Do you only lose at Tetris because you don't play optimally or make mistakes? Even with any amount of time to think about it, an unfavorable sequence of stones cannot be saved. The decisive factor is five narrow lanes and an odd number. We consider a playing field with \(10\) columns and \(20\) rows without a clipboard (hold): Each new stone is drawn independently and equally distributed from the seven tetrominoes . Each shape consists of four connected squares. The entire episode may be known in advance.
The fact that a game almost certainly ends does not mean that every conceivable sequence of stones ends. A pure sequence of O-stones can be played endlessly: five squares placed next to each other delete two rows and empty the board. In 7-Bag, each randomly mixed bag contains all seven shapes exactly once; our chance argument does not apply to this. In How to Lose at Tetris, Heidi Burgiel shows a sequence that even perfect play doesn't help: \(S,Z,S,Z,\ldots\) .
S and Z are mirror images, not different rotations of the same stone. The accompanying explanation describes the geometric idea that we present here as a proof by contradiction. The obvious claim “at some point only Z stones appear and no row disappears with them” is wrong. Just five identical, vertically placed S-stones provide the following arrangement on an initially empty board. The same letters indicate the same stone; Dots mark free spaces:
a.b.c.d.e.
aabbccddee
.a.b.c.d.eThe middle row is complete and deleted. If you mirror the arrangement, you get the same counterexample for Z-stones. The problem is not deleting individual rows, but the question of whether you can control all the imbalances that arise across an unlimited number of stones. Let's number the columns from \(1\) to \(10\) . By \(b_i\) we denote the total number of all new cells placed in column \(i\) , including those later deleted.
\(h_i\) counts horizontal S or Z stones with middle column \(i\) , \(v_i\) vertical stones in columns \(i\) and \(i+1\) . The counters record the course of the game, not just the currently visible stack. A horizontal stone distributes its cells in the ratio \(1,2,1\) over three columns, a vertical stone in the ratio \(2,2\) in two columns. Both mirror images have the same column balance. If you add up these contributions across all placements, you get for each column:
$$b_i=2v_{i-1}+2v_i+h_{i-1}+2h_i+h_{i+1}.$$
Outside the board we set all counters to \(0\) ; in particular \(v_0=v_{10}=h_0=h_1=h_{10}=h_{11}=0\) . This edge convention prevents the balance from counting stones that would extend beyond the edge of the playing field. Such placements are not allowed in our model. If the S/Z sequence under consideration begins on an already filled board, let \(a_i\) be the initially occupied cells and \(c_i\) the currently occupied cells in column \(i\) . After \(R\) deleted rows is:
$$c_i=a_i+b_i-R,$$
$$b_i-b_j=(c_i-c_j)-(a_i-a_j).$$
As long as the game is running, all \(a_i,c_i\) are between \(0\) and \(20\) . Therefore \(|b_i-b_j|\leq40\) applies. If the board is initially empty \(20\) is enough. Each deleted row takes away exactly one cell from each column. This is why the number of deleted rows is reduced when comparing two columns, even if the counters of the set cells continue to grow. You count occupied cells instead of stack heights; The two sizes differ when it comes to holes. If we write this limit generally as \(D\) , it results on the left or right edge:
$$2v_2+h_2+h_3=b_2-b_1\leq D,$$
$$2v_8+h_8+h_9=b_9-b_{10}\leq D.$$
For the inner column pairs we add two restricted column differences: \(b_4-b_3\) and \(b_2-b_1\) on the left, \(b_7-b_8\) and \(b_9-b_{10}\) on the right. After inserting the column balance, the counters in between cancel each other out exactly. This leaves only non-negative contributions on each left-hand side, which are jointly bounded by \(2D\) . This also gives us an upper limit for the respective placement numbers for the two inner pairs:
$$\begin{aligned}2v_4+h_4+h_5&=(b_4-b_3)+(b_2-b_1)\\&\leq2D,\\2v_6+h_6+h_7&=(b_7-b_8)+(b_9-b_{10})\\&\leq2D.\end{aligned}$$
If we add the four inequalities, each horizontal stone is counted exactly once. The four types of cross-track vertical placements also have positive coefficients. The horizontal stones take up three columns and therefore do not fit completely into one of the two-column paths considered later. For the vertical stones, we distinguish the tracks according to the left column. Since all counters are non-negative, none of these contributions can be canceled by another:
$$\sum_{i=2}^{9}h_i+2(v_2+v_4+v_6+v_8)\leq6D.$$
Horizontal placements and vertical placements with a straight left column can therefore occur a maximum of \(120\) times on an empty board or \(240\) times on any pre-filled board. This is an upper bound, not a claim that it can actually be achieved. In any case, these evasive moves cannot be used infinitely often. Now let's assume that you could play the alternating sequence indefinitely.
