text:
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."
https://twitter.com/tobyelliott/status/1410662063228211206

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.

@JoeW that's what I assumed too, but the person that writes the guidelines for tournament Magic play said in their tweet that you could shortcut it after a certain point

Maybe once you have 15 dice and the lowest you can roll is 15.

@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.

Well I am confused as it appears you are rolling an additional dice each time around as the success chance goes up.

Yes, you roll an additional die for each Pixie Guide, and ignore all but the greatest result. As the number of dice increases, the expected value of the greatest result increases.

Corollary question: At what point does executing this loop become Slow Play and/or Stalling?

@toroid I believe you roll three dice the first time since you copy the Pixie Guide before rerolling, and so on thereafter, ie your second column of numbers is off by one.

@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.

@toroid Ah got it, I see what you mean.

I'd suggest that anyone running this kind of deck should have multiple d20s on hand so you can roll them all at once instead of one at a time.

Wow, so, the guy in the tweet is assuming that anyone would care about short-circuiting the die rolls after their opponent already conceded? With the assumption of already having the concession, that tweet becomes such a total non-answer that it makes me hope there's some way for players to downvote inane statements made by judges in public.

@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.

Every iteration is dependent on succeeding in the iterations before them, thus the probability of successfully completing the 2nd iteration (if you start from the beginning) would be .51*.66, the 3rd would be .51*.66*.76, etc. Your maths is only correct if you're already at the iteration directly before it. If you're at, say, the 4th iteration, there's only a 65% (.83*.88*.92*.96) chance that you'll get to the 8th iteration. So the probability that you'll actually keep going for particularly long (or indefinitely) is much lower than it appears with the probabilities in the question.

@JoeW If you role 1 die and need a 15 or higher, you have a 6 in 20 chance (30%). If you roll 2 dice and need one 15 or higher, you have a 6 in 20 chance for each, that works out to 30% for the first die, and 21% (30% of the 70% first die failure) for a total chance of 51%. A third die would be another 14.7% (30% of 49% chance the first two failed) for a total of 65.7% - there's always a chance every die will fail, but it gets less likely each cycle.

@Joe There are ways that an opponent could be advantaged by their opponent creating infinite tokens in magic, or at least not disadvantaged, even a few ways to win because of it ("quicken" spells like stronghold discipline or pact of the serpent). And if a mandatory effect like this does go on infinitely (if it had ever reached 100%) the rules do cover entering into a loop that neither player can break out of, it forces a draw.

@Andrew I fully understand that the chance of getting something goes up the more times you roll a die. What I didn't understand from the question is what exactly it meant by dice rolled. At first I was thinking it was total dice rolled and not number of times a die is rolled.

@nick012000 there is no slow play problems if you have demonstrated a loop you choose an arbitrary amount of iterations of the loop and the game continues. A thing that few MTG players realise is that you cannot go infinite in this game.

@NeilMeyer Except that in this case, you don't choose an arbitrary amount of iterations - if you want to execute the loop 100 times, you need to actually execute the loop 100 times.

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.

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.

If you can please see my answer and make a comment. I would appreciate a seasoned rules experts view on my answer.

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?

@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.

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.

In particular, what would be your answer if before you went off with your Pixie combo, I had gone off with my Spike Feeder / Heliod combo and gained Graham's number life?

"The point where your chances of ever failing are under 1 in a google is after 667 pixies." Did you mean a Googol?

@Tim yes I did.

@murgatroid99 Removed typo of / 1.

@PhilipKendall I added a possible way a judge could rule based on "stalling", but who am I kidding, this would be unprecedented.  :)

I wonder if a call of stalling would be appropriate, since as you point out, the person stalling is not actually doing anything other than passing priority in a timely manner - they're not taking too long to do anything they should be doing, and they can't do anything to make the loop play out any faster by accepting a shortcut. Such a call would mean that players could be penalized for failing to concede in situations they're unlikely to win, but there's no threshold for how unlikely that should be before mandating that a player concede.

@NuclearHoagie The wording of "stalling" implies it isn't your actions, it is your motivations.  And if the other party offers a fair way to shorten the near-infinite set of dice rolls and actions, refusing it would be the stalling.  But, as I said, this would be both unprecedented, and extremely unlikely to ever occur in a serious match.  You'd need a near infinite-life combo that doesn't use it to win, which then comes up against this near-infinite pixie combo before it wins, and the near-infinite life player not accepting their loss.

@Yakk The problem is, the combo player has no option to offer a valid shortcut since there is no defined end state. Suppose you have X pixies and roll Xd20 to see if you can achieve "infinite" pixies, and fail the roll - the board state is now undefined, since it's impossible to say how many pixies you actually have. Your fair method is indeed pretty fair, but it's not a valid shortcut because it's conditional on the dice roll. I don't see any way the combo player can propose a valid shortcut, and if no proposal is made, the non-combo player can't stall by rejecting it.

Basically, rejecting a valid shortcut would be stalling. Rejecting a series of actions intended to approximate actual gameplay in a manner explicitly disallowed by the rules wouldn't be. The suggested "shortcut" is invalid can't be legally proposed or accepted. Unless the combo player proposed a valid shortcut, there's literally nothing the non-combo player can do within the rules to make the combo play out any faster.

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

I also want to add my voice pushing back on the claim that it is "stalling" if the opponent does not concede. I have never seen an official ruling that requires someone to concede the game, and I do not believe it would ever be a fair requirement. The basic principle here is that if a player chooses to play a slow, non-deterministic combo, the responsibility for finishing the combo in a timely manner lies with that player, and it would be unfair to punish their opponent for making them play by the rules. This is why you lose instead of winning for running out your clock in MTGO.

Related discussion on why a different but similar combo is stalling: https://magicjudge.tumblr.com/post/152521707334/why-isnt-the-four-horsemen-combo-a-loop-like-why

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.

Stalling requires both an intent to run out the clock as well as actions to serve that intent. You can't penalize someone for stalling simply because they'd like to run out the clock if they haven't taken any action to do so. The non-combo player can take no action whatsoever to run out the clock - they can't accept an illegal shortcut or a "replacement" die roll to speed things up. Literally all they can do is pass priority, but you're suggesting they could be penalized if they don't accept an illegal shortcut or some other illegal approximation of gameplay.

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.

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

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.

I'm not talking about judges. I am talking about what you are communicating to the reader of this answer.

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

For official tournaments this may not work, but for kitchen table, this is excellent!

@L.ScottJohnson, I assume you used double precision arithmetic to compute these results?

@John The question is about Competitive REL. Kitchen table is out of scope for answers.

@John double precision, yes. The tables are created in MS Excel, which uses IEEE 754 double precision.


text:
How does Bolt Bend interact with Cycling triggers?
I was playing Lotus Field Combo in pioneer, and I cycled Vizier of Tumbling Sands:

Cycling {1}{U} ({1}{U}, Discard this card: Draw a card.)
When you cycle Vizier of Tumbling Sands, untap target permanent.

Targeting Lotus Field to untap.
My opponent cast Bolt Bend:

This spell costs 3 less to cast if you control a creature with power 4 or greater.
Change the target of target spell or ability with a single target.

Targeting the cycling ability on Vizier of Tumbling Sands. My question is if this is possible, since the ability did not untap the field, rather the trigger did. Cycling the card triggered the text "When you cycle Vizier of Tumbling Sands, untap target permanent", which untapped my Lotus Field.

...the ability did not untap the field, rather the trigger did. I can see how this might be confusing, especially for a new Magic player. People often do talk about 'triggers', but this is just short for 'triggered abilities'.

