Scrambleverse 6RR
Sorcery (R)
For each nonland permanent, choose a player at random. Then each player gains control of each permanent for which he or she was chosen. Untap those permanents.
At a high-level event, suppose the following sequence of actions happen:
(A simpler example would be for one player to make lots of copies and then cast Scrambleverse himself. But this version is more likely to actually happen.)
Assume neither player wishes to concede or intentionally draw, what happens at this point? Do the players keep flipping coins until time is called? What happens then, since the current turn won't end?
If the number of a certain kind of token is significantly high, if each player agrees then you can divide those tokens evenly among the players, then flip a coin/roll a die/etc. for the remainder of those tokens and the other nonland permanents.
First I'll assume that for some reason the original Deciever Exarch can't make any more copies for some reason, or else it doesn't matter who gets which exarch (except the "real" one) since whoever gets the enchanted one can just make more.
I don't know how they'd rule it in an event (they might not know this stuff), but according to math if you have enough Exarch tokens (doesn't work with 100000000000000, but it does with "arbitrarily high" AKA the limit as # of tokens approaches infiniti), then you have exactly 50% chance of one player having an arbitrarily higher quantity of tokens than the other one, even if that number is really really small compared to the total amount of tokens. So, 50% chance that player B can one-shot kill his opponent no matter how much life he has (like normal Exarch combo), and 50% chance that player A has enough tokens to block all of player B's tokens and any creatures he might have, as well as kill an arbitrarily high number of them by multi-blocking, but far far far from all of them. Though that won't make player A win since the tokens then die.
The problem is that you can't shortcut through probabilistic processes (where it applies, this usually involves repeated shuffling with other things happening in-between them, but making random choices also counts). I'd be interested in hearing what someone familar with the floor rules at high level events has to say about this.
The game will either end in a draw or it will be a mutually-agreed upon intentional draw.
"715.2a 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. "
The result of the Scrambleverse cannot be predicted. DCI policy demands that a loop be 100% predictable to be shortcutted on. This action of randomizing that many tokens is nowhere near 100% predictable, and therefore cannot be shortcutted on.
That means the players can either agree to randomize all of the tokens, which will take too long and force the round to time, or they can just say the game is a draw and move on.
This is like the old "one million coins" and "lantern of insight" threads.
This is like the old "one million coins" and "lantern of insight" threads.
I've read those threads, but I don't think they provide any answer in this situation.
In those situations, a player decides how many times to execute a loop with a random event. As I understand the official policy, the player cannot choose a number that would take an unreasonable amount of time to resolve.
However, in this case there is no such decision. Scrambleverse's instructions are mandatory and the only choice the player is making is to play the card.
I'd have a suggestion:
* Take a number of tokens equal to the largest number on a die you have minus one. (Usually 20).
* Roll the die. Separate the tokens into two piles, one according to that number and the rest (maximum value is 0).
* The roll the die again with even/odd to decide which player gets which pile.
Repeat until no tokens remain.
1. This doesn't really solve the issue, as even with this shortcut it would still take way too long to complete all the randomization.
2. This would give a different result. Consider the simple case of 19 tokens. With your method, the chances of a 10/9 split and a 19/0 split are the same. If you flipped a coin for each, the 10/9 split is much more likely.
This would give a different result. Consider the simple case of 19 tokens. With your method, the chances of a 10/9 split and a 19/0 split are the same. If you flipped a coin for each, the 10/9 split is much more likely.
There isn't any provision for making sure the outcomes of any event have a specific distribution. In his example, every token is being randomly assigned per the result of 2 coin flips each, they're just being batch-processed as they're grouped into the results of HH, HT, TH, and TT.
Alternatively given X Exarchs, by using Base Y (Y can be 6, 20 or whatever) you can roll ZDY ordered dice, where Z is given by the equation Y^Z - 1 = X.
So;
Y^Z = X - 1
Logy(Y^Z) = Z = Logy(X-1)
So using D20s, with X = 100,000,000,000,000,000,000,000 [number copied from OP]
Z = Log20(99,999,999,999,999,999,999,999) = 17.68 D20 dice rolls. (rounded to two sf)
So, ignoring the inconvenient fact that you need to roll the dice .68th of a time, roll a d20 that many times, work out what the value is in base 20, then flip a coin to see who gets that many exarchs - the other player gets the remainder.
