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  • เผยแพร่เมื่อ 15 ส.ค. 2024
  • What happens if you have a situation that you can't resolve with normal means? Judge destroyer combos, crazy amounts of coin flips, what happens if you can't just plow through normally?
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ความคิดเห็น • 268

  • @edoardospagnolo6252
    @edoardospagnolo6252 2 ปีที่แล้ว +522

    On this day, a new warrior was born: one adamant in bringing a modern deck including both scrambleverse and an infinite squirrel nest combo to a tournament and resolving it.

    • @daverapp
      @daverapp 2 ปีที่แล้ว +20

      I want to be that guy.

    • @bradjones7491
      @bradjones7491 2 ปีที่แล้ว +26

      honestly this would probably result in a DQ for the scrambleverse player because you would run out the clock trying to resolve it.

    • @PomesCollege
      @PomesCollege 2 ปีที่แล้ว +10

      I immediately though "Ah, its some sort of hilarious RG brew that doesn't have a way out from under Ensnaring Bridge or a mana generator to pay-through a Propaganda effect"

    • @ZoltarDeathNnja
      @ZoltarDeathNnja 2 ปีที่แล้ว +30

      Wrong - two warriors were born. One playing an infinite token deck in modern and another playing Scrambleverse. It doesn't matter WHO casts Scrambleverse as long as it does!
      These two warriors have one goal - play each other in the finals (or just at some point in the tournament)

    • @TeChNoWC7
      @TeChNoWC7 2 ปีที่แล้ว +21

      I’m confused why he didn’t see the more (well it appears to me) more feasible example of player 1 makes a million tokens, they have summoning sickness, player 2 plays scrambleverse with a chance now of winning.

  • @jonlamoreaux2228
    @jonlamoreaux2228 5 หลายเดือนก่อน +59

    This man is secretly goading us into creating ridiculous situations at tournaments

  • @SnugglesConquererofWorlds
    @SnugglesConquererofWorlds 2 ปีที่แล้ว +181

    The interaction is simple. After they create the million tokens and cast scrambleverse, you respond by ending the friendship

    • @EnderPryde
      @EnderPryde 2 ปีที่แล้ว +12

      Sorry, I already resolved my Nightmare Moon flip ability - by the rules, they are friends with me until the end of the game ;P

    • @psymar
      @psymar ปีที่แล้ว +2

      ​@@EnderPrydethe game ended when they flipped the table

    • @laytonjr6601
      @laytonjr6601 5 หลายเดือนก่อน +2

      In response to Scramble verse, I cast Shaarazade (I have a leyline of anticipation in play)

  • @sapphosfriend9558
    @sapphosfriend9558 2 ปีที่แล้ว +329

    The solution to mathematician-proof the scrambleverse thing is to have cathars crusade in play. Then every token has a different power so there's no reasonable way to simulate or approximate how it would resolve.

    • @daverapp
      @daverapp 2 ปีที่แล้ว +5

      You are a monster.

    • @nicholasfoster5065
      @nicholasfoster5065 2 ปีที่แล้ว +65

      Strictly speaking, the power and toughness at that point while different for each creature follows an easy ascending pattern. Since each token is n+1 larger than the last, we could use some calc to sum from 1 to n, where n is the number of creatures to get the total power [T].
      From there, the fair scrambleverse solution when unsolvable is to assume we are using a truly fair random assignment, thus giving each player equivalent probability. From there we use the judge shortcut rule to say "each player gets an equivalent number of tokens".
      Taking these two facts, the average power of a player's creatures can be determined. It can be assumed that an arbitrarily large creature can be blocked by 2 smaller creatures controlled by the defending player to match or exceed that large creature, thus killing it.
      From here, just assume board sizes are roughly equivalent and from there play such that combat advantages win the board. Nobody can profitably attack, and if 1 player attacks a 2nd, the third then controls the highest power and will thus win. Trample does nothing since everyone can block with a theoretically equivalent amount of toughness, but any other evasion-like effect wins if unanswered/unblockable.
      Of course, you could always write up about 20 lines of your favorite programming language and solve this randomly every time, but it won't be true random since there isn't such a thing unless assumed mathematically. =D Also idk why I wrote this up, fun puzzle I guess.

