1 = 1) = 48/52 (no ace first) * 4 / 51 (Ace second)
P(X1 = 2) = 48/52 * 47/51 * 4/50 (no ace, no ace, Ace)
P(X1 = 3) = 48/52 * 47/51 * 46/50 * 4/49 (no, no, no, Ace)

and so on up to P(X1 = 48) = 48/52 * 47/51 * 46/50 * ..... * 1/5 * 4/4

We can write those as
4/52 * 48/51 * 47/50 * ... * however many factors but simply rearranging the numerators, since multiplication is commutative and associative.
So we have

For the above that gives:
k = 0: 4/52 * 48! / (48!) * 51! / 51! = 4/52 = 1/13
k = 1: 4/52 * 48! / (47!) * 50! / 51! = 4/52 * 48 / 0! * 50! / 51! = 4/52 * 48/51
k = 2: 4/52 * 48! / (46!) * 49! / 51! = 4/52 * 48 * 47 / (51 * 50) = 4/52 * 48/51 * 47/50
and so on

b)

The 4 aces divide the deck of 52 cards into 5 groups:
g1 ..A1 ... g2 ... A2 ... g3 .... A3 ....g4 .... A4 .... g5 (where A* = aces and g* = other cards (0 or more) )

On average, the length of g[i] = 48/5 = 9.6 so no matter where the Aces fall, and all places are equally likely, for each g[i] which is less than 9.6, there will be a compensating length
in one of the others to make up for it.