Expected Value and Indicator Variable - Deck of cards If two people each draw $n$ cards out of a deck of 52 distinct cards with replacement, find $n$ such that the expected number of common cards that both people drew is at least 1.
Since each card is replaced immediately after it's drawn, I am not sure how to compute this. 
I was thinking that each card in drawing $n$ cards would have $\dfrac{n}{52}$ chance to be the same as one of the cards that are drawn by the first person. Then using properties of expectation, can I just sum up different $n$ values starting from 1 until the sum is greater than 1? Using this approach, I got $n$ should be 10 but I don't think that's right. Also, since it is at least 1, am I supposed to calculate the complement instead somewhere and subtract it by 1? 
Thank you!
 A: HINT:
The choices are:
For n=0 it is not possible, hence implies 0, over one case: the null case.
For n=1 trivially implies 52, over $52^2 $ choicss.
For n=2, at least one common implies $52 \ 52^2$, over $52^4$ choices.
For n=3, at least one common implies $52\ 52^4$, over $52^6$ choices.
For n=k, at least one common implies $52\ 52^{2k-2}$, over $52^{2k}$ choices.
Hence the expectation of n when having at least one common choice is:
$$
\mathbb{E}\{n\}=\sum_{k=1}^{n} k p(k)=\sum_{k=1}^{n} {1 \over 52} k ={1 \over 52} {n(n+1) \over 2}
$$
Which must be equal to 1, so:
$$
n^2+n-104=0\\
n=\frac 12 (-1 \pm \sqrt{1+4 \ 104})= \frac 12 (-1 \pm \sqrt{417})\\
n = 10
$$
A: $N$=52. Let $X_i, Y_i$ be the indicator that the $i$-th card was picked at least once by the first and second persons, respectively. By symmetry, 
$$\mathbb{E}[X_i]=\mathbb{E}[Y_i] = 1 - \mathbb{P}(i\textrm{-th card not picked in }n \textrm{ draws}) = 1 - \left(1 - \frac{1}{N}\right)^n$$ for all $i\in[N]$. The expected number of common cards is $$\mathbb{E}\left[ \sum_{i \in [N]} X_iY_i\right] = \sum_{i \in [N]} \mathbb{E}\left[X_iY_i\right] = \sum_{i \in [N]} \mathbb{E}\left[X_i\right] \mathbb{E}\left[Y_i\right] = N\left[1 - \left(1 - \frac{1}{N}\right)^n\right]^2.$$
For this to be at least 1, we need
$$ N\left[1 - \left(1 - \frac{1}{N}\right)^n\right]^2 \geq 1 \Rightarrow n \geq \left\lceil
\frac{\log\left( 1 - \frac{1}{\sqrt{N}}\right) }{ \log \left( 1-\frac{1}{N}\right) }
\right\rceil = 8.$$