Vizier of Tumbling Sands' ability "When you cycle Vizier of Tumbling Sands, untap target permanent." is a triggered ability, as defined in rule 603.1:

Triggered abilities have a trigger condition and an effect. They are written as “[When/Whenever/At] [trigger condition or event], [effect]. [Instructions (if any).]”

Like any triggered ability, it is an ability that goes on the stack after it triggers, and can be responded to and interacted with. And that ability has a single target, the permanent to untap, making it a valid target of Bolt Bend. The opponent cast Bolt Bend targeting that triggered ability, and redirected it to target something else.

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sort: votes


text:
The 'speed' of conceding a game
During a game of Commander at my local store last week some of the folks I was playing with stated that conceding the game was a 'sorcery speed' action. The playgroup made some valid points about this preventing collusion and also the somewhat-irritating situation where one player takes control of someone else's creature and swings for lethal damage with it, only to have that creature's owner leave the game, taking the creature with them.
At the time I just accepted that this could well be the case and got on with the game (I had no intention of conceding) but in a moment of curiosity decided to do a quick google search which brought me to the below entry from the comprehensive rules:

104.3a A player can concede the game at any time. A player who concedes loses the game immediately.

This makes it pretty clear to me that by normal rules a player can concede at any time, they don't even need to have priority.
Does anyone know whether this supposed sorcery speed concession is just a myth/common misconception or is it some sort of rule enforced only in multiplayer tournaments?

Interesting practical application: 3-way multi player game... Player 1 is at 1 life, Player 2 gains control of all Player 1's creatures and attacks Player 3 and Player 1 for lethal damage. Player 1 likes Player 3 so he concedes before Declare Blockers taking all his creatures controlled by Player 2 with him. Player 2 is then wide open to counter attack by Player 3.

You answer yourself in your question.

As you have shown in your question, it is a house rule. That said, it's not a particularly surprising house rule as generally the kind of concession this would arise in could easily be a jerkish move, though not necessarily.

104.3a A player can concede the game at any time. A player who concedes leaves the game immediately. He or she loses the game.
101.1. Whenever a card’s text directly contradicts these rules, the card takes precedence. The card overrides only the rule that applies to that specific situation. The only exception is that a player can concede the game at any time (see rule 104.3a).
712.6. The controller of another player can’t make that player concede. A player may concede the game at any time, even if he or she is controlled by another player. See rule 104.3a.
405.6. Some things that happen during the game don’t use the stack.
405.6g A player may concede the game at any time. That player leaves the game immediately. See rule 104.3a.

Conceding isn't covered by the rules governing timing and priority. It is not listed in the rules under special actions. Rules as written; it can be done at any time, even when not in control of your own turn.

Does rule 101.1 apply to the whole of 104.3a? or just that they can concede at any time. Just thinking about Platinum Angel..."I play Platinum Angel and concede, but I don't lose..."

Actually I suppose, you leave the game, as do your permanents, therefore you lose. Fun to think about though.

@Cameron your comment reminds me of this joke; https://m.reddit.com/r/magicTCG/comments/24lrp1/standoff_in_honolulu/

Isaac - Absolutely hilarious. Thank you for sharing!

As far as official rules are concerned, it's certainly a misconception. The right to concede is unaffected by the state of the game or the number of remaining players. A player conceding can absolutely affect the game for the remaining players in a multiplayer match, but there still is no official rule anywhere that restricts your right to concede at any time. Of course, house rules can change any official rule - if your game store uses house rules for their (inofficial) tournaments, then that's the way it is.
Note that collusion is a valid concern in a tournament setting and forbidden by the official tournament rules. However, while conceding a game can be the result of collusion, the act of collusion is forbidden and punishable, not the act of conceding. If the tournament you have mentioned was a sanctioned event, changing the rules of conceding to be sorcery speed was not consistent with the official rules of the game, but if the Head Judge made that call, it still stands, as the Head Judge is the final rules authority at an event.

The game in question wasn't a tournament game, just a friendly so there was no Head Judge ruling per se. I was less querying the official rules on the matter and more curious as to whether this misconception arose from rules in some tournament, a third-party format rule (e.g. it originated in EDH but didn't translate into the official Commander format) or some other source that could be traced. If that's not clear in my question I can edit it to try and make that more prominent, great answer anyway though.

@Aiken Tis not really a misconception, just a rule put into place to prevent dickishness.

Ah. That's a slightly different question - I didn't catch that nuance. Seems more likely to me that it was a house rule put in place a while ago, then taught to new players who took it as gospel. But I don't think you will get certainty; baring knowing that it isn't from an official, central source. That said it isn't an uncommon thing to put house rules in place about - though this isn't the only solution I've seen.

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text:
If I exile Rout with Narset's ability, when can it be cast?
Let's say I control Narset, Enlightened Master and attack, exiling the top four cards of my library, including Rout (alt link). Rout's first ability states that I can cast it any time I could cast an instant by paying an additional {2}.
I think I've narrowed this down to two possible scenarios:

Rout's first ability defines an alternative cost (like Bestow or Overload) and thus can't be paid as I'm already casting it using an alternative cost ("without paying its mana cost").
Rout's first ability is an optional additional cost (like Kicker) and can be paid to allow me to cast the spell any time I could cast an instant.

Which, if either, of the above interpretations is correct?

@Gendolkari Thanks for the edit, but I deliberately didn't use the autocard markdown for this question since it doesn't open Rout properly, it only opens a search where Rout itself doesn't actually appear until the 4th item down. I'm guessing this is an issue with Gatherer where short searches aren't automatically matched but rather used as a search term, thus returning all names containing "Rout" instead of the only card with that as its full name. Is there a particular reason the autocard markdown should be used as opposed to a direct link to the card?

In general you should use the autocard so that if for some reason they change Gatherer so links break only the autocard needs to be fixed rather than having to individually fix every link. Though in cases like Rout where it doesn't link directly to the card I generally use a direct link instead.

It's a cost increase, so you may use it in combination with Narset's alternative cost.

The cost to cast a spell or activate an ability is calculated as follows[CR 601.2e]:
total cost to cast
= mana cost or alternative cost

additional costs and other cost increases

cost reductions

Rout's ability adds to the total cost, so it's a cost increase. It's not an additional cost since it doesn't use the word "additional"[CR 117.8], so it's just an generic cost increase[1]. It's not an alternative cost since those replace the mana cost. Alternative costs usually include the phrase "rather than pay" or "without paying its mana cost"[CR 117.9].
So, the total cost to cast is {2}. (Alternative cost of free, plus a cost increase of {2}.)

Another example of a cost increase: Thalia, Guardian of Thraben.

To my knowledge, there's nothing that distinguishes between additional costs and other cost increases. They're the same except in name.

601.2e [...] The total cost is the mana cost or alternative cost (as determined in rule 601.2b), plus all additional costs and cost increases, and minus all cost reductions. [...]

117.8. Some spells and abilities have additional costs. An additional cost is a cost listed in a spell’s rules text, or applied to a spell or ability from another effect, that its controller must pay at the same time that player pays the spell’s mana cost or the ability’s activation cost. A cost is an additional cost only if it’s phrased using the word “additional.” [...]

117.9. Some spells have alternative costs. An alternative cost is a cost listed in a spell’s text, or applied to it from another effect, that its controller may pay rather than paying the spell’s mana cost. Alternative costs are usually phrased, “You may [action] rather than pay [this object’s] mana cost,” or “You may cast [this object] without paying its mana cost.” [...]