Not sure how to deal with that fraction of a dice roll though.
Still ~18 dice rolls is a feasible number
This doesn't give the proper answer. Let's choose X=100 for simplicity. What you're basically doing is choosing a random integer from 0 to 100 and giving that many tokens to the first player.
The chance that I get all the tokens is 1/101. However, if I were to choose a random player for each of the 100 tokens, the chance that I get them all is (1/2)^100=1/1 267 650 600 228 229 401 496 703 205 376.
Quite a big difference.
There's potentially an additional problem with the general technique of determining how many each player gets: each token has a different timestamp, which may matter in the future.
You can use a computer to model a very large number of distributions monte carlo style, but the person who made the 'infinite tokens' could easily have chosen a number so large as to be computationally infeasible. For example, if he specificed that he made one googolplex of tokens. Computation can't solve this problem
Yes, my method doesn't give you the timestamps exactly, but if that's not important (and with arbitrarily high amount of tokens, when will it be actually?) then you DO arrive with player b having 50% chance of killing player a, and that no matter how much life player a has (unless player a can gain an arbitrarily high amount of life after the exarches are created).
I'm not that good at explaining, look up the Law of large numbers if you don't understand why.
That won't work because the rules demand an exact accounting of the game state. You need to determine exactly how many tokens each player controls; you aren't allowed to use an approximate answer. So using the law of large numbers won't work.
A solution which doesn't calculate timestamps is also problematic because even if it's irrelevant at the time it's possible that it will matter at some point in the future.
That won't work because the rules demand an exact accounting of the game state. You need to determine exactly how many tokens each player controls; you aren't allowed to use an approximate answer. So using the law of large numbers won't work.
A solution which doesn't calculate timestamps is also problematic because even if it's irrelevant at the time it's possible that it will matter at some point in the future.
Exactly. This is why the game must be a draw. The players will have two options. They can either 1) Go through each individual coin flip to resolve the spell (which means the round will go to time and be a draw), or 2) Choose to accept a mutual draw and move on to the next game. This was the conclusion that was reached when the Worldgorger Dragon player milled an Oath player for 100000000+, but the Oath player had 2 Blessings in his deck. There was no way to determine the exact card order without going through the motions, so the players were given the two options I have said above.
Mathematical modeling is awesome, but it is not part of the rules. Any solution that involves such modeling is not going to work.
Exactly. This is why the game must be a draw. The players will have two options. They can either 1) Go through each individual coin flip to resolve the spell (which means the round will go to time and be a draw), or 2) Choose to accept a mutual draw and move on to the next game. This was the conclusion that was reached when the Worldgorger Dragon player milled an Oath player for 100000000+, but the Oath player had 2 Blessings in his deck. There was no way to determine the exact card order without going through the motions, so the players were given the two options I have said above.
Got a reference for this ruling?
The problem I see with option 1 is that the end of match procedure specifies 5 additional turns. But the current turn won't be ending anytime soon, so the players won't get to the point where the game reaches the end of the additional turns and becomes a draw. So the entire tournament will be paralyzed and unable to get to the next round.
This would also be really annoying in practice. If this occurs in game 2, whoever won game 1 would obviously choose option 1, since that would (assuming that it does lead to a game draw) given that person the match win.
Similarly, let's say upon resolution of all the flips, one player or the other will have a creature advantage; they could attack for the win, due to having (for example) 1 million+20 attackers, vs. 1 million potential blockers. Odds are high that the split would not be perfectly 50/50, but enough of a difference that the attack from whoever has more would be for the win. If you think this situation has a chance of happening in a given tournament, I'd check with your local judges how they would solve the issue, barring an official answer from Wizards.
If you think this situation has a chance of happening in a given tournament, I'd check with your local judges how they would solve the issue, barring an official answer from Wizards.
The point of having rules is so what happens in a given situation (where the facts are clear) is not up to the discretion of individual judges.
That is a similar situation. However, the closest thing that thread has to an answer is self-contradictory:
There is no way to resolve a situation like this. The game will end in a draw.
