An impossible sequence of Tetris pieces. [duplicate]

Is there a finite sequence of pieces of Tetris such that for every way of play you always lose?

A game of Tetris (on a 10×20 grid) consisting of alternating S and Z tetrominoes will necessarily end before 70000 tetrominoes are played.

http://euclid.trentu.ca/aejm/V4N1/Tsuruda.V4N1.pdf — Lemma 2.

Not anymore.

Older versions of Tetris had a uniform stateless randomizer. This would allow a repeating SZ-sequence that eventually causes a loss. This is possible yet unfathomably unlikely in practice because a competent player can survive this sequence for 150 Tetriminos on a standard 10x20-cell matrix, and the odds of an ideal randomizer producing this sequence are 1 in (7/2)¹⁵⁰ = 4×10⁸¹, roughly one in the number of atoms in the observable universe.

But since 2001, the randomizer rule in Tetris Worlds, Tetris DS, and other authentic Tetris games has been the "bag" rule, called "Random Generator" by Tetris developer Blue Planet Software. It deals a permutation of the seven distinct Tetriminos (5,040 possibilities), then reshuffles, then deals another permutation, etc. An SZ-sequence longer than length 4 is forbidden because a sequence of length 4 has to be sandwiched between two permutations of IJLOT.

It turns out that if you keep a pile of S, T, and Z in columns 1-4 of a 10x20-cell matrix, and J, L, and O in columns 7-10, you can enter a pattern that repeats every 140 Tetriminos and thus extend the game indefinitely. In 2007, Colour Thief released a proof of this, titled "Playing forever". The following tiling of a 10x56-cell rectangle should help you grasp the pattern used by the proof.

If the board has odd width (any length is fine) then yes, there is trivially such a sequence. Just assume all pieces are the $2 \times 2$ squares. So maybe you should insist that the board width is even?