# Probability of a successful DT Cannon

In multiplayer Tetris, there are a number of opening setups that let you send powerful attacks very quickly. One popular opening is the DT Cannon, which allows the player to very quickly send a T-Spin Double followed by a T-Spin Triple.

The standard DT Cannon setup looks like this:

Here's a video of a guy completing a DT Cannon

# The Puzzle

In most modern Tetris games, blocks are generated using the "7-bag" random generator. For our purposes, this means that the first 7 pieces in a Tetris game will always contain one of each type.

In the DT Cannon setup, these 7 blocks are always placed first:

You might notice that certain pieces must be placed before others: The J piece must be placed before the Z piece, the L before the S piece, and the S before the T piece. Because of this, the setup is not always possible depending on the permutation of the first 7 pieces. For example, if your first two pieces are S and Z, you are unable to place either piece in it's proper position.

However, if you have a S piece followed by a L piece, you're still able to continue the setup! Modern Tetris games allow you to hold a single piece, allowing you to swap your current piece with the piece in the hold. In this case, you can place your S piece in your hold, place the L piece, and swap out the next piece with the S piece to place down. Note that if your first two pieces are S and Z, it's still impossible to complete the setup, even if you hold the S piece.

After the first 7 pieces are placed successfully, it's always possible to complete the rest of the DT Cannon setup.

What percentage of all the possible 7-piece permutations are able to complete the DT Cannon setup?

As this is , your answer must contain some form of logical explanation or reasoning. However, you may use computers to verify your solution.

Tetris pieces color guide: I - light blue, J - blue, L - orange, O - yellow, S - green, Z - red, T - purple

In the following, I've made quite a few assumptions, but to me they all seem somewhat sensible. Looks like there's a

$$96.\overline{6} \%$$

chance we can use the first 7 blocks to build a shape with the outline given by OP, or the mirror image of it, which works just as well for the purposes of building a DT gun.

To see how this works, let's first improve OP's design by swapping some pieces around:

This stack has the exact same shape and the same pieces as before, but there are a couple of important advantages:

1. as long as the I piece hasn't fallen, the L and J pieces can be tucked.
2. either of the pieces under the T can support the T, the other can be tucked. (There's a possible complication of tucking the J under the T after the L has fallen, but seeing all the insane wiggling going on in the linked video, it'll probably work just fine.)

Also, it's possible to build the DT cannon in mirror image, so there's a variant where the T lies on top of the LS pair instead of the JZ:

With these setups in mind, we can construct the following strategy:

1. Hold the I until the L and J have both dropped
2. Hold the T until there's a piece on which to land it. Any of LJSZ will do.
3. When the first piece out of LJSZ drops, choose the variant where that piece goes under the T, so that the hold time of point 2 is minimised. (We may have to choose sides earlier, if the O piece drops. To do that, we'll look at the "next 3 pieces" preview to determine which of the LJSZ will be the first to drop. In the linked video, the game has a 5-piece preview, so that should work for OP.)

This strategy will work every time there's at least one of LJSZ before both I and T have fallen.

Even in that single worst case, we will still succeed as long as the first piece out of LJSZ is either L or J.

For the purpose of keeping the demonstration simple, let's assume it's an L:

1. Hold the T
2. Look at the preview to determine the L is going to be the key piece to drop first
3. Choose the second variation, because in it the T is on top of the L, and place the I accordingly
4. Place the L
5. Release the T, put it on top of the L, which leaves room for tucking the S (see picture below) even if the O has dropped
6. Hold the Z until the J drops

Sadly, there's no way to place the I piece flat on the bottom and still build the exact shape given by OP, so the other cases seem to be forced failures.

To calculate the probability of success, we can ignore the O piece altogether, it can drop anytime, and it won't affect the result in any way.

This means we can enumerate the cases in which we fail: it happens only when the first three pieces (ignoring any Os) are exactly "ITS", "ITZ", "TIS" or "TIZ"

Given that there are $$6 \times 5 \times 4 = 120$$ ways to choose the three first blocks out of the six options (again ignoring any Os), the final probability of success is

116 cases out of 120, or $$96.\overline{6} \%$$

• My original intention was for people to solve specifically for the setup in the question (the success rate is about 50% by the way) Nonetheless the wording of the question was ambiguous and your solution does work, with a significantly higher success rate! I'll accept your answer :D Mar 7, 2021 at 4:18
• Good work! Also, @thesilican, the video you link also show the alternative setup, so I was thinking that we can adjust the positioning too. Mar 7, 2021 at 5:36