I agree with the first sentence. I don't see anything in the rules which allows this kind of situation to be resolved. The second sentence contradicts this by saying the game ends in a draw. I don't see anything in the rules which causes this situation to be resolved by the game being declared a draw (unless the players mutually agree). The failsafe in the rules which causes a game which enters an infinite loop to end as a draw does not apply since this is not actually an infinite loop.
I think the only thing that needs to be determined is who gets the original Exarch. (+other permanents)
The exarch will remain untapped (unless the combo user is dumb) so more tokens can be produced after scrambleverse resolves.
Situation A. Player A gets Exarch, he makes more tokens to block any that are sent his way, they're all exiled at the end of turn, Player A does the combo again on his turn and wins.
Situation B. Player B keeps Exarch, makes more tokens, sends them over for the win.
If for some reason Exarch IS tapped, then I think a single coin flip to determine "who gets more" would be sufficient. You still have a 50/50 outcome of winning the duel. If you end up with more, you can block them all and die next turn when he does the combo again. Or he wins the flip, and likely has more than 20 Exarchs with that arbitrarily large number.
Edit: I forgot the Exarch wouldn't gain haste if it changed control. In that case just flip once for the winner. Of course this short cut doesn't need to be agreed on, but your chances are the same either way.
Sorcery (R)
For each nonland permanent, choose a player at random. Then each player gains control of each permanent for which he or she was chosen. Untap those permanents.
At a high-level event, suppose the following sequence of actions happen:
1. Player A starts game with Leyline of Anticipation in play.
2. Player B casts Deceiver Exarch, then enchants it with Splinter Twin.
3. Player B creates 100000000000000000000000 copies of the Exarch using the combo.
4. Player A casts Scrambleverse.
(A simpler example would be for one player to make lots of copies and then cast Scrambleverse himself. But this version is more likely to actually happen.)
Assume neither player wishes to concede or intentionally draw, what happens at this point? Do the players keep flipping coins until time is called? What happens then, since the current turn won't end?
Practice for Khans of Tarkir Limited:
Draft: (#1) (#2) (#3) (#4) (#5)
Real world?
It'll never happen due to the high cost of Scrambleverse.
If player A has Leyline and 8 red mana open he should have already won before Player B's Exarch/ST combo takes him out.
I don't know how they'd rule it in an event (they might not know this stuff), but according to math if you have enough Exarch tokens (doesn't work with 100000000000000, but it does with "arbitrarily high" AKA the limit as # of tokens approaches infiniti), then you have exactly 50% chance of one player having an arbitrarily higher quantity of tokens than the other one, even if that number is really really small compared to the total amount of tokens. So, 50% chance that player B can one-shot kill his opponent no matter how much life he has (like normal Exarch combo), and 50% chance that player A has enough tokens to block all of player B's tokens and any creatures he might have, as well as kill an arbitrarily high number of them by multi-blocking, but far far far from all of them. Though that won't make player A win since the tokens then die.
I'm Mike, from The Mana Pool.
Check out my Tapped Out profile and comment on my decks!
"715.2a 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. "
The result of the Scrambleverse cannot be predicted. DCI policy demands that a loop be 100% predictable to be shortcutted on. This action of randomizing that many tokens is nowhere near 100% predictable, and therefore cannot be shortcutted on.
That means the players can either agree to randomize all of the tokens, which will take too long and force the round to time, or they can just say the game is a draw and move on.
This is like the old "one million coins" and "lantern of insight" threads.
-ktkenshinx-
I've read those threads, but I don't think they provide any answer in this situation.
In those situations, a player decides how many times to execute a loop with a random event. As I understand the official policy, the player cannot choose a number that would take an unreasonable amount of time to resolve.
However, in this case there is no such decision. Scrambleverse's instructions are mandatory and the only choice the player is making is to play the card.
1. This doesn't really solve the issue, as even with this shortcut it would still take way too long to complete all the randomization.
2. This would give a different result. Consider the simple case of 19 tokens. With your method, the chances of a 10/9 split and a 19/0 split are the same. If you flipped a coin for each, the 10/9 split is much more likely.
