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So Wizards of the Coast recently released a Dungeons & Dragons-themed Magic: The Gathering set. In this set they have the following two creatures: Pixie Guide and Delina. With those two cards attacking, a funny thing happens:

  1. A 20-sided dice is rolled for each Pixie Guide on the field + 1.
  2. If at least one of those dice rolls 15 or greater, a copy of Pixie Guide is made, then the process is repeated starting from 1.

Question: What are the chances that this goes infinite?

Partial solution:

Given $i$ dice, the probability of low-rolling all rolls is $p^i$, where $p=14/20$. The probability of high-rolling at least one dice is thus $1-p^i$, and the probability of going infinite is thus

$$P=\prod_{i=2}^{\infty}(1-p^i)$$

Note that the product starts at $i=2$, because the very first time we roll two dice.

Does $P$ have a closed expression for $0 < p < 1$? A simulation suggests it converges to 0.14105299 for 100 terms.

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    $\begingroup$ Your $P = \phi(p) / (1-p) $ where $\phi(p)$ is called the Euler function. I doubt there is a known closed form except for things like $p=e^{-\pi}$ $\endgroup$
    – Henry
    Jul 21, 2021 at 0:17
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    $\begingroup$ Wow, the Wikipedia article on the Euler function that @Henry links to ends with "[This result needs concrete references, as I have been unable to verify it]". I haven't seen something like that in the main text of a Wikipedia article in many years. $\endgroup$
    – David
    Jul 21, 2021 at 0:23
  • $\begingroup$ I should note that Wizards of the Coast avoided this problem of mathematical convergence by changing the text of Delina to say "You may roll again" rather than "Roll again". And they did so even before the set's release date. $\endgroup$
    – Peter O.
    Jul 21, 2021 at 0:31
  • $\begingroup$ What if the player always wants to roll again? Is there a rule that forces a player to eventually stop doing something recursive? @PeterO. $\endgroup$ Jul 21, 2021 at 0:35
  • $\begingroup$ @Timkinsella: A question like that is better asked on Board & Card Games, since it has relatively little to do with math. $\endgroup$
    – Peter O.
    Jul 21, 2021 at 0:41

1 Answer 1

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I initially, and incorrectly posted the following plausible looking answer.

This will never go infinite. Although it can create impractically large numbers of attacking Pixie Guides. This is because the number of dice being rolled is always finite. So the chances of the process ending are always strictly greater than $0$ (and are equal to the chances that all of your rolls due to the Pixie Guide(s) plus the one from Delina's roll (again) all roll 14 or lower). That said, if you get past the first few iterations, it may take quite a while before you manage to roll poorly enough that you'll be forced to stop.

This is incorrect, for a less than obvious reason. Namely that at each step the chances of continuing increase, and their limit is $1$. Therefore the question of this continuing indefinitely is a question converge of the product $P$ given by the original poster.

It turns out the product converges to a non-zero value. I used a result from here: https://en.wikipedia.org/wiki/Infinite_product Specifically the part that refers to $a_n=1-q_n$, which motivated me to consider the converge of the related series

$$\sum_{i=1}^{\infty}q_n$$

which for our purposes is the geometric series

$$\sum_{i=1}^{\infty} \left( \frac{14}{20}\right)^i$$

which converges to $\frac{20}{6}$ by the basic theorem on geometric series. As a consequence of the sum being convergent and cited result, we get that the infinite product is convergent to a non-zero value. Since all the terms of the product are positive, we know the product itself must be non-negative and since it's non-zero we know that $p>0$. As for an expression for the exact value, I have no clue (and it might even be an intractable problem).

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