    • @user-et3xn2jm1u
      @user-et3xn2jm1u 2 ปีที่แล้ว +13

      @@nicholasfoster5065 There are some probability heuristics that could shortcut in a more elegant way, imo. For instance, you know that there is a normal probability distribution. Rather than giving equal tokens, you could roll a d6 (or any size die) to decide where on the probability distribution you land, between giving most tokens to one player, or most ot the other.
      You could do something similar to break up each player's mass of tokens into regimes of power and toughness, as well as assigning a "polarity" within those regimes (so one player might get high 300's, another low 300's, the low 300's have to double-block the high ones to destroy them). Although, in many games the size and polarity part would be irrelevant, because the player with more creatures in play almost always have enough spare creatures with large enough power for lethal damage, assuming sufficiently large numbers. Having one dozen extra 20/20's isn't that different from having two or three dozen in most games.
      edit: honestly, if you really wanted to, you could flip a coin for "who gets the bigger army" and leave it there, treating all other creatures as equivalent.

    • @FinetalPies
      @FinetalPies 2 ปีที่แล้ว +2

      Could still write a quick program to resolve that

    • @Ashrubel
      @Ashrubel 2 ปีที่แล้ว +11

      @@FinetalPies before it goes to clock? No you couldn’t.

  • @joshuahudson2170
    @joshuahudson2170 2 ปีที่แล้ว +36

    Playing scrambleverse on your own million tokens has a ruling: "unreasonably slow play"

  • @FinetalPies
    @FinetalPies 2 ปีที่แล้ว +87

    The game instructs to "choose at random" but doesn't specify how, as long as everyone agrees that a computer program is sufficiently random, that could quickly distribute a million squirrels to any number of players

    • @florasplace3404
      @florasplace3404 2 ปีที่แล้ว +12

      Oh. Right. Yeah. That's probably the play 🤣

    • @redstonepro5412
      @redstonepro5412 5 หลายเดือนก่อน +8

      It is easy to pick a number that a computer couldn't handle (assuming you have a cathars crusade so all the tokens are different)
      But it is also possible to make a turing-complete board state, and the outcome of a turing machine is proven to be undecidable.

    • @InterloperBob
      @InterloperBob 5 หลายเดือนก่อน +7

      ​@@redstonepro5412Even if you picked 52 factorial, a computer could still do a random distribution, as long as the players agree on some common sense parameters, like, divide the number by 2, then choose from 1 out of 1000 points on a bell curve away from 52 factorial /2.

    • @redstonepro5412
      @redstonepro5412 5 หลายเดือนก่อน +1

      @@InterloperBob what if i pick 52 factorial to the power of 52 factorial?
      but yes, it is possible to compute that for very large quantities. it isn't possible to calculate the outcome of a turing machine tho, and it is not that hard to alter the mtg turing machine in a way that one person wins if the turing machine halts, and the game is a draw if the turing machine doesn't halt.

    • @Idran
      @Idran 5 หลายเดือนก่อน +2

      @@redstonepro5412 Are you sure? I know that you can make a Turing-complete board state in the sense that you can make a board state that simulates a Turing machine in such a way that determining the _optimal play_ is equivalent to determining if a given algorithm halts, and so it's impossible to design a Magic AI that can always determine the optimal play in a given situation. But as far as I know, it's never been shown that you can set up a board state in such a way that _resolving_ that board state according to the Comp Rules is uncomputable. I'd imagine that resolving an arbitrary board state wouldn't even be NP-hard.

  • @dragonkingxd29
    @dragonkingxd29 2 ปีที่แล้ว +127

    I love the line "As a person, I don't identify as a computer" in response to questions like this. If the scenario isn't actually happening in a game, a judge doesn't owe anyone their time for unreasonable questions like that replenish example.

