It is impossible to have gotten a duplicate on the first draw. It is impossible to have not gotten a duplicate by the 1297'th draw by pigeon-hole principle.
To have gotten your first duplicate on the $k$'th draw, you need the first $k-1$ draws to all be distinct and the $k$'th to be a duplicate.
The first draw will always be distinct. The second will be distinct from the first with probability $\frac{1295}{1296}$. The third will be distinct from the first two with probability $\frac{1294}{1296}$ and so on... the $(n)$'th will be distinct from the earlier $n-1$ with probability $\frac{1296-n+1}{1296}$. Multiplying these, we get for $n$ draws to all be distinct, this will occur with probability $\frac{1296\frac{n}{~}}{1296^n}$ where $x\frac{n}{~}$ represents a falling factorial $x\frac{n}{~}=\underbrace{x(x-1)(x-2)\cdots (x-n+1)}_{n~\text{terms in the product}}=\frac{x!}{(x-n)!}$.
Next, supposing $k-1$ distinct values have all been taken, for the $k$'th to duplicate one of the earlier results, this will occur with probability $\frac{k-1}{1296}$
We have then the probability distribution function for $X$, the number of draws until the first duplicate:
$$Pr(X=k)=\frac{(k-1)1296\frac{k-1}{~}}{1296^k}$$
Applying the definition of expected value for a pdf: $E[X]=\sum\limits_{k\in\Delta} kPr(X=k)$ we have then the expected value is
$$\sum\limits_{k=2}^{1297}\frac{k(k-1)1296\frac{k-1}{~}}{1296^k}\approx 45.7889$$
wolfram link: http://www.wolframalpha.com/input/?i=sum+from+n%3D2+to+1297+of+n(n-1)(1296!%2F(1296-n%2B1)!)%2F1296%5En
