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Suppose we have $52$ cards and we want to pick cards with replacement until an ace has appeared or we've chosen $5$ cards and we want to determine the expected number of cards we have picked. Let $X(s)$ be the random variable points the outcomes to number of cards we have picked in $s$.

a. What are the possible values of $X(s)$?

b. For each value $k$ that we've listed in (a), calculate $P(X = k)$.

c. What is $E[X]$?

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The possible values are $1,2,3,4,5$.

Clearly $\Pr(X=1)=\dfrac{4}{52}$.

To have $X=2$, we need first card non-Ace, second card Ace. Thus $\Pr(X=2)=\dfrac{48}{52}\cdot\dfrac{4}{52}$. This is because of the hypothesis of replacement.

Similarly, $\Pr(X=3)=\left(\dfrac{48}{52}\right)^2\cdot\dfrac{4}{52}$.

Similarly, $\Pr(X=4)=\left(\dfrac{48}{52}\right)^3\cdot\dfrac{4}{52}$.

For $X=5$, the situation is a little different, since we don't care about the fifth card. So $\Pr(X=5)=\left(\dfrac{48}{52}\right)^4$ (four "failures" in a row)

Now expectation is a routine calculation. It is $\sum_{i=1}^5 i\Pr(X=i)$.

Remark: In most card games, we do not have replacement. (By replacement I assume you meant replacement and shuffling, not replacement on top of the deck!)

Without replacement, the probabilities would be a little different. For example, for $X=3$ we would have $\frac{48}{52}\frac{47}{51}\frac{4}{50}$.

If you prefer to use counting, let us calculate for example the probability that $X=4$. There are $52^4$ sequences of $4$ cards, all equally likely. Exactly $(48)(48)(48)(4)$ of them consist of $3$ non-Aces followed by an Ace. So $\Pr(X=4)=\frac{(48)^3(4)}{52^4}$. Earlier, we obtained the same answer in a somewhat different way.

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Thanks, I had no idea how to solve it, thank you –  Hooman Nov 22 '12 at 7:44
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