    • @raygun2180
      @raygun2180 2 ปีที่แล้ว +8

      Me on ask a judge: “Sooo… squirrel best scramble verse?”
      The judges:” I DONT OWE YOU MY TIME”

  • @epicnessbrian5295
    @epicnessbrian5295 2 ปีที่แล้ว +33

    Anyone who actually gets the scrambleverse scenario in a tournament would forever be known as a legend.

  • @Elekester
    @Elekester 5 หลายเดือนก่อน +6

    I think the Modern Squirrel Nest + Scrambleverse has a somewhat well known precedent in the Legacy Four Horseman deck. Four Horseman was a deck built around a non-deterministic combo that could take infinitely many iterations to resolve in a favorable state. Statistically, this would never happen, but the combo still took along time to resolve since the boardstate potentially changes in a meaningful way each iteration and couldn't be shortcut. It came to be consistently ruled that the combo could be considered slow play by the player of the Four Horseman deck, with the rulings that go along with that.
    As the combo from the video also cannot be shortcut, it seems reasonable to me that that it should be considered slow play by the Scrambleverse player.

  • @supermarth64
    @supermarth64 2 ปีที่แล้ว +40

    I like how Dave is a lot lower in his chair because he was dreading this question being asked and it aligns with his state of "wow I really don't want to answer this question". Just thought that was a funny coincidence.

    • @JonReid01
      @JonReid01 2 ปีที่แล้ว +1

      I noticed that too lol

    • @JudgingFtW
      @JudgingFtW  2 ปีที่แล้ว +23

      This is probably part of it, but I think another important factor is that I'm not disciplined enough to sit up straight for more than the length of a normal DDR.

  • @SpitefulAZ
    @SpitefulAZ 2 ปีที่แล้ว +53

    Speaking of questions that actually happened, here is one is that may go over a topic you haven't talked about: what if a player does a play based on an incorrect ruling they received from a judge?
    At MTG Vegas, i got a called to a table where a player asked If Grafdigger's cage will stop a player from casting a spell with cascade. I said no, it does not. The player responded with "well a judge told me earlier that it does, and that's the only reason i searched for the cage using urzas saga final chapter ability, i would like to search for a different artifact now if that's the case."
    What would you do? (This event was regular REL, i would love to hear what you would do in competitive, since that sounds a whole lot messier.)

    • @voltcorp
      @voltcorp 2 ปีที่แล้ว +22

      funnily enough I just came to this video from a newer one called "What happens when a judge makes a mistake?" so anyone wondering the same I suggest you go to that one.

    • @axvd2105
      @axvd2105 2 ปีที่แล้ว +10

      I had a similar situation happen to me, the head judge ruled that because the incorrect previous call did not lead to any illegal gamestates or game actions, I was not allowed to rewind or change what had happened. He merely informed me that I was given the wrong ruling and I had to live with the consequences. If the prior call HAD led to an illegal gamestate, I assume that the next ruling would be similar to what happens when an illegal gamestate is normally discovered: rewind if nothing too consequential has happened, carry on if otherwise (or something to that effect, I'm not a judge myself).
      In any case, a previous ruling being wrong seemingly will never absolve any player of any action they made as a result of it. They just say too bad and we just all shrug and move on.

    • @SpitefulAZ
      @SpitefulAZ 2 ปีที่แล้ว +10

      @@axvd2105 that's terrible! But i think that is the best course of action at competitive REL; otherwise players could get free taksie baksies. However, i remember once reading something about allowing a player to reverse a decision if the interaction is not what they thought, but i can't find it now so it might be an outdated policy.

    • @TheJacklikesvideos
      @TheJacklikesvideos 4 หลายเดือนก่อน +2

      @@axvd2105 of course, because "a judge said so" is an empty and infallible appeal to authority. inB4 "i only mulliganned down to four because some other judge told me before that i get ten more starting life for each card i put back. can i draw three more cards now?"

