Average time for a tetris field to fill I have a tetris field, infinitely wide (irrelevant) and N blocks high. Now random blocks start falling down, and I do not move them.
Given that the field is N blocks high, the blocks aren't moved, the blocks fall one block per second and the blocks are the default blocks (they always "spawn" with the same orientation), how long does it take on average before the game is over?

There are $7$ different kinds of blocks. The height of the stack will generally increase by $2$ in each step, but only by $1$ in the following cases: A green block preceded by any block; a purple block preceded by a red block; and a blue block preceded by either a yellow or a gray block. I'll treat the events of certain pairs following each other as independent; they're not, but it doesn't make a difference for the average height after time $t$ (by linearity of expectation) and only a small difference for the average time $t$ for height $N$.
So we have $49$ different pairs that are assumed to occur with equal and independent probabilities of $1/49$; for $1\cdot7+1\cdot1+1\cdot2=10$ of these pairs the second block increases the height by $1$, and for $49-10=39$ of them by $2$. Thus the average height after time $t$ will be $\frac{10+2\cdot39}{49}t=\frac{88}{49}t$. A good estimate for the average time to reach height $N$ is therefore $\frac{49}{88}N$. But we can be a bit more precise than that.
The average time $t_N$ to reach height $N$ satisfies:
$$t_N=\frac{10}{49}t_{N-1}+\frac{39}{49}t_{N-2}+1\;,$$
since there is a probability of $\frac{10}{49}$ that the next step will have height $N-1$ left and a probability of $\frac{39}{49}$ that it will have height $N-2$ left, and in either case the current step takes time $1$.
We can solve this inhomogeneous linear recurrence relation by adding a specific solution of the inhomogeneous relation to the general solution for the homogeneous relation and then imposing initial conditions. We can guess a specific solution for the inhomogeneous relation from our earlier estimate: It is simply $t_N=\frac{49}{88}N$. For the general solution of the homogeneous equation, we have to solve the characteristic equation
$$\lambda^2-\frac{10}{49}\lambda-\frac{39}{49}=0\;,$$
with solutions
$$\lambda_{1,2}=\frac{5\pm\sqrt{1264}}{49}\;,$$
$$\lambda_1\approx 0.8276, \lambda_2\approx -0.6235\;.$$
Then the general solution of the inhomogeneous relation is
$$t_N=\frac{49}{88}N+c_1\lambda_1^N+c_2\lambda_2^N\;.$$
The initial conditions are $t_0=0$ (if there is no height left, the game is over) and $t_1=10/49$ (if there is only one row left, the game is over unless the next block only increases the height by $1$, with probability $10/49$). Substituting these into the general solution yields
$$c_2=-c_1=\frac{10-49^2/88}{\sqrt{1264}}\approx 0.2431\;.$$
Thus, the correction terms proportional to $c_1$ and $c_2$ are quite small to begin with and die off quickly, so by the time the height reaches $N$ the result is well approximated by the initial estimate $t_N=\frac{49}{88}N$. The height in the image seems to be $N=20$, so this would lead to an average time $t_N=\frac{245}{22}\approx11.1$ until the game is over.
