# Probability of Getting Aces

A standard deck of 52 cards is shuffled and dealt. Let $X_1$ be the number of cards appearing before the first ace, $X_2$ the number of cards between the first and second ace (not counting either ace), $X_3$ the number between the second and third ace, $X_4$ the number between the third and the fourth ace, $X_5$ the number after the last ace. It can be shown that each of these random variables $X_i$ has the same distribution, $i=1,2,...,5$, and you can assume this to be true.

a) Write down a formula for $P(X_i=k),0\le k \le 48$.

b) Show that $E(X_i) = 9.6$ [Hint: Do not use your answer to a).]

c) Are $X_1,...,X_5$ pairwise independent? Prove your answer.

I could get some numerical answers for specific $X_i$, but don't know how to get a formula for the general case.

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a) Since you know they're all the same, you can focus on $X_1$. Ignoring the permutations of the non-ace cards, how many ways are there to arrange $4$ aces in the deck? And how many ways are there to arrange $3$ of them in the last $52-(X_i+1)$ cards?
b) What's the sum of the $X_i$?
c) What's the probability to find $X_5=48$ given $X_1=48$? Is that joint probability the product of the marginal probabilities $P(X_i=48)$?