Is there any functional difference between additional costs and cost increases? Thalia increases the cost by {1}, but it could have imposed an additional cost of {1}.  Brutal Suppression has an additional cost of "sacrifice a land", but it could have increased the cost by "sacrifice a land" (although that would look funny written on the card). I'd be interested in a rule or example that highlights the functional difference between the two mechanics. If it isn't worth adding to your answer, but you want to respond, hit me up in chat and I'll delete my comment.

@Rainbolt, Additional costs are cost increases ("plus all cost increases (including additional costs)"), so you're asking if anything distinguishes between additional costs and other cost increases. I don't know of any, and I can't imagine why anything would want to do distinguish between the two. I actually meant to note that in my answer.

That makes sense (and now I see more clearly the careful wording you used in the answer). Thanks.

It is actually neither of those options. It isn't an alternate cost because it isn't worded like "You may [action] rather than pay [this object's] mana cost," or "You may cast [this object] without paying its mana cost."[117.9] and it isn't an additional cost because it doesn't contain the word "additional"[117.8]
What it is is a cost increase.

117.8e Some effects increase the cost to cast a spell or activate an ability without using the word "additional." Those are not additional costs, and are not considered until determining the total cost of a spell or ability as described in rule 601.2e.

And we can use the cost increase with Narset's alternate cost because

117.9d If an alternative cost is being paid to cast a spell, any additional costs, cost increases, and cost reductions that affect that spell are applied to that alternative cost. (See rule 601.2e.)

According to Comprehensive Rules 117.8:

A cost is an additional cost only if it’s phrased using the word “additional.”

Therefore, Rout's extra cost of {2} is not additional.
Further in 117.8e:

Some effects increase the cost to cast a spell or activate an ability without using the word “additional.” Those are not additional costs, and are not considered until determining the total cost of a spell or ability as described in rule 601.2e.

And 117.9d:

If an alternative cost is being paid to cast a spell, any additional costs, cost increases, and cost reductions that affect that spell are applied to that alternative cost. (See rule 601.2e.)

To sum up, you are using an alternative mana cost of {0} due to Narset's ability, and you may increase this alternative cost by {2} to play Rout as an instant.

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text:
How many units and resources will the Germans get when they research logistics?
Is there any official/de facto interpretation for this topic?

After setup, each time the Germans research a tech that upgrades or unlocks a unit, they build one of that unit for free and gain one resource of their choice from the market."

It seems that this description is ambiguous and I think it can be interpreted as below:

They get three units and three resources.
They get three units and one resource.
They get one unit and one resource.

There is not (at least yet) an official, explicit statement for this. For the time before a new FAQ version is released, I would suggest to stick into the Official and Unofficial rules and clarifications made by Itchi (BGG). He was one of the playtesters of the Wisdom & Warfare expansion, and this UFAQ is used in our playgroup if no help is offered by the official rules or the FAQ.

Germans get 3 resources and 3 units by researching or learning Logistics if they havent already unlocked these levels of units. For
example, if they already had Democracy and research Logistics, they
will gain 2 resources and 2 free units (Mounted and Artillery). For
example, if they already have Railroad and Mathematics and learn
Logistics, they will gain 2 resources and 2 free units (Infantry and
Mounted).

Remember, that this is true also even if you have level 3 or level 4 units unlocked. From the official FAQ v.2.0:

Q: If the Germans research a tech which unlocks a level 3 or 4 unit
and later researches a tech which unlocks the level 2 unit of the same
type, do they still get the free unit and resource?
A: Yes


text:
In Pandemic: on the brink, can I remove multiple bioterrorist cubes at once for a cured strain?
In Pandemic On the brink we ran into a situation where both the black strain and bioterrorist purple strain were cured. At one point one of the players moved into a territory that had multiple black and purple cubes.
My question, is it one action to remove all black and purple cubes because they are both cured or does it require one action each?

In regular pandemic, after a cure is found, you can use 1 action to remove all cubes of a single color, just like the medic can do normally. It is still 2 actions to remove 2 colors. But don't know if the bioterrorist cubes have special rules.

It doesn't in the aspect of removing the cubes, they work in the same way.

You need 2 actions, 1 per color. The ability to treat a cured disease is described in the original rules, under the list of actions you can perform:

Remove 1 disease cube from the city you are in, placing it in the cube
  supply next to the board. If this disease color has been cured (see
  Discover a Cure below), remove all cubes of that color from the city you
  are in.

So it is 1 action to remove all cubes "of that color" if that color is cured. Unless On the Brink rules describe a special exception for bio-terrorist cubes, then this still applies.

This applies in On the Brink as well.  Unless of course the player was the medic, in which case they would treat all the cubes without taking any actions.

Awesome thank you, this is what we thought but wanted to double check!


text:
Can I tap Rupture Spire to pay for its own ability with an Amulet of Vigor in play?
Assuming I stack the abilities correctly can I get the Rupture Spire untapped in time to pay for its own ability with an Amulet of Vigor?

Yes, you can.
When Rupture Spire enters the battlefield, 2 different abilities trigger. The next time someone (usually you) you would gain priority, you put both of those abilities on the stack, in an order you choose.

603.3b If multiple abilities have triggered since the last time a player received priority, each player, in APNAP order, puts triggered abilities they control on the stack in any order they choose.

Choose to put Rupture Spire's ability on the stack first, with Amulet of Vigor's on top. Amulet of Vigor's ability will resolve first, untapping Rupture Spire.
Then Rupture Spire's ability will resolve, at which point you are free to tap Rupture Spire for the 1 mana you need to pay.
A key point here is that Rupture Spire's ability doesn't do anything until it resolves; it does not require mana payment ahead of time.
You also have priority between Amulet of Vigor's ability resolving and Rupture Spire's ability resolving. You could just tap Rupture Spire at this time and float the 1 mana until Rupture Spire's ability resolves. Either of these would result in the same net effect. It's just like the difference between tapping a Mountain and then casting Lightning Bolt; or beginning to cast Lightning Bolt and tapping a mountain to pay for it when you get to the pay costs part of casting.

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text:
Can The Card "Suppression Bonds" enchant a creature with Shroud
So the description of Suppression Bonds states:

Enchant non land permanent. Enchanted permanent can't attack or block,
and it's activated abilities can't be activated.

The card doesn't use the word target so basically I was confused on whether or not I could use it on a creature with shroud. Any clarification is welcome and appreciated.
P.S. If it can enchant a creature with shroud using this method, could I also use other cards that do not specifically target a creature with shroud?

It depends on how you are enchanting the creature.
If you are casting the spell then no, you cannot enchant a creature with shroud. This is because aura spells target while they are on the stack.

114.1b Aura spells are always targeted. These are the only permanent spells with targets. An Aura’s target is specified by its enchant keyword ability (see rule 702.5, “Enchant”).

If the aura is entering the battlefield due to some ability or spell just putting it there (like Starfield of Nyx's triggered ability) you can put it on a creature with shroud, because the aura is not on the stack, therefore it doesn't target.

303.4f If an Aura is entering the battlefield under a player’s control by any means other than by resolving as an Aura spell, and the effect putting it onto the battlefield doesn’t specify the object or player the Aura will enchant, that player chooses what it will enchant as the Aura enters the battlefield. The player must choose a legal object or player according to the Aura’s enchant ability and any other applicable effects.

Suppression Bonds is an Aura and as such, it targets as per the following rule:

303.4a An Aura spell requires a target, which is defined by its enchant ability.

