Expected number of tosses until winner in game with two players with two different coins 
Two players each have a coin which gives heads with probability $0.7$
  (player 1) and $0.3$ (player 2), respectively. Player 1 goes first,
  and the players alternate until someone gets heads. What is the
  expected number of tosses?

The geometric distribution would model for each player separately the number of tosses until the first heads ($1.43$ and $3.33$), but I can't figure out the solution in this alternating case. I also tried an approach using Markov chains but to no avail. This is not my homework; I saw this problem online and couldn't figure out a solution.
Simulations place the expected number of tosses near $1.65$. 
set.seed(142857)

res <- replicate(1e5, {
  which.max(rbinom(100, 1, c(0.7, 0.3)))
})

mean(res)
[1] 1.6479

 A: Let $X$ be the number of tosses. Let $E[X]=M$ be the expected value of it. $X$ is  $1$ with $0.7$ probability, and $2$ with $0.3\times 0.3=0.09$ probability. For all other cases, i.e. with $0.7\times 0.3 = 0.21$ probability, the whole process is repeated. So, we pay the two toss price and still expect $M$ tosses for the play to finish, which gives us the following recursive formula:
$$M=1\times0.7+2\times 0.09+(2+M)\times0.21\rightarrow M\approx1.65$$
A: Here is a solution using the machinery of markov chains. The canonical form of a transition matrix with $r$ absorbing states and $t$ transient states is
$$
\mathbb{P} = \pmatrix{\mathbf{Q} & \mathbf{R} \\ \mathbf{0} & \mathbf{I}}
$$
where $\mathbf{I}$ is an $r\times r$ identity matrix, $\mathbf{0}$ is an $r\times t$ zero matrix, $\mathbf{R}$ is a nonzero $t\times r$ matrix and $\mathbf{Q}$ is a $t\times t$ matrix. Hence, the $4\times 4$ transition matrix of the four states "P1 turn", "P2 turn", "P1 wins", "P2 wins", in its canonical form is given by:
$$\mathbb{P} = \pmatrix{0 & 3/10 & 7/10 & 0 \\ 7/10 & 0 & 0 & 3/10 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1}$$
The corresponding fundamental matrix is given by $\mathbf{N}=(\mathbf{I}_2 - \mathbf{Q})^{-1}$. In our example:
$$
\mathbf{N} = \pmatrix{100/79 & 30/79 \\ 70/79 & 100/79}
$$
The expected number of steps to absorption when starting in transient state $i$ is the $i$th entry of the vector
$$
\mathbf{t} = \mathbf{N}\mathbf{1}
$$
where $\mathbf{1}$ is a length-$t$ column vector with all $1$s. Applied to our example:
$$
\mathbf{t} = \pmatrix{100/79 & 30/79 \\ 70/79 & 100/79}\pmatrix{1 \\ 1} = \pmatrix{130/79 \\ 170/79} \approx \pmatrix{1.646 \\ 2.152}
$$

Hence, if player 1 starts, the expected number of tosses until a win is roughly $1.65$ and
  if player 2 starts, the expected number of tosses is roughly $2.15$.

We can also get the variance on the number of tosses which is given by:
$$
\left(2\mathbf{N} - \mathbf{I}_{2}\right)\mathbf{t} - \mathbf{t}\circ\mathbf{t}
$$
where $\circ$ denotes the Hadamard product (element-wise product). For the example this evaluates to:
$$
\pmatrix{9030/6241 \\ 9870/6241}\approx\pmatrix{1.447 \\ 1.581}
$$