After the last of the finite number of evasive moves, only vertical placements in five fixed lanes would be possible: \((1,2),(3,4),(5,6),(7,8),(9,10)\) . Identical shapes fit into each other in a step-like pattern: the bottom meets the matching top of the previous stone. Changing from S to Z or vice versa leaves a gap one cell wide and two cells deep. The detail shows a Z-stone over an S-stone; both “o” mark the gap:
.Z
ZZ
Zo
So
SS
.SAnother vertical stone in the same lane cannot fill this gap: its second column would collide with the neighboring occupied cells. Deleting rows doesn't help either as long as the gap is open, because both rows are not complete. A repair requires one of the already excluded cross-track or horizontal placements. After the last evasive move, such gaps are permanent. An infinite number of new, disjoint gaps cannot arise on a finite board.
As a result, there can only be a finite number of changes in form on the tracks. An initially empty lane doesn't change this: your first S or Z stone already leaves a single-column gap on the floor that cannot be closed by further pure lane play. In a supposedly infinite game, each lane must also hold an infinite number of stones. If a lane were to remain untouched at some point, only a finite number of cells could be accommodated in the remaining lanes due to the limited column balances.
We therefore reach a point in time at which all five lanes are permanently occupied and firmly assigned to a form. From this point on we count \(t\) more stones and with \(n_j(t)\) the number in track \(j\) . Each of these stones contributes two cells to each column of its track. Because row deletions treat all columns equally, the differences \(n_j(t)-n_k(t)\) remain limited. In the long term, all five lanes must each hold a fifth of the stones:
$$\frac{n_j(t)}{t}\longrightarrow\frac15.$$
If \(k\) of the five lanes are reserved for S-stones, their share should be around \(k/5\) . However, in the alternating sequence it is \(1/2\) : in each prefix the numbers of the S and Z stones differ by at most \(1\) . The division of the lanes would have to fulfill both parts at the same time. However, a share of half cannot be represented as the sum of whole fifths. This is where the frequency of alternating forms and the odd number of equally used tracks come together. That would be mandatory:
$$\frac{k}{5}=\frac12\quad\Longrightarrow\quad k=\frac52.$$
The number of lanes is an integer. This contradiction ends the assumption of an infinite game. The argument also shows why the board width is part of it: With four trains, a division into two S and two Z trains would not be ruled out by this contradiction. For the random proof we need a finite block length \(N\) that leads to loss regardless of the current board. “Every strategy loses at some point” alone is not enough.
The uniformity follows from the finite number of possible states. After each completed placement, there are at most \(2^{200}\) occupancy patterns. There is also the information as to whether S or Z appears next, i.e. a maximum of \(2^{201}\) states. If you were to survive \(N=2^{201}\) stones, the \(N+1\) states before and after these moves would have a repetition. Because of the identical board and the same next shape, the legal moves in between would be repeatable any number of times: the infinite game just excluded.
The limit also counts unattainable board patterns and does not predict practical game length. Burgiel's finer geometric counting gives smaller bounds; here a finite \(N\) suffices. It applies to all permissible initial assignments and both phases of the S/Z sequence. Let us decompose the infinite random sequence into disjoint blocks of length \(N\) . Each block is exactly the specified sequence \(S,Z,S,Z,\ldots\) with probability \(q=7^{-N}>0\) . The blocks are independent. It is therefore probable that none of the first \(m\) blocks contain this sequence:
$$\mathbb P(\text{kein Treffer in }m\text{ Blöcken})=(1-q)^m\longrightarrow0.$$
So with probability \(1\) a deadly block will appear at some point, unless the game ends before then. This also applies with any amount of time to think about it and complete knowledge of the entire sequence of stones. This does not result in a short practical playing time. An unfocused, bad start alone doesn't mean defeat; surprising comebacks are certainly possible. So – don’t be discouraged and enjoy the timeless game.