As an answer to your main question, see Diego's answer.
In response to your P.s., yes shroud only prevents the creature from being targeted - nothing else.
For example Clone, Wrath of God, Diabolic Edict, Prahv, Spires of Order, Sovereigns of Lost Alara, Quest for the Holy Relic and Bolster do not target creatures and all work against or with a creature with shroud.
Generally things that specify 'all' of something, or things that target players can effect creatures with shroud.
There are some cards, like Clone that say 'choose' rather than target, or that do things to specific cards without using the word 'target' like Sovereigns of Lost Alara - for these, if the card is older, it is generally worth double checking, as it may have had errata like Blue Elemental Blast which on the printed card doesn't mention targeting, but has had errata to include that word*, but other than these older cards with errata you can generally choose or select a creature with shroud, so long as the card doesn't target and you aren't casting an aura or equipping an equipment.
N.b. Generally these cards also get around protection - Clone or Wrath of God for example ignore protection just as they ignore shroud, that said Diego's specific example (getting an enchantment on to a battlefield without casting it) doesn't get around protection like it does shroud. (Protection prevents Damage, Enchanting, Blocking and Targeting - providing the handy acronym DEBT - from anything that meets the criteria it has protection from; The classic example is that even Progenitus is killed by Day of Judgement)
Example
Cyclonic Rift cannot effect a creature with shroud because it tries to target the creature, unless it is Overloaded, in which case it no longer targets and it instead effects 'each creature you don't control', rather than 'target creature you don't control'. You can see this verified in the first and second rulings on Gatherer for Cyclonic Rift.
*Generally done to preserve the original intended functionality following a rule change.

Do you know of a card that said "choose" in an older printing, but says "target" in oracle? Wizards generally avoids errata that would cause functional changes without good reason.

Older as in lots older. Cards like this were generally errata'd to preserve original functionality. I believe there are cards out there printed specifically with the word choose that were errata'd to target (It's not obvious how to search printed text on Gatherer, if it is possible at all). But for the moment I'll direct you to cards like Blue Elemental Blast from Alpha - it has the word target in Oracle, but the printed version doesn't mention targeting at all.

I'll also edit my answer so that it references Blue Elemental Blast, rather than making an assertion without sources.

Interestingly the card Intuition had errata to make it choose rather than target, and then a second errata to revert it to the original text. http://magic.wizards.com/en/articles/archive/latest-developments/power-level-errata-b-gone-2006-07-14


text:
Seedborn Muse & Sands of Time on the stack
If an opponent has Seedborn Muse out in play, I cast Sands of Time, then the opponent's turn comes back around.
How would the stack resolve between these two cards on their turn?
Does Sands of Time trigger resolve first or does Seedborn Muse? Or does the opponent pick which happens first since both trigger onto the stack at the same time?

Seedborn Muse would have no effect, because Seedborn Muse modifies the untap step, but Sands of Time makes all players skip their untap step.
The ability of Seedborne Muse is a static ability that modifies the rules of the game. It does not trigger and does not go on the stack. It just modifies how the untap step would play out, by making you untap during every player's untap step instead of just your own.
The first ability of Sands of Time is also a static ability. It creates a replacement effect.

614.1b Effects that use the word “skip” are replacement effects. These replacement effects use the word “skip” to indicate what events, steps, phases, or turns will be replaced with nothing.

Whenever an untap step would occur for any player, the game instead goes into the upkeep step. The untap step simply doesn't happen.
It doesn't matter that Seedborn Muse modifies the untap step, because that step simply doesn't happen. Even if Seedborn Muses's ability was a triggered ability, it still wouldn't happen because the trigger event (untap step) doesn't happen.
Therefore, Sands of Time completely nullifies Seedborn Muse's ability.

Sands of Time makes everyone skip their untap step.  Period.  It is a continuous effect, it doesn't trigger.  Seedborn Muse will do nothing in this scenario as there are no untap steps during which to untap it's controller's permenants.

Sands of Time has 2 separate abilities; the first is not a triggered ability, but a continuous ability that creates a replacement effect:

614.1b Effects that use the word “skip” are replacement effects. These replacement effects use the word “skip” to indicate what events, steps, phases, or turns will be replaced with nothing.

The replacement effect replaces "Untap step" with "nothing". Because the Untap step never happens; Seedborn Muse will simply not do anything at all; there is no untap step during which its ability would get to be used.
Also, even if Sands of Time did keep the untap step around and simply change what happens during the untap step; it would not interaction with Seedborn Muse. This is because Sands of Time only changes what happens to a player on their own turn; it does not do anything to you if it isn't your turn. Seedborn Muse only affects you on your opponent's turn; it doesn't do anything at all during your own turn. So if Sands of Time were worded the way you seem to have interpreted it; then on your own turns you would do the Sands of Time tap-switching effect; and on your opponent's turns you would simply untap all permanents you control.


text:
Thawing Glaciers return to hand interaction
I have a Thawing Glaciers, a Thespian's Stage, and  Deserted Temple on the battlefield in addition to four other mana producing lands. In the previous turn I tapped Thespian's Stage to make it a copy of Thawing Glaciers.
I tap the Thespian's Stage (Thawing Glaciers Copy) to search my library for a basic land.
I then use Deserted Temple to untap Thespian's Stage
Finally, I use Thespian Stage's copy ability to copy the basic land that just entered the battlefield.
When I end my turn and enter the cleanup step, the text from Thawing Glacier's ability attempts to take effect, "Return Thawing Glaciers to its owner's hand at the beginning of the next cleanup step"
Does my Thespian's stage get returned to my hand, or does nothing happen because the card that activated the ability is no longer a Thawing Glaciers?

Thespian's Stage will return to your hand.
It doesn't matter that it became a copy of a different land. The effect "Return Thawing Glaciers to its owner's hand at the beginning of the next cleanup step." is still around. The "Thawing Glaciers" wording in that text only means "this permanent", it doesn't matter that it isn't actually named Thawing Glaciers.

706.11. If an effect refers to a permanent by name, the effect still tracks that permanent even if it changes names or becomes a copy of something else.

and

201.4. Text that refers to the object it's on by name means just that particular object and not any other objects with that name, regardless of any name changes caused by game effects.


text:
Standstill, Phyrexian Tyranny and both players wanting to cast instants immediately after standstill is played. Who gets damage?
A question regarding Standstill; I have Phyrexian Tyranny on the table and we both have 2 life left. I play Standstill and immediately play an instant so that my fellow opponent would have to draw 3 cards, lose life and die. But, as I play Standstill my opponent also immediately plays an instant. Who draws the cards and loses?

do you play the instant in response to the stnadstill? standstill has an ability that only does something when the enchantment is on the battlefield.

You would have to draw the 3 cards from Standstill and lose the game.
Standstill has a triggered ability. It triggers whenever its trigger condition is met, i.e. whenever a player plays a spell. It can keep triggering as long as it's on the battlefield.

603.2. Whenever a game event or game state matches a triggered ability’s trigger event, that ability automatically triggers. The ability doesn’t do anything at this point.

After you played Standstill, you automatically received priority, i.e. the right to cast spells and activate abilities. You used that priority to cast your own instant, on which Standstill triggered.

603.3. Once an ability has triggered, its controller puts it on the stack as an object that’s not a card the next time a player would receive priority.

Putting something on the stack is not the same as that thing resolving. It's only in queue to resolve at a later point.
Since you are the active player, you receive priority after you cast your instant. So you first put your instant on the stack, followed by Standstill's ability.
The stack, from top to bottom, is:

Standstill triggered by your instant
Your instant

You receive priority again, but pass to your opponent. Your opponent casts his own instant. Standstill can trigger again because it's still on the battlefield. The opponent puts his instant on the stack, then another Standstill trigger:

Standstill triggered by opponent's instant
Opponent's instant
Standstill triggered by your instant
Your instant

Then you both pass priority in sequence because neither of you take any more actions. The stack starts resolving:

608.1. Each time all players pass in succession, the spell or ability on top of the stack resolves. (See rule 609, “Effects.”)