    • @ich3730
      @ich3730 4 หลายเดือนก่อน

      @@TheJacklikesvideos i will yoink that as a fun mulligan rule for casual commander

  • @florasplace3404
    @florasplace3404 2 ปีที่แล้ว +32

    Thinking through how to resolve the scrambleverse problem is pretty interesting.
    One potential solution I came up with is to evenly dole out the vast majority of the tokens (say, for example, each player gets 499,975), then flip coins for the remaining few (in this case 50). To me, this captures the spirit of the interaction in a reasonable mathematical approximation that doesn't take all day. By flipping for 50 squirrels, you're scaling it down to an imbalance that is within an order of magnitude of your life totals, so it doesn't just instantly set it up as a "winner take all" situation (like flipping for sets of 50,000 squirrels would for example.)

    • @thomassynths
      @thomassynths 5 หลายเดือนก่อน +5

      I would just make a bell curve and have my phone generate a position on this curve. Distribute the tokens in accordance to the curve point. The non-tokens get normal coin flips.

    • @ericfaulk2204
      @ericfaulk2204 5 หลายเดือนก่อน +8

      Or just simulate a million coin flips in Python. Should only take a few milliseconds.

    • @Userkaffe
      @Userkaffe 4 หลายเดือนก่อน

      If you distribute most tokens evenly, then only flip for 50, the chance of either player being able to attack for lethal is nonexistent. If you flip for all million, that chance is greater than 50%. The shortcut you are suggesting changes the state of the game significantly.

  • @joshuahudson2170
    @joshuahudson2170 2 ปีที่แล้ว +17

    I once proposed a very long shortcut. In a commander game, I had played Progenetor Mimic on Platnium Angel (my win condition was supposed to be play that on Verdent Force but I digress). I was into negative life due to being pounded. I proposed the shortcut. "I have a counterspell in my hand and nothing useful in my deck. You need two sources of infinite artifact busting to win, otherwise I deck you in four hours." (He couldn't have counterspell due to playing mono green.) He did not, but he could create an infinite loop so the game was drawn.

    • @presterjack9764
      @presterjack9764 ปีที่แล้ว +1

      I don't see how the game ended in a draw. It sounds like your opponent should have conceded. Did he have a way of stopping you from dealing combat damage? What was the infinite loop?

    • @joshuahudson2170
      @joshuahudson2170 ปีที่แล้ว +2

      @@presterjack9764 He was generating tokens from something else so combat would never go my way. He just said he had an unbreakable infinite loop in his deck and I saw no reason to not believe him, so that was that. (He had several infinite gears on the board already so it was probably true.)

    • @matthewgagnon9426
      @matthewgagnon9426 8 หลายเดือนก่อน

      @@presterjack9764 If you kick off an infinite loop and neither player has any way of stopping it the game ends in a draw.

    • @jakx2ob
      @jakx2ob 5 หลายเดือนก่อน

      ​@@matthewgagnon9426 but if they can stop it at any point, they are just stalling and need to take the L

    • @ryanjensen1945
      @ryanjensen1945 5 หลายเดือนก่อน +2

      @@jakx2ob Depending on your meaning of 'if they can stop it' this may not be true. If the loop only loops because of an active choice they make (for example, choosing a target for an ability) then they must stop. If they have a way to disrupt the loop actively (for example, an instant that could remove one of the loop pieces) they're under no obligation to do so, and can let the draw go through.

  • @xdavid00
    @xdavid00 2 ปีที่แล้ว +59

    I'm curious to what extent digital assistance is allowed, although as you said it's not really a situation where it matters. Like the for the Scrambleverse situation, a computer can relatively easily provide the results by simulating a million random selections (other solutions are possible even if each token was distinctly different). If an "official" result is required, would computers be able to help?

    • @JudgingFtW
      @JudgingFtW  2 ปีที่แล้ว +92

      If it's at an actual tournament, the judge staff can use whatever means they deem appropriate to answer a question. If it's kitchen table, the players can use whatever means they deem appropriate. There's no reason an official answer couldn't rely on digital assistance.

    • @magemanne7723
      @magemanne7723 ปีที่แล้ว

      Solution, make so many tokens that computer cannot solve it.

    • @Idran
      @Idran 5 หลายเดือนก่อน +2

      @@magemanