The probability that you lose at every stage in the process is a product of probabilities:
$$
\frac{9}{10} \cdot \frac{99}{100} \cdot \frac{999}{1000} \cdot \frac{9999}{10000} \cdots
$$
or, in infinite product notation, $$\prod_{k=1}^\infty \left(1 - \frac1{10^k}\right).$$ The answer we want is $1$ minus this infinite product.
This infinite product does not have a closed form in terms of familiar functions. (Mathematica "simplifies" the infinite product for me to the $q$-Pochhammer symbol $(\frac1{10}; \frac1{10})_{\infty}$, but that's less of a simplification and more of "giving this infinite product we can't evaluate a name".)
Approximately, $1 - (\frac1{10}; \frac1{10})_{\infty} \approx 0.1099899$, so the answer of $\frac19$ is very close, even though it's not correct.
To see why the answer is close to $\frac19$, imagine that at the very start of the lottery, you were given a ticket for every stage of the lottery, and - crucially - promised that at most one of those tickets is a winning ticket. In this case, the probabilities $\frac1{10}, \frac1{100}, \frac1{1000}, \dots$ would be probabilities of disjoint events: that is, there would be no overlap between them. We could then simply add them together for a final answer of $$\frac1{10} + \frac1{100} + \frac1{1000} + \cdots = 0.111\dots = \frac19.$$ Unfortunately, the actual lottery is a bit less nice than this; there are some outcomes where multiple of these tickets could be winning tickets, so some of the wins are "wasted". This brings the probability down from $\frac19$ by a small amount.
Even though we don't have a formula for the answer better than the unhelpful "$1 - (\frac1{10}; \frac1{10})_{\infty}$", we can actually compute many decimal digits of this number easily! (Credit to Don Hatch in the comments for helping me notice this.) This comes down to a powerful result known as Euler's pentagonal theorem: the identity $$\prod_{k=1}^\infty (1 - x^n) = 1 + \sum_{k=1}^\infty (-1)^k (x^{k(3k+1)/2} + x^{k(3k-1)/2}).$$ In our case, if we set $x = \frac1{10}$, we get $$1 - \prod_{k=1}^\infty \left(1 - \frac1{10^k}\right) = \sum_{k=1}^\infty (-1)^{k+1} \Big(10^{-k(3k+1)/2} + 10^{-k(3k-1)/2}\Big).$$ If we take the first few terms of this infinite sum, it looks like $$10^{-1} + 10^{-2} - 10^{-5} - 10^{-7} + 10^{-12} + 10^{-15} - 10^{-22} - 10^{-26} + \dots$$ which are in fact adjustments to individual digits past the decimal, increasingly far apart in the decimal expansion.
It takes a bit of work to turn this into something practical, but here is the pattern for the decimal digits of the final answer. We can divide the digits into blocks of lengths $4, 7, 10, 13, 16, \dots$, so that:
- The first block is $1\,0\,99$.
- The second block is $8\,99\,0000$.
- The third block is $1\,000\,999999$.
- The fourth block is $8\,9999\,00000000$.
- In general, for odd $k$, the $k^{\text{th}}$ block is $1\, \underbrace{00\dots00}_k\, \underbrace{99\dots99}_{2k}$...
- ...and for even $k$, the $k^{\text{th}}$ block is $8\,\underbrace{99\dots99}_k\, \underbrace{00\dots00}_{2k}$.
This lets us very quickly write down many many digits of the probability of winning the multi-stage lottery!