The topmost item is the Standstill trigger triggered by your opponent's instant. It resolves, Standstill is sacrificed. You have to draw 3 cards, Phyrexian Tyranny triggers 3 times (the triggers become the topmost items on the stack), and you lose the game.

In other words, barring other interactions, the last player to play an instant is the one who wins. If you would've played an instant in response to your opponent's instant, then a third trigger would be on the stack and, assuming that your opponent did nothing else, they would lose.

The rulings below the card say this:

It only works once. If someone casts another spell after it triggers, but before that trigger resolves, then it will trigger again. The first time one of these triggers resolves, it will be sacrificed for the full effect. Any additional triggers on the stack will do nothing when they resolve because you will be unable to sacrifice it additional times.

That means the instant your opponent casts (if they do it before the Standstill trigger resolves) puts the last Standstill trigger on the stack, so that resolves first, and you have to draw three cards and lose the game (unless you pay the mana to prevent the Tyranny effect).

I agree that this is correct, however the question doesn't seem really clear to me if the OP played his instant by retaining priority and the opponent followed in priority order or if they didn't know about priority rules and are playing a game of who can say they're playing their spell first.  It might be a good idea to note that with the priority system here, the person who blinks first is going to be on the receiving end of Phyrexian Tyranny.

@Becuzz, unless they have another instant to add on the stack. (Or rather, one more in total than their opponent, since you could go deeper indefinitely. Except that at some point just paying the mana to Tyranny would be cheaper.)


text:
Order of enter the battlefield effects in multiplayer (Warp World)
In a four player EDH game, Player 1 casts Warp World. Each player begins by shuffling their permanents into their library. Player 2 has a Banisher Priest on the battlefield, which currently has a Yarok, the Desecrated, owned by Player 3, under it. Next, everyone reveals an amount of cards from the top of their library equal to the amount of permanents they owned. They put all artifacts, creatures, and lands onto the field, and after that, enchantments. Then they put all cards not put onto the field onto the bottom of their library. Player 1 flips, among other cards, Grenzo, Dungeon Warden, and Faithless Looting. Player 2 hits Solemn Simulacrum and Panharmonicon. Player 3 flips Murderous Redcap, Panharmonicon, and Solemn Simulacrum. Player 4 hits Harvestguard Alseids, Doomwake Giant, and 4 enchantments.
What is the order of resolution on abilities?
Can Enchantment Creatures be put out AFTER other creatures during the enchanment phase of flipping? Or must they be put on the battlefield at the same time as other creatures?
Assumed:
If possible, the Murderous Redcap will target Grenzo, Dungeon Warden, to attempt to stop him from putting the Faithless Looting in his graveyard after it goes to the bottom of his library.
Player 4, if possible, will use Harvestguard Alseids to keep Player 1's Grenzo alive.
Player 4, if possible, will have Doomwake Giant enter the battlefield as an enchantment
The order of play is Player 1, Player 2, Player 3, Player 4.

What exactly are you asking? The order of triggered abilities is a different question from when enchantment creatures have to be put on the battlefield during Warp World resolution.

It is a different question, agreed. I chose not to make a seperate question because the order of resolution of abilities, which is my main question, relies directly on whether or not you are forced to put an enchantment creature down as a creature.

Can you narrow this question down to focus on one aspect? There are a lot of different objects here and the question seems to be about 3 or 4 interactions at once, plus multiple rules questions.

The multiple questions are down to a rules misunderstanding, they do boil down to one question. I'm writing it up.

But yes, the multitude of irrelevant card links is confusing. Please try to focus on the cards directly relevant to the question.

I only included cards that would effect the order of resolution of abilities, or put more abilities on the stack.

During Warp World's resolution, you have to put enchantment creatures on the battlefield as creatures. However, during Warp World's resolution, no triggered abilities enter the stack either, so for Murderous Redcap's ability, this specific order doesn't matter. Ultimately, you can't prevent Grenzo from fetching the Faithless Looting.
Warp World gives clear instructions on the order in which to place permanent cards on the battlefield: you put all artifact, creature, and land cards down in any order, then enchantments (so that auras have something to attach to), and the rest to the bottom of the library. Enchantment creatures are creatures, so you put them down together with the other non-enchantments.
During the resolution, ETB effects trigger as normal. However, they do not enter the stack yet, so you don't to choose any targets yet:

117.2a Triggered abilities can trigger at any time, including while a spell is being cast, an ability is being activated, or a spell or ability is resolving. (See rule 603, “Handling Triggered Abilities.”) However, nothing actually happens at the time an ability triggers. Each time a player would receive priority, each ability that has triggered but hasn’t yet been put on the stack is put on the stack.

Therefore, by the time you have to choose a target for Murderous Redcap, all permanents will be on the battlefield to choose a target from, so it would not matter whether you put down enchantment creatures as creatures or enchantments. Consequently, it also does not matter whether, in the non-enchantment part of WW, you put down the Redcap or the Grenzo first.
When the Warp World has finished resolving, the order of triggered abilities is determined by priority: The player with priority orders their abilities first, then all other players in turn order.

603.3b If multiple abilities have triggered since the last time a player received priority, each player, in APNAP order, puts triggered abilities they control on the stack in any order they choose. [..]

That means that only when Warp World has finished resolving will the abilities go on the stack, and player 1 has an opportunity to activate Grenzo before Redcap's ability resolves. You have no way of preventing that in your scenario.

I hate to say this on an answer like this, but my question is not whether the Redcap can stop Grenzo, it is the full order of resolution of all cards listed. Redcap and Grenzo is just one part of the fallout from Warp World.

As you can read from my answer, the order of abilities depends entirely on how players choose to order their ETB triggers after Warp World. During WW, the order of creatures does not matter, that's the whole point. I'm sure if you understand the principle behind 117.2a and 603.3b, you can work it out for yourself

I think it might be worth addressing the fact that the creature see each other enter, so you don't actually lose anything here from not being able to delay enchantment creatures entering.


text:
What spells affect Planeswalkers in MTG?
Does anyone know which spells can affect or Target Planeswalkers in MTG?

What exactly do you mean by "affect"? After all, that could include anything from Merciless Eviction, Despise or even Call the Gatewatch. Maybe provide more info as to what kind of information you're actually looking for? As in, are you looking to handle your opponents' planeswalkers once they're on the battlefield, or are you wondering for different reasons altogether?

There are a ton of such spells, in several categories. A spell might affect (or target):

any permanent (e.g. search for "target permament")
nonland permanents, noncreature permanents, etc (e.g. search for "target nonland permanent")
just planeswalkers, but this is rare (and there are things that say "target creature or planeswalker", so it's a little trickier to search for - this works, but has false positives)

Along with those searches, you can also search for things like "all nonland permanents" to find plenty of things that don't target.
And if a source you control deals non-combat damage to an opponent, you can redirect that damage to a planeswalker they control.

You can redirect any non-combat damage that would be dealt to an opponent to a planeswalker they control. The source doesn't have to be a spell.

Yeah, sorry - I was stuck saying "spell" since the OP asked about spells, didn't mean to exclude other things.

There's a lot of them. Here's a list of things to search for

Things that deal damage to players; at resolution the controller of the spell/ability about to damage a player can redirect the damage to a planeswalker that player controls instead. There are way too many spells/abilities in magic that do this to list them all. If you do go looking, don't forget that the spell/ability doesn't need to target.
Creatures can attack planeswalkers instead of players. Again, far too many to list.
Anything that targets planeswalkers. Scratch that, actually; anything that has the word "planeswalker" in its rules text "affects" planeswalkers in some way.
Anything that targets permanents, or otherwise affects them.

