# Aces. A standard deck of 52 cards is shuffled and dealt. Let X1 be the number of cards appearing before the f?

Aces. 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: Do not use your answer to a).]

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

### 1 Answer

- MathMan TGLv 77 years agoBest Answer
Part a)

Well, the first one, X1, is probably the easiest to tackle, so let's give that a try.

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

P(X1 = k) = 4/52 * 48! / (48 - k)! * (51-k)! / 51! ← Formula (answer to part a)

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

Part 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.

Hence E(Xi) = 9.6

Part c)

Sorry, I cannot help with this part.

I suspect they are not independent, since,

for any Xi above average, there has to be another one (at least)

which is below average, since their sum is always 48.

To take an extreme example,

if X1 = 48 (all the Aces are the last 4 cards in the deck),

then X2 = X3 = X4 = X5 = 0 necessarily.

Similarly, if X1 = 0 (Ace is first card dealt),

then necessarily one of the others will be above the average

to "make up for the shortfall".