Practice for Khans of Tarkir Limited:
Draft: (#1) (#2) (#3) (#4) (#5)
There isn't any provision for making sure the outcomes of any event have a specific distribution. In his example, every token is being randomly assigned per the result of 2 coin flips each, they're just being batch-processed as they're grouped into the results of HH, HT, TH, and TT.
I'm Mike, from The Mana Pool.
Check out my Tapped Out profile and comment on my decks!
This doesn't give the proper answer. Let's choose X=100 for simplicity. What you're basically doing is choosing a random integer from 0 to 100 and giving that many tokens to the first player.
The chance that I get all the tokens is 1/101. However, if I were to choose a random player for each of the 100 tokens, the chance that I get them all is (1/2)^100=1/1 267 650 600 228 229 401 496 703 205 376.
Quite a big difference.
There's potentially an additional problem with the general technique of determining how many each player gets: each token has a different timestamp, which may matter in the future.
Practice for Khans of Tarkir Limited:
Draft: (#1) (#2) (#3) (#4) (#5)
I'm not that good at explaining, look up the Law of large numbers if you don't understand why.
A solution which doesn't calculate timestamps is also problematic because even if it's irrelevant at the time it's possible that it will matter at some point in the future.
Practice for Khans of Tarkir Limited:
Draft: (#1) (#2) (#3) (#4) (#5)
Exactly. This is why the game must be a draw. The players will have two options. They can either 1) Go through each individual coin flip to resolve the spell (which means the round will go to time and be a draw), or 2) Choose to accept a mutual draw and move on to the next game. This was the conclusion that was reached when the Worldgorger Dragon player milled an Oath player for 100000000+, but the Oath player had 2 Blessings in his deck. There was no way to determine the exact card order without going through the motions, so the players were given the two options I have said above.
Mathematical modeling is awesome, but it is not part of the rules. Any solution that involves such modeling is not going to work.
-ktkenshinx-
Got a reference for this ruling?
The problem I see with option 1 is that the end of match procedure specifies 5 additional turns. But the current turn won't be ending anytime soon, so the players won't get to the point where the game reaches the end of the additional turns and becomes a draw. So the entire tournament will be paralyzed and unable to get to the next round.
This would also be really annoying in practice. If this occurs in game 2, whoever won game 1 would obviously choose option 1, since that would (assuming that it does lead to a game draw) given that person the match win.
Practice for Khans of Tarkir Limited:
Draft: (#1) (#2) (#3) (#4) (#5)
EDIT: At least, I guess it would be similar.
http://forums.mtgsalvation.com/showthread.php?t=201049
GX Tron XG
UR Phoenix RU
GG Freyalise High Tide GG
UR Parun Counterspells RU
BB Yawgmoth Token Storm BB
WB Pestilence BW
The point of having rules is so what happens in a given situation (where the facts are clear) is not up to the discretion of individual judges.
That is a similar situation. However, the closest thing that thread has to an answer is self-contradictory:
I agree with the first sentence. I don't see anything in the rules which allows this kind of situation to be resolved. The second sentence contradicts this by saying the game ends in a draw. I don't see anything in the rules which causes this situation to be resolved by the game being declared a draw (unless the players mutually agree). The failsafe in the rules which causes a game which enters an infinite loop to end as a draw does not apply since this is not actually an infinite loop.
Practice for Khans of Tarkir Limited:
Draft: (#1) (#2) (#3) (#4) (#5)
The exarch will remain untapped (unless the combo user is dumb) so more tokens can be produced after scrambleverse resolves.
Situation A. Player A gets Exarch, he makes more tokens to block any that are sent his way, they're all exiled at the end of turn, Player A does the combo again on his turn and wins.
Situation B. Player B keeps Exarch, makes more tokens, sends them over for the win.
If for some reason Exarch IS tapped, then I think a single coin flip to determine "who gets more" would be sufficient. You still have a 50/50 outcome of winning the duel. If you end up with more, you can block them all and die next turn when he does the combo again. Or he wins the flip, and likely has more than 20 Exarchs with that arbitrarily large number.
Edit: I forgot the Exarch wouldn't gain haste if it changed control. In that case just flip once for the winner. Of course this short cut doesn't need to be agreed on, but your chances are the same either way.