Needless to say, your question is incredibly broad.


text:
Do on-death triggered abilities of creatures happen in multiplayer when a creature's owner loses at the same time?
So my friend had gained control of my Wurmcoil Engine and attacked me. When damage resolved Wurmcoil Engine and myself had lethal damage, so both myself (player) and my creature with a "when this creature dies" effect die at the same time. Would my friend controlling the Wurmcoil Engine still receive the two tokens from Wurmcoil Engine dying? my understanding is that when I die, everything I own is removed from game thus when the check is made on the ability of Wurmcoil being in my graveyard to generate the tokens, it would not be in my graveyard but instead be in exile along with everything else.

Related: https://boardgames.stackexchange.com/questions/4749/in-mtg-what-happens-when-a-player-loses-in-multiplayer?rq=1

Possible duplicate of In MTG, what happens when a player loses in multiplayer?

@NathanHase, I read through that question and answer, and can't find an answer to this scenario within it.

@GendoIkari I thought that this would cover it: "When a player leaves the game, all objects (see rule 109) owned by that player leave the game ... This is not a state-based action. It happens as soon as the player leaves the game"

@NathanHase after reading that a couple times, I actually still don't know which way this answer goes. The question here is whether or not the Wurmcoil's triggered ability ever triggers. If it does, then it doesn't matter that the Wurmcoil has left the game; it's trigger will still exist and resolve. But I don't know if it triggers or not when it dies at the same time as a player.

@GendoIkari: retracting the close vote. The answer below is the correct one (a quote from the rules would help it :)). The Wurmcoil trigger will be on the stack when the cards owned by the dead player are exiled.

@NathanHase Minor correction, the cards aren't exiled; they leave the game.

@NathanHase Added relevant rules quotes :)

When you lose the game, you and everything you own (including permanents under other player's control and triggered abilities) leaves the game with you.
That said, a Wurmcoil Engine's trigger is not owned by you, but simply controlled by your opponent, even though the card that has the ability is owned by you - that is part of what a control-changing effect does for objects that aren't represented by cards (such as triggered abilities).

603.3a A triggered ability is controlled by the player who controlled its source at the time it triggered (...)

Therefore, the Wurmcoil Engine will be put into your graveyard by state-based actions at the same time you will lose the game. Its ability triggers at this point, and is put on the stack the next time a player would receive priority, which will be after you and your Wurmcoil Engine have left the game.

603.2a Because they aren’t cast or activated, triggered abilities can trigger even when it isn’t legal to cast spells and activate abilities.

Even though losing the game for having 0 or less life is the first item on the list of state-based actions, the order matters not in this case , since all SBAs are performed at the same time.

704.3. Whenever a player would get priority (see rule 116, “Timing and Priority”), the game checks for any of the listed conditions for state-based actions, then performs all applicable state-based actions simultaneously as a single event. (...)

On a side note, you can easily concede the game before the damage is dealt. This way, Wurmcoil Engine doesn't die and its ability doesn't trigger.

@M.Herzkamp While that is accurate and perfectly possible, it is textbook kingmaking territory and something I wouldn't recommend players actually doing in practice.

A good resource in this situation is http://chat.magicjudges.org/mtgrules/ - their answer follows:

Both them losing and Wurmcoil dying happens as part of the same "set" of State Based Actions.  Wurmcoil will hit the graveyard and they will lose at the same time.  Wurmcoil will still trigger in that case since it did, in fact, die.

Edit: Relevant rules

118.6. If a player has 0 or less life, that player loses the game as a state-based action. See rule 704.
704.3. Whenever a player would get priority (see rule 116, “Timing and Priority”), the game checks for any of the listed conditions for state-based actions, then performs all applicable state-based actions simultaneously as a single event. If any state-based actions are performed as a result of a check, the check is repeated; otherwise all triggered abilities that are waiting to be put on the stack are put on the stack, then the check is repeated. Once no more state-based actions have been performed as the
  result of a check and no triggered abilities are waiting to be put on the stack, the appropriate player gets priority. This process also occurs during the cleanup step (see rule 514), except that if no state-based actions are performed as the result of the step’s first check and no triggered abilities are waiting to be put on the stack, then no player gets priority and the step ends.
704.5. The state-based actions are as follows:
      704.5a If a player has 0 or less life, he or she loses the game.
      704.5g If a creature has toughness greater than 0, and the total damage marked on it is greater than or equal to its toughness, that creature has been dealt lethal damage and is destroyed. Regeneration can replace this event.
      704.5h If a creature has toughness greater than 0, and it’s been dealt damage by a source with deathtouch since the last time state-based actions were checked, that creature is destroyed. Regeneration can replace this event.


text:
How does flicker relate to a card like Ulamog's dispoiler
I was wondering how a card like Flicker and a card like Ulamog's Despoiler would work together. Since Ulamog's Despoiler returns a card from exile and flicker exiles a card snd then returns it would flicker return the card that was put into a graveyard to the battlefield or would it just stay in the graveyard? I tried to look at rule 400.7 but I got confused on if it would work. If anyone could help me I would be very grateful.

What are you trying to flicker here?  Something other than the Despoiler?

Yes, I am trying to flicker a permanent a opponent controls

They won't interact in any way.  You have to completely finish resolving Flicker then the Despoiler's trigger can resolve (or in the opposite order).  Basically you can't pick the two cards to move from exile to the graveyard while Flicker is resolving.  For Flicker, you move the card to exile then back to the battlefield before anything else can do anything.

Ok, thank you, do you know how it would work with a card like Admonition Angel?

@Bootskat If you want to know about cards which have significantly different effects from Flicker, please ask that in an entirely new question not as comments to this question. Thanks!

You can't cast or resolve anything whilst the target of Flicker is in exile, as it is returned to the battlefield as part of the spell.
You could do it to the target of Otherworldly Journey, as that returns the card at the next end step. If you did, it would stay in the graveyard, the object in the exile zone that Journey is looking for is no more.

Flicker and the Despoiler won't interact in the way you want them to.  You can't Flicker something to exile and then move it with the Despoiler while it is there.  The stack won't let you do that.
The stack defines the order things are done in.  You have to completely finish resolving one spell / ability before you can start resolving another or put something else on the stack.
So if you cast Flicker targetting an opponent's Hill Giant, the stack looks like this:

(top of stack)
Flicker
(bottom of stack)

If you start resolving Flicker without anyone responding, it will move the Hill Giant to exile then immediately back to the battlefield before you can do anything else.
If you were to cast Despoiler and respond with Flicker you still end up in the same situation.  The stack would look like this:

(top of stack)
Flicker
Despoiler
(bottom of stack)

You'd resolve Flicker (moving the Hill Giant to exile and immediately back to the battlefield).  Then you'd go to resolve the Despoiler.  At that point the Hill Giant wouldn't be in exile so you couldn't touch it with the Despoiler.
And just to be clear, you couldn't put them in the other order on the stack under normal circumstances (you can't cast a creature while the stack isn't empty).  Even if you could with some shenanigans, the Despoiler would resolve before Flicker while the Hill Giant is still on the battlefield.


text:
Does a Venomthrope gets an extra attack for using multiple weapons?
This question is related to the 7th edition core rules and Codex Tyranids 6th edition:
The Codex has a "designer note" in the Melee Weapons section stating:

Tyranid Melee weapons come as pairs. For game purposes, each pair is treated as a single Melee weapon. This means that Tyranid models must fight with two pairs of any Tyranid Melee weapons to gain a bonus Attack in close combat for fighting with two weapons.
  [...]

