How Many Rounds Will I Last?

Let $E_i$ be the expected number of rounds that can be played when you have $i$ lives. We need to solve for $E_3$ .

Then the rules lead to the following equations: $\left\{\begin{array}{rl} E_1 = & 1 + \tfrac9{10}E_2 \\ E_2 = & 1 + \tfrac9{10}E_3 + \tfrac1{10}E_1 \\ E_3 = & 1 + \tfrac9{10}E_3 + \tfrac1{10}E_2 \end{array}\right.$

To solve this, I noticed the similarity between the last two lines: $E_3 = E_2 - \tfrac1{10}E_1 + \tfrac1{10}E_2 = \tfrac{11}{10}E_2 - \tfrac1{10}E_1.$ Substitute into the equation for $E_2$ : $E_2 = 1 + \tfrac9{10}(\tfrac{11}{10}E_2 - \tfrac1{10}E_1) = 1 + \tfrac{99}{100}E_2 + \tfrac1{100}E_1\ \therefore\ E_2 = 100 + E_1.$ Substitute this into the equation for $E_1$ : $E_1 = 1 + \tfrac9{10}(100 + E_1) = 91 + \tfrac9{10}E_1\ \therefore\ E_1 = 910.$ Backsubstitute: $E_2 = 100 + E_1 = 100 + 910 = 1010;\ \ \ E_3 = \tfrac{11}{10}E_2 - \tfrac1{10}E_1 = 1111 - 91 = \boxed{1020}.$

[meta solution]

Arjen's solution already shows an elegant way to solve the problem.

Here is how we could use a Markov chain model in Mathematica to solve the problem:

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chain = DiscreteMarkovProcess[
   4, {{1, 0, 0, 0}, {0.1, 0, 0.9, 0}, {0, 0.1, 0, 0.9}, {0, 0, 0.1, 
     0.9}}];

Mean[FirstPassageTimeDistribution[chain, {1}]]

The technique is described in detail here