How many KOs to achieve Tetris Maximus?

Tetris 99, Nintendo’s battle royale spin on the classic line-clearing phenomenon, has players chasing “Tetris Maximus” by outliving their fellow 98 competitors. As you clear lines, perform combos or get tetrises, you send garbage lines to competitors. Hopefully that garbage is enough to push them over the edge of the screen, scoring you a KO. Naturally, this begs the question:

What is the expected number of KOs for a Tetris 99 winner?

Extinction Events

We define the extinction event $X_{a\to b}$ as the period in which the number of remaining players drop from $a$ to $b$, with $b<a$. We denote $\underset{–}{X}$ as the set of all extinction events in a game:

$$\underset{–}{X}=X_{\left(1\right)},X_{\left(2\right)},...,X_{\left(k\right)}$$

with $X_{\left(k\right)}=X_{a\to 1}$ for some $a>1$.

We assume that all $b$ remaining players have an equal share in the $(a-b)$ KOs that happened during $X_{a\to b}$. ¹ In other words, every remaining player has an equal probability of scoring a KO that happened during this event.

If we denote a given player’s expected number of KOs during $X_{a\to b}$ as $E\left[X_{a\to b}\right]$, we have

$$E\left[X_{a\to b}\right]=\frac{a-b}{b}$$

Denoting the winner’s KOs as $E\left[W\right]$, we have:

$$E\left[W\right]=\sum _{i=1}^{k}E\left[X_{\left(i\right)}\right]=\sum _{X\in \underset{–}{X}}E\left[X_{a\to b}\right]$$

Some Models

It’s clear that our choice of $a$ and $b$ for every $X_{a\to b}\in \underset{–}{X}$ will have an effect on our predicted $W$. Let’s play around with a few different $\underset{–}{X}$ and see how it affects $E\left[W\right]$.

1. Taking $b=\lceil \frac{a}{2}\rceil$

Event $i$	a	b	$E\left[X_{\left(i\right)}\right]$	${\sum }_{j=1}^{i}E\left[X_{\left(j\right)}\right]$
$X_{\left(1\right)}$	99	50	$\frac{99-50}{50}=0.98$	0.98
$X_{\left(2\right)}$	50	25	$1$	1.98
$X_{\left(3\right)}$	25	13	$\frac{12}{13}$	2.90
$X_{\left(4\right)}$	13	7	$\frac{6}{7}$	3.76
$X_{\left(5\right)}$	7	4	$\frac{3}{4}$	4.51
$X_{\left(6\right)}$	4	2	$1$	5.51
$X_{\left(7\right)}$	2	1	$1$	6.51

In this model, $E\left[W\right]$ is 6.51. Initially, this result may seem a bit counterintuitive. By the end of $X_{\left(k\right)}$ 98 players have been eliminated, so it’s natural to expect the winner to have knocked out more than 6.51 of them. We correct our intuition by realizing that each of those 6.51 players eliminated a bunch of other players in turn, similar to how it takes only 7 wins to win a 128-player tournament such as Wimbledon.

2. 10 at a time up to Top 10, and then $b=\lceil \frac{a}{2}\rceil$

Event $i$	a	b	$E\left[X_{\left(i\right)}\right]$	${\sum }_{j=1}^{i}E\left[X_{\left(j\right)}\right]$
$X_{\left(1\right)}$	99	90	$\frac{9}{90}=0.1$	0.1
$X_{\left(2\right)}$	90	80	$\frac{10}{80}$	0.225
Some time passes…
$X_{\left(9\right)}$	20	10	1	2.82
$X_{\left(10\right)}$	10	5	1	3.82
$X_{\left(11\right)}$	5	3	$\frac{2}{3}$	4.48
$X_{\left(12\right)}$	3	2	$\frac{1}{2}$	4.98
$X_{\left(13\right)}$	2	1	1	5.98

Interestingly enough, $E\left[W\right]$ is close to our estimate in the first model.

3. 1 at a time

This is an interesting (and quite an extreme) case, implying that every player plays a part in every KO.

In this model, we have

$$E\left[W\right]=\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}+1=H_{98}=5.17$$

with $H_{98}$ being the 98th Harmonic number. The expected number of KOs when reaching the top 10 is $H_{98}-H_{11}$ = 2.14.

4. Whatever else

Clearly we can only speculate on how these extinction events play out in reality. Messing around with a few more handcrafted guesses, $E\left[W\right]$ tends to end up between 5 and 7.

Takeaways

• It’s expected to achieve “Tetris Maximus” with only $\approx$ 6 KOs - a result which is lower than I intuitively expected.
• It’s expected to reach the top 10 with only $\approx$ 3 KOs. You have no need to panic when the fear-inducing Top 10 music starts playing and you notice that you only have 3 KOs.

Fun fact: When I eventually got my first Tetris Maximus, I did it with 6 KOs - making me a perfectly average winner. The fact that my estimations played out (and that I finally won!) made being perfectly average, perfectly OK.