(emphasis mine)
The Venomthrope model has 4 lash-whips, which seems to indicate it should get the bonus attack for multiple weapons.
But, the army list entry for Venomthrope simply states he has

Lash-whips
Toxic miasma

Unlike the Ravener entry, for example, which specifically states:

Two pairs of scything talons

So, does the Venomthrope gains the bonus attack? Given its base Attack attribute of 2, how many melee attacks can it make in a 'normal' assault phase (disregarding charge etc.)?
If possible, please support your answer with the relevant rules reference.
Thanks!

Updated note: as there are contradicting answers at the moment, I'm waiting to see some more up-votes or support from the rules before accepting any of them (both make sense IMO, but reach different conclusions)

The reason why the entry for Venomthropes does not list off "two pairs of scything talons" is because Venomthropes--unlike Raveners--have options for how to arm up that are zero cost.
Looking at the Fast Attack Ravener Brood entry, it lists under options:

Any Ravener may exchange one pair of scything talons for rending
  claws - 5 pts/model

In other words, a Ravener will have two pairs of scything talons--all the time--unless you pay 5 pts/model to switch it out, using this optional upgrade.
Let's focus on the Designer's Note you referenced:

Designer’s Note: Tyranid Melee weapons come as pairs. For game
  purposes, each pair is treated as a single Melee weapon. This means
  that Tyranid models must fight with two pairs of any Tyranid Melee
  weapons to gain a bonus Attack in close combat for fighting with two
  weapons. For example, a Hormagaunt armed with a single pair of
  scything talons does not gain a bonus Attack, but a Ravener with two
  pairs of scything talons, or a Genestealer with a pair of rending
  claws and a pair of scything talons, does.

If your model has four (4) lash whips, that is very straightforwardly two pairs of weapons.  Each pair is treated as a single melee weapon.  Your model therefore has two melee weapons and will gain a bonus attack in close combat for fighting with two weapons.
Or put another way:

set aside two lash whip models.  two bits, one pair, for game purposes single melee weapon.
set aside two lash whip models.  two bits, one pair, for game purposes single melee weapon.
does the venomthrope support holding all of the physical weapon bits?  If the venomthrope has all four weapon bits, the venomthrope has two melee weapons.

So in a normal assault-phase attack (non-charge), you would get 3 attacks: 2 (base attack score) +1 (double melee weaps)

Can you please add how many melee attacks can it make in a 'normal' assault phase (disregarding charge etc.)? I'm guessing you mean it'll have 3 (base A2 + 1 for two sets of melee weapons) - right? I want to have the answers as easy to compare as possible.

(added at bottom)

You've answered your own question. If the entry for them listed 2 pairs of lash whips, then you would indeed get an extra attack (per the Tyranid paired melee rule). Since it doesn't you don't get any. So 2 attacks, 3 on the charge, barring some other circumstance.

So, the model's 4 weapons are purely for esthetic value (or, rather, they are already reflected in its Attack 2 attribute)?

@G0BLiN The base attacks are in addition to the bonus attacks you get from multiple weapons. You'd get the 2 basic attacks + 1 for multiple weapons for 3 attacks. You would also gain any bonuses from charging as well.


text:
Mage Knight The Board Game : Where Reputation plays a role?
In Mage Knight, reputation is important when spending influence at some sites for reducing or increasing the influence costs. I am looking for a list of sites where the player's reputation is used as a modifier during the interaction with the site and which sites the reputation track is ignored and instead you only use raw influence.
The list of sites:

Buying healing from a village
Buying healing from a Monastery
Buying AA Cards from a Monastery
Buying Spells from a Mage Tower
Green City : Buying AA Cards
Blue City : Buying Spells
White City : Buying Units
Red City : Buying Artefacts

At all cities you gain, Once per turn, +1 Influence for each of your own shields on city.


text:
How the forced capture rule works with crowned pieces( in international checkers or...)
If there is a possible capture two or more fields away from a king like b--w- or b-----w- are those count as a compulsory/necessary jumps or not?

It depends on the checkers/draughts variant that you are playing:

For American checkers / Britisch draughts (which are two names for the same game) and Italian checkers, kings are short-ranged and cannot jump over empty squares.
For all other variants (International/Russian/Spanish/Czech/Frisian/Thai draughts and Pool checkers), kings are long-ranged and can jump over empty squares as you indicated.

NOTE: in fact, for all but one (Thai draughts) of the long-ranged king variants, kings can also land on a sequence of adjacent empty squares behind a captured piece. In your notation, this would be patterns of the form B- - -w- - -, where the black king can land on any of the three - squares behind the white man.


text:
Bizzaro Playing Cards
Here’s a problem a friend of mine posed to me a long time ago, that I was recently reminded of. I figured, with modern computing, there might be some way to figure it out:
Assume that there are two set of 52 cards, 13 of each of four sets. Now, for the purpose of a game, each card of the first deck becomes associated with one card in the other card such that, say, A♠ = 5♣.
The second set follows these conditions:
1.) No card may be it’s own twin
2.) No card may appear twice
3.) Four of a kind in the normal deck may not repeat suits in the second deck.
4.) Any poker run in a normal deck (straight, flush) must be disrupted in the second.
Is there a set of cards that matches these conditions? And if so, what is it?

This question appears to be off-topic because it is about a puzzle that doesn't have anything to do with an actual game.  Try the Puzzling Stack Exchange.

Your third and fourth conditions are unclear. From the first two conditions, it seems like you are looking for a partition of the 52 cards into 26 disjoint pairs. However, last two conditions' mentions of the "second deck" make it sound more like you are looking for a bijective mapping from the 52 card deck to another 52 card deck. Basically, do you want to pair cards up and consider those pairs as a single unit, or do you want to replace each card in the first deck with another card in the second deck?

There are two decks, each of 52 cards, the pairing those of one deck to another. They are paired, not replaced. Sorry was unclear.

The Puzzling Stack Exchange is a good idea. Thanks!

math.SE is probably an entirely better source than puzzling.SE - though this could easily be a math homework problem (in abstract algebra...)

Note that (1) and (2) are satisfied by any derangement of the cards. There are probably many "mappings" that satisfy your conditions, but since all you seek is one possible answer, I propose the following solution.
For the values, map them in this way:
A -> 2, 2 -> 4, 3-> 6, 4 -> 8, 5 -> 10, 6 -> Q, 7 -> A, 8 -> 3, 9 -> 5, 10 -> 7, J -> 9, Q -> J, K -> K
Then: A234567890JQK -> 24680QA3579JK
So clearly any straight will not map to any straight. And because any 4 of a kind maps to a 4 of a kind, there won't be any suit duplication (your condition 3).
For the suits, consider some order of the suits [clubs, diamonds, hearts, spades]:

For values A-4: map to the same suit
For values 5-8: map to the next suit (e.g. clubs -> diamonds, spades -> clubs)
For values 9-Q: map to the previous suit
For K: map to the "opposite suit"

Then any flush (which consists of 5 cards of the same suit) cannot map to a flush, because you cannot obtain more than 4 cards of the same suit if you start with a flush.
Note that changing the suit of the K resolves the issue where K -> K, since it is required that no card be its own "twin".


text:
Creatures Dying At Same Time Controller Loses
Player A, B, and C are in a Multiplayer Game. It is player A's turn.
Player A has: Sword of Vengeance attached to Kobolds of Kher Keep, a Mystic Crusader, and 1 life.
Player B has: Memnite, a tapped Darksteel Colossus and 1 life
Player C has: Blood Artist enchanted with Rancor and 2 life.
No one has a hand, graveyard or any lands.
Player A attacks Player B with the Kobolds and he attacks player C with the Mystic. Player B blocks with the Memnite and takes 1 damage in the first strike step. Does Player C's Blood Artist trigger from the memnite's death, allowing Player C to win by targeting Player A with the ability. Or does memnite never make it to the graveyard due to Player B losing the game, allowing Player A to win when the Mystic deals combat damage?

