# After how many iterations can I shortcut the Delina, Wild Mage / Pixie Guide combo?

I'm playing at competitive rules enforcement with Delina, Wild Mage and Pixie Guide. If I attack with Delina, Wild Mage targeting Pixie Guide with the triggered ability, I'll continue creating copies of Pixie Guide for as long as I successfully roll `15—20`.

For my first roll, I'll roll two dice and ignore the lowest roll from the Grant an Advantage ability. If I succeed on that roll, I'll have an extra copy of Pixie Guide and roll again, but with three dice and ignoring the two lowest. As I continue to succeed, it will become more and more likely that I succeed:

Iterations Dice Rolled Approx. Prob. Success Approx. Aggregate P(Success)
1 2 51% 51%
2 3 66% 33%
3 4 76% 25%
4 5 83% 21%
5 6 88% 19%
6 7 92% 17%
7 8 96% 16%

As I continue to succeed, the probability of success approaches 100%, but is never certain. At any point during the combo, I can choose to stop as Delina, Wild Mage was errata'd to say "you may roll again".

Normally in Magic, once you demonstrate a loop, you can normally shortcut it by saying that you'd repeat the iteration until you reach a desired result. However, at each iteration, the probability of success is not guaranteed. Since the loop is non-deterministic, it seems like I shouldn't be able to shortcut it - I'm not able to definitively say where the game state will be after each iteration.

However, Toby Elliot (a level 5 judge who writes the Magic Tournament Rules and Infraction Procedure Guide for judges) spoke about the combo on Twitter saying:

"Once you hit a certain point, your opponent probably just concedes, so the question is mostly moot.

Beyond that, once you hit a point where the odds are basically 1, announce how many you’re making and stop. But your opponent already conceded."

It seems like at after a certain number of iterations, I am able to shortcut this combo, as long as the odds are "basically 1".

What is the number of iterations in which I can stop rolling handfuls of d20s and start shortcutting this combo to say something like "I make a number of Pixie Guides equal to twice your life total"?

• I would guess since it can fail and stop at any time you can't shortcut it at all. Commented Jul 20, 2021 at 0:44
• @JoeW Pixie Guide's ability doesn't add dice together, but rather makes you ignore the lowest results. There's a small chance during each iteration that you'll fail to roll 15 or above on every die you roll. Commented Jul 20, 2021 at 0:48
• Corollary question: At what point does executing this loop become Slow Play and/or Stalling? Commented Jul 20, 2021 at 9:12
• @DavidZ The first iteration is the first time I resolve Delina's ability - I don't have any copies of Pixie Guide yet beyond the original. The first iteration defines the loop, the second iteration repeats that loop. Commented Jul 20, 2021 at 16:16
• @nick012000 Pokemon judge, not MTG, but there's no reason it would be stalling. Each action increases the chance of the player winning; and the other player is free to concede at any point. If it's effectively 100%, then the other player should concede, unless they think they're advantaged by not doing so.
– Joe
Commented Jul 20, 2021 at 21:39

The statement in that tweet is not backed up by the rules. Rule 722.2a describes how taking shortcuts works:

At any point in the game, the player with priority may suggest a shortcut by describing a sequence of game choices, for all players, that may be legally taken based on the current game state and the predictable results of the sequence of choices. This sequence may be a non-repetitive series of choices, a loop that repeats a specified number of times, multiple loops, or nested loops, and may even cross multiple turns. It can’t include conditional actions, where the outcome of a game event determines the next action a player takes. The ending point of this sequence must be a place where a player has priority, though it need not be the player proposing the shortcut.

in addition, section 4.4 of the Magic Tournament Rules includes a paragraph that says

Non-deterministic loops (loops that rely on decision trees, probability, or mathematical convergence) may not be shortcut. A player attempting to execute a nondeterministic loop must stop if at any point during the process a previous game state (or one identical in all relevant ways) is reached again. This happens most often in loops that involve shuffling a library.

