# Graham's Number versus another large number

I recently read this article about the most damage you can do in a single turn in Magic the Gathering. According to the current version of the deck, that damage is about

a) $2 \rightarrow 17 \rightarrow 417$

This got me thinking about whether this number is larger than Graham's Number (sorry, not allowed to put more than 2 links in a post). I've tried to work it out some, but I'm having trouble understanding the chained arrow notation, let alone comparing it to Graham's number.

If it helps, I noticed that Graham's number is between

b) $3 \rightarrow 3 \rightarrow 64 \rightarrow 2$

and

c) $3 \rightarrow 3 \rightarrow 65 \rightarrow 2$

But I don't know how that compares to

a) $2 \rightarrow 17 \rightarrow 417$

Thank you for your time, and any answers.

Edit: corrected typo in damage estimate "copied" from MTG site.

• Well, you can any arbitrary amount of damage in a single turn in magic the gathering using infinite mana combos and fireball, for example. – JacksonFitzsimmons Jul 1 '15 at 7:27
• The article you linked seems to end with the conclusion that the damage is $2 \rightarrow 17 \rightarrow 417$. Is the $4$ a typo, or something outside of the article? – Sloan Jul 1 '15 at 13:43
• @Sloan Good eye, that was a typo. Corrected it in original comment. – Saurfon Jul 1 '15 at 19:09
• @JacksonFitzsimmons The problem stated on that site is that it is a finite amount of damage in the first turn (with perfect luck). – Saurfon Jul 1 '15 at 19:12
• Sorry, I didn't read the article. Silly me. – JacksonFitzsimmons Jul 1 '15 at 20:50

## 2 Answers

Intuitively, Grahams number should be much larger. Let $M=2 \rightarrow 17 \rightarrow 417$; in Knuth's up-arrow notation we have $M=2 \uparrow^{417} 17$. Let $G$ represent Graham's number. That is,
$\left. \begin{matrix} G &=&3\underbrace{\uparrow \uparrow \cdots\cdots\cdots\cdots\cdots \uparrow}3 \\ & &3\underbrace{\uparrow \uparrow \cdots\cdots\cdots\cdots \uparrow}3 \\ & &\underbrace{\qquad\;\; \vdots \qquad\;\;} \\ & &3\underbrace{\uparrow \uparrow \cdots\cdot\cdot \uparrow}3 \\ & &3\uparrow \uparrow \uparrow \uparrow3 \end{matrix} \right \} \text{64 layers}$
This representation is both unwieldy and, I fear, not enlightening. Informally, we can think of $G$ as $3 \uparrow^{\text{massive}}3$. The heuristic argument in comparing two numbers in Knuth notation is that the one with more arrows is larger. This is not always the case (as explored here and briefly here). For us, the number of arrows in $G$ ridiculously overwhelms the number of arrows in $M$ (the former I could never feasibly write out in decimal form and the latter is $3$-digits long). Hence, I think it's fair to say $M < G$.

• I also had the same thought about the numbers, but I wasn't sure how the 17 affected it, compared to the 3. Though, I suppose the ridiculously larger number of up arrows in G overwhelms the 14 larger number after the arrows. – Saurfon Jul 2 '15 at 16:53
• That's essentially what I've argued here! Very interesting article :) – Sloan Jul 2 '15 at 16:54
• Since $2 < 3$ and $17 < 3^3$, can we say that $3\uparrow^{2\times417}3$ is larger than M ? – BusyAnt Jul 27 '15 at 9:54
• I find it funny when we deal with numbers and we say something like $$3\uparrow^{massive}3$$Where we just say that spot has a very big number. – Simply Beautiful Art Jan 7 '16 at 23:42
• Already $G_2=3\uparrow^{3\uparrow \uparrow\uparrow \uparrow 3}\ 3$ is much larger than $2\uparrow^{417} 27$ – Peter Feb 16 '16 at 16:54

In case you wonder about an actual proof: $$2 \uparrow^{417} 17 < 3 \uparrow^{417} 17 < 3 \uparrow^{417} 27 < 3 \uparrow^{417} 3 \uparrow^{417} 3 = 3 \uparrow^{418} 3 < 3 \uparrow^{3 \uparrow\uparrow\uparrow\uparrow 3} \; \; 3$$

The fact that Knuth's arrow notation is monotonous in all arguments is proven in Saibian, Sbiis. A theorem for Knuth arrows.

• $3 \uparrow^{417} 27 < 3 \uparrow^{417} 3 \uparrow^{417} 3 = 3 \uparrow^{418} 3$ ?? Do you mean $3 \uparrow^{417} 27 < 3 \uparrow^{417} 3 \uparrow 3 = 3 \uparrow^{418} 3$ ? – BusyAnt Jul 27 '15 at 9:58
• @BusyAnt No, $3 \uparrow^{417} 3 \uparrow^{417} 3 = 3 \uparrow^{418} 3$. Please look to the definition again very carefully. – wythagoras Jul 27 '15 at 10:00
• Sorry, I was wrong, I misread the definition – BusyAnt Jul 27 '15 at 10:06