Is the Rancor relevant? Or the Darksteel Colossus?

Not super relevant. It gives the other players ways to kill A on their turns, so A has to go all out this turn.

Note that Player B could choose to concede at any point prior to Memnite dying if they wanted A to win over C for whatever reason.

@Samthere That is correct, but most playgroups I have played with considered "strategic concession" like that to be unsportsmanlike.

Player C can win with Blood Artist's triggered ability.
After First Strike damage, Player A would gain priority, so state-based actions are checked. The following ones actually happen:

704.5a If a player has 0 or less life, he or she loses the game. (Player B, in this case)

704.5g If a creature has toughness greater than 0, and the total damage marked on it is greater than or equal to its toughness, that creature has been dealt lethal damage and is destroyed. Regeneration can replace this event. (Player B's Memnite, in this case).

Then rule 704.3 says

Whenever a player would get priority (see rule 116, “Timing and Priority”), the game checks for any of the listed conditions for state-based actions, then performs all applicable state-based actions simultaneously as a single event. If any state-based actions are performed as a result of a check, the check is repeated; otherwise all triggered abilities that are waiting to be put on the stack are put on the stack, then the check is repeated. Once no more state-based actions have been performed as the result of a check and no triggered abilities are waiting to be put on the stack, the appropriate player gets priority. This process also occurs during the cleanup step (see rule 514), except that if no state-based actions are performed as the result of the step’s first check and no triggered abilities are waiting to be put on the stack, then no player gets priority and the step ends.

So, simultaneously, Player B loses and Memnite goes to the graveyard. At this point, Blood Artists ability triggers. Rule 800.4 has a complete description of what happens when Player B loses; essentially, they take all of their cards and leave. Now, SBA's are checked again, and since no action needs to be taken, Blood Artist's triggered ability goes on the stack, with Player C presumably choosing player A as the target. Eventually, the ability resolves, putting player A at 0 life. The check described above repeats, and player A loses.
The important thing here is that there is no mention in rule 800.4 or any of its subrules that anything happens to abilities that other players own and control that happen to have trigger conditions associated with the losing player.


text:
Cards that untap lands on opponents' turns
I used to use Prophet of Kruphix in my commander deck, but it is now banned.
While giving all creatures flash is nice, but I really used it for it's land untap ability. Are there any other cards that let me untap my lands on other player's turns?

You can do a quick search for keywords like 'untap' + 'other player' to find some cards.
This'll put you on Seedborn Muse. She's the original Prophet of Kruphix, but hasn't been hit by a ban because she doesn't have that super-juicy "flash" clause. (And because being mono-green means she props up more diverse archetypes than a UG card does.)
Is that your only option? Not quite: try other keywords (maybe filter down to your colors and search 'untap' + 'land' or 'untap' + 'permanent').
Some highlights that you're likely to miss unless you try a lot of search permutations:

Awakening is a green enchantment that untaps everyone's stuff. (And allows heavily-instant-speed decks to double-dip mana on their own turn.)
Patron of the Orochi is a big green creature that you can use to untap all your green creatures and forests each turn. (And can also be used to double up your mana on your own turn.)

Damn... Seedborn muse is actually so much better because your creatures essentially gain a better version of Vigilance.

Here are 2 other options for you:

http://www.mtgsalvation.com/cards/tenth-edition/13990-seedborn-muse
http://www.mtgsalvation.com/cards/stronghold/4946-awakening

Another option (though unfortunately just for a single land) is Urban Burgeoning from Return to Ravnica. I haven't looked it up myself, but im pretty sure there are other cards that untap on opponent turns if you widen your search for "land", "untap", "upkeep", and "player".


text:
When resolving "destroy X when Y enters the battlefield", must I choose a target if the only available one is my own?
Say that I cast a Viridian Shaman and there is only one artifact on the battlefield - my Sol Ring. Must I destroy my own artifact due to Viridian Shaman's trigger?

Yes you must destroy your own artifact if it is the only one on the battlefield. If it were optional it would have the word 'may' in the ability like Acid Web Spider.
When an ability triggers and it has a legal target it will always be put on the stack and a legal target chosen. Then when the ability resolves if it has a 'may' in it you can choose not to have its effect occur otherwise it does.

603.2. Whenever a game event or game state matches a triggered ability’s trigger event, that ability automatically triggers. The ability doesn’t do anything at this point.
603.3. Once an ability has triggered, its controller puts it on the stack as an object that’s not a card the next time a player would receive priority.
603.5. Some triggered abilities’ effects are optional (they contain “may,” as in “At the beginning of your upkeep, you may draw a card”). These abilities go on the stack when they trigger, regardless of whether their controller intends to exercise the ability’s option or not. The choice is made when the ability resolves.

Thanks for the reply. It kind of made sense to be so, but I couldn't easily find a reference online...

You bolded "The choice is made when the ability resolves", but Viridian Shaman targets, so the choice is made when the ability is placed on the stack (after it triggers, not when it resolves).

Unfortunately yes.
Upon resolution of the enter the battlefield trigger it checks for a valid target and if there is one you have to choose it.

Care to reference the official rules? I would really like to know for sure.


text:
2 Hg Team damage vs player damage
So trying to read the 2HG format rules, and i'm not sure i understood how does it works.
Any team has a shared life pool starting from 30. Any damage to a player is taken from the shared hp, so when it reach 0 the team lose the game.
But the damage is dealt to a single player. So there could be a case where the team lose the game even if the shared health pool is not zero (excluding special cards).
For example:
if Team A cast a card that says i.e. "Deal 10 damage to target player", and now the team B has 20 life points. Team A deals 10 damages to player 1 in Team B, and he can't block this action (neither is teammate), it means that since the single player life is the (total life)/2 = 10 Player 1 of Team B reach 0 and then the team lose the game.
Correct?

Dealing 10 damage to a single player will not kill a team that has 20 life. This is where you are going wrong:

the single player life is the (total life)/2

This might have been true years ago (more on that later), but by current rules, individual players in two-headed giant do not have life totals. You cannot divide the total life in half and assign that to each player. The shared life total is the only life total that a team has.

Correct: The team has 30 life.
Incorrect: Each individual player has 30 ÷ 2 life.

On the other hand, if the opposing team had 20 life, and you somehow dealt 10 damage to each opponent (e.g., with Fall of the Titans), then you would kill the opposing team.
Another user discovered that, prior to April 23, 2010, a rule for life totals in two-headed giant stated:

If an effect needs to know the value of an individual player's life total, that effect uses the team's life total divided by two, rounded up, instead.

With the release of Rise of the Eldrazi, the rule changed:

If a cost or effect needs to know the value of an individual player's life total, that cost or effect uses the team's life total instead.

This could explain part of your confusion. I omitted rule numbers intentionally, because they don't line up exactly with the current rules.

Part of the asker's confusion could be that prior to around 2010 (I can't find the exact date of the change) if something cared about an individual player's life total it would use half the team total. But even then this still wouldn't happen

yeah, i just realized that i was reading a 2009 article, and during the reading i missed also that important sentence: "Despite that, if an effect lets a player pay any amount of life, that player may pay as much life as the team has, not as much as his individual life total.", Thank you! :)