The shortcut described only works on the condition that you get the 15-20 effect every time your roll, so it is not allowed by those rules as written. I think it's worth noting that in the current Magic Tournament Rules, none of the predefined tournament shortcuts have any conditional actions.

It's not really clear how seriously that tweet is supposed to be taken. A later tweet by the same person in the same thread, in response to one asking where the cutoff is for rounding up to 1, says

It’s the point after which your opponent has conceded, so we never actually have to get into the math.

I think you could easily read this as saying that you can never actually take the described shortcut, because it's only possible after the game is already over.

Mathematically, there is no independently defined cutoff where one number is "basically 1" and another isn't. Of course, it is possible to define a cutoff such as "whenever the probability of failing to get the desired number of consecutive successes is <1%", but until the Magic rules provide such a cutoff, there's no consistent way to resolve this.

• I forget what the combo was, but many years back there was a combo deck in legacy that had similar issues. People would demonstrate a loop that theoretically allowed them to stack their deck to an exact sequence and then stack their deck in the desired way. The problem was that the chance of this exact sequence occurring wasn't 1. As I remember people were no longer allowed to stack their deck as desired without going through the motions, and going through the motions was prohibitively slow. Commented Jul 21, 2021 at 8:56
• You might be thinking of "Four Horsemen", a combo in which players would try to mill a certain set of cards into their graveyard in a certain order, and if they failed, they would mill for an Emrakul, the Aeons Torn to start over. The problem with that combo as opposed to this one is that the probability of success never increased, so they never got any closer to winning. After each attempt, they either won, or were no better off than when they started. Commented Jul 21, 2021 at 14:37
• If you can please see my answer and make a comment. I would appreciate a seasoned rules experts view on my answer. Commented Sep 5, 2021 at 9:50

The probability it stops when you have N pixies .7^N. As .7 is very close to 1/sqrt(2), this means the probability of stopping is very close to 1 in 2^(N/2).

The probability of it ever stopping after pixie N is thus basically equal to the sum of i from N+1 to infinity of 1/2^(i/2). As 1/2^((i+1)/2) is less than 1/2^(i/2), we can bound this above by the sum from (N+1)/2 to infinity of 2 * 1/2^i, which is simply 4 * 1/2^((N+1)/2), which is less than 4 / 2^(N/2).

So after 20 rolls, the odds you will ever fail are less than 0.4%.

This is depressingly slow. I suspect most players won't concede if their chance of surviving is over 1%. It is merely an upper bound; the actual value is going to be a touch lower, but not that much (under 25% difference, if my napkin math is correct).

The point where your chances of ever failing are under 1 in a million is after 40 pixies. The point where your chances of ever failing are under 1 in a trillion are after 80 pixies. The point where your chances of ever failing are under 1 in a googol is after 667 pixies.

Your opponent has probably conceded before you have 667 pixies. But you'd probably stop making pixies before you hit 667 as well.

# TL;DR

The point, in number of Pixies, where you would reasonably be able to state "I am almost certainly able to make as many Pixies as I want" is almost always going to be at a point after you have enough Pixies to win.

There is no rule in MtG that translates "you are almost certainly going to be able to do X" into "you can do X". However, in this case it is pretty much moot, because the rate at which certainty grows isn't really fast enough to matter much.

It is reasonable for the rules to allow your opponent to ask you to roll 13d20 and ensure you get a 15+ if you want another pixie, as that has a 1% chance of failure.

At 20 pixies, getting at 15+ roll has a 0.08% chance of failure, and the accumulated chance of a failure at any point in the future is more like 0.25%; a 1 in 400 chance.

MtG doesn't say "this is almost certain, you win" at any point, and even if it does it would probably be well after 20 pixies.

It is, however, possible that if the defending player has Graham's number life and insisted that you continue to roll d20s to spawn more pixies after you have 100 of them. If this ever happens, I suspect the judge's decision will be scrutinized heavily.

"Unsporting conduct -- Stalling" may be appropriate here. The player with the infinite pixie combination is almost certainly going to win, and the decision not to concede and require the (near infinite) dice to be rolled (with a near zero chance of failure) is almost entirely aimed at running the round time out. This is going to be a strange case, because the player doing the actions is not the one possibly stalling. "Stalling" is about intent to use the time constraint, not actions, and the intent of requiring all the dice to be rolled is "I am going to lose, but I tie if I make all the dice be rolled due to time constraints".

# A fair way to handle it

It isn't hard to calculate the cumulative chance of failure after having X pixies at any point from there to infinity. Pre-calculating that would provide a "fair" way to handle the Graham's (or other insane life totals; 100 and Graham's number chance of failing is not practically different) cases.

If F_i is the chance of failure when rolling id20 for a 15+, then an upper bound on the chance of ever getting a failure after X pixies is sum i from X to infinity of F_i.

(This is actually the expected total number of failures, which is higher than the chance of ever failing).

F_i is .7^i, the sum is .7^(X) + .7^(X+1) + ..., which is .7^X * (1 + .7^2 + .7^3 + ...) which is .7^X * (1/1-.7), which is 3.33 * .7^X.

Which in turn is less than rolling Xd20 4 times and passing all 4 tries.

The actual chance of any failure is: .7^X + (1-.7^X).7^(X+1) + (1-(1-.7^X).7^(X+1)) * .7^(X+2) + ...

When .7^X is almost 0, this is very close to the above calculation (it is off by a factor of less than (1-.7^X)). 3.333/4 < (1-.7^X) when X>=5; so this approximation is fair (gives an advantage to your opponent) if you have at least 5 pixies already.

So at any point in this combo, when you have X (5 or more) pixies, asking "I want to continue until I have a near infinite number of pixies. The chance of failure is less than the chance of rolling Xd20 4 times and failing on any of the 4 rolls. How about I just do that instead?"

That gives the defender a greater chance of you failing than doing it manually, and it takes far less time. You, personally, choose to manually roll up to X pixies, where X is large enough that you feel comfortable with the advantage you are giving your opponent by not rolling forever.

(4 rolls seems unfair to say "I get to win an infinite number of rolls", but it works because each roll you win would make the chance of a later failure drop further; the improvement is so rapid that the 4 rolls without improvement is a fair handicap to match up against infinite rolls with improvement.)

• Are you saying that at Competitive REL, you can or cannot use this shortcut? If you can, after how many iterations are you saying that the shortcut can be taken? Commented Jul 20, 2021 at 19:17
• @murgatroid99 I am stating that the point where the shortcut is almost certainly going to work indefinitely is past the point where you have almost certainly already won. So the point is moot; keep rolling until you have enough fairies to win.
– Yakk
Commented Jul 20, 2021 at 19:25
• Your answer seems to be assuming that the number of pixies needed to win is about 20. Does your answer change if it isn't? What if the opponent has gained 100 life? 1000? Also, I think there's an error in the last expression in your second paragraph. It looks like you're inverting the whole exponential expression, as compared to the previous expression. Or maybe the notation is just confusing. Commented Jul 20, 2021 at 19:41
• Your proposed solution of rolling some finite number of dice to simulate all future rolls is probably not legal. The tournament rules (section 5.2) say "The result of a match or game may not be randomly or arbitrarily determined through any means other than the normal progress of the game in play. Examples include (but are not limited to) rolling a die..." Since you are suggesting rolling those dice in lieu of taking the actual game actions, doing so may be in violation of the tournament rules Commented Jul 21, 2021 at 20:29
• A critical difference between this combo and four horsemen is that in this combo, every roll either ends the combo in failure, or makes meaningful progress towards ending the combo in success. On the other hand, in the four horsemen combo, every iteration either wins the game immediately, or resets to the original state, no closer to winning. That reset disallows the combo under the loop rules I quoted in my own answer. There is also a big difference between calling a combo stalling because it makes no progress, and saying that the opponent is stalling for not conceding to the combo. Commented Jul 22, 2021 at 15:21

You might try the infinite series approach and just build a table for the expected number of Pixie Guide Tokens, then map that to a roll of permillenial dice (6 d10s read as digits of a number from 000000-999999, with all zeroes being read as 1,000,000). For fairness, you'll end up reducing your token in count in some cases so as to never have a case where your odds for a given "token count or more" are increased:

(The table for the odds used to calculate the roll follows the first table)

Perhaps the judge will be familiar enough with infinite series to allow this as the shortcut.

Roll Tokens
1-490,000 1
490,001-664,930 2
664,931-745,380 3
745,381-788,173 4
788,174-813,094 5
813,095-828,486 6
828,487-838,373 7
838,374-844,895 8
844,896-849,276 9
849,277-852,256 10
852,257-854,300 11
854,301-855,711 12
855,712-856,689 13
856,690-857,369 14
857,370-857,842 15
857,843-858,172 16
858,173-858,402 17
858,403-858,563 18
858,564-858,675 19
858,676-858,753 20
858,754-858,808 21
858,809-858,846 22
858,847-858,873 23
858,874-858,891 24
858,892-858,904 25
858,905-858,913 26
858,914-858,919 27
858,920-858,923 28
858,924-858,926 29
858,927-858,928 30
858,929-858,929 31
858,930-858,930 32
858,931-858,931 33
858,932-858,932 34
858,933-858,933 35
858,934-858,934 36
858,935-858,935 37
858,936-858,936 38
858,937-858,937 39
858,938-858,938 40
858,939-858,939 41
858,940-858,940 42
858,941-858,941 43
858,942-858,942 44
858,943-858,943 45
858,944-858,944 46
858,945-858,945 47
858,946-858,946 48
858,947-1,000,000 Infinity

Calculation table:

N Dice Success Aggregate Success Odds for N Odds of "N or More"
1 2 0.51 0.51 49.0000% 100%
2 3 0.657 0.33507 17.4930% 51.00000000%
3 4 0.7599 0.254619693 8.0450% 33.50700000%
4 5 0.83193 0.211825761 4.2794% 25.46196930%
5 6 0.882351 0.186904672 2.4921% 21.18257612%
6 7 0.9176457 0.171512269 1.5392% 18.69046722%
7 8 0.94235199 0.161624928 0.9887% 17.15122688%
8 9 0.959646393 0.155102779 0.6522% 16.16249278%
9 10 0.971752475 0.150721509 0.4381% 15.51027790%
10 11 0.980226733 0.147741253 0.2980% 15.07215094%
11 12 0.986158713 0.145696324 0.2045% 14.77412526%
12 13 0.990311099 0.144284686 0.1412% 14.56963235%
13 14 0.993217769 0.143306114 0.0979% 14.42846863%
14 15 0.995252438 0.14262576 0.0680% 14.33061142%
15 16 0.996676707 0.142151772 0.0474% 14.26257597%
16 17 0.997673695 0.141821084 0.0331% 14.21517725%
17 18 0.998371586 0.141590141 0.0231% 14.18210841%
18 19 0.99886011 0.141428744 0.0161% 14.15901407%
19 20 0.999202077 0.141315894 0.0113% 14.14287436%
20 21 0.999441454 0.141236963 0.0079% 14.13158944%
21 22 0.999609018 0.141181742 0.0055% 14.12369629%
22 23 0.999726313 0.141143102 0.0039% 14.11817418%
23 24 0.999808419 0.141116062 0.0027% 14.11431021%
24 25 0.999865893 0.141097137 0.0019% 14.11160618%
25 26 0.999906125 0.141083892 0.0013% 14.10971371%
26 27 0.999934288 0.141074621 0.0009% 14.10838917%
27 28 0.999954001 0.141068131 0.0006% 14.10746207%
28 29 0.999967801 0.141063589 0.0005% 14.10681315%
29 30 0.999977461 0.14106041 0.0003% 14.10635892%
30 31 0.999984222 0.141058184 0.0002% 14.10604097%
31 32 0.999988956 0.141056626 0.0002% 14.10581842%
32 33 0.999992269 0.141055536 0.0001% 14.10566263%
33 34 0.999994588 0.141054772 0.0001% 14.10555358%
34 35 0.999996212 0.141054238 0.0001% 14.10547724%
35 36 0.999997348 0.141053864 0.0000% 14.10542381%
36 37 0.999998144 0.141053602 0.0000% 14.10538640%
37 38 0.999998701 0.141053419 0.0000% 14.10536022%
38 39 0.99999909 0.141053291 0.0000% 14.10534189%
39 40 0.999999363 0.141053201 0.0000% 14.10532906%
40 41 0.999999554 0.141053138 0.0000% 14.10532008%
41 42 0.999999688 0.141053094 0.0000% 14.10531380%
42 43 0.999999782 0.141053063 0.0000% 14.10530940%
43 44 0.999999847 0.141053042 0.0000% 14.10530632%
44 45 0.999999893 0.141053027 0.0000% 14.10530416%
45 46 0.999999925 0.141053016 0.0000% 14.10530265%
46 47 0.999999948 0.141053009 0.0000% 14.10530159%
47 48 0.999999963 0.141053003 0.0000% 14.10530085%
48 49 0.999999974 0.141053 0.0000% 14.10530034%
49 50 0.999999982 0.141052997 0.0000% 14.10529997%
50 51 0.999999987 0.141052995 0.0000% 14.10529972%
51 52 0.999999991 0.141052994 0.0000% 14.10529954%
52 53 0.999999994 0.141052993 0.0000% 14.10529942%
53 54 0.999999996 0.141052993 0.0000% 14.10529933%
54 55 0.999999997 0.141052992 0.0000% 14.10529927%
55 56 0.999999998 0.141052992 0.0000% 14.10529923%
56 57 0.999999999 0.141052992 0.0000% 14.10529920%
57 58 0.999999999 0.141052992 0.0000% 14.10529918%
58 59 0.999999999 0.141052992 0.0000% 14.10529916%
59 60 0.999999999 0.141052991 0.0000% 14.10529915%
60 61 1 0.141052991 0.0000% 14.10529914%

(At this point, any result N>60 has the same results as N=60.)

• Listing probabilities as 0 and 1 when they are not actually 0 and 1 is misleading, because a probability of 1 has a very different meaning than a probability of "very nearly 1" when talking about deterministic vs nondeterministic processes. I suggest instead "10^-n" and "1-10^-n" to show how close they actually are to 0 or 1. Commented Jul 22, 2021 at 15:33
• Also, as I posted on the other answer, your proposed solution of rolling some finite number of dice to simulate all future rolls is probably not legal. The tournament rules (section 5.2) say "The result of a match or game may not be randomly or arbitrarily determined through any means other than the normal progress of the game in play. Examples include (but are not limited to) rolling a die..." Since you are suggesting rolling those dice in lieu of taking the actual game actions, doing so may be in violation of the tournament rules Commented Jul 22, 2021 at 15:34
• Well, I did say "familiar enough with infinite series", which should cover any confusion about the rounding. If the rounding trips them up, this approach is destined to fail anyhow. Commented Jul 22, 2021 at 17:06
• I'm not talking about judges. I am talking about what you are communicating to the reader of this answer. Commented Jul 22, 2021 at 17:09
• Judges would not allow this as a shortcut no matter how familiar they are with the mathematics involved. The rules for shortcuts simply don't allow for it. We've seen plenty of precedent in decks that use infinite random events, like Four Horsemen. Even though their outcome is virtually guaranteed eventually, judges do not allow the random events to be shortcut past and require them to be played out. Players are only granted shortcuts for deterministic sequences shortcuts, like infinite scry 2 to rearrange a deck arbitrarily. Read more here Commented Jul 22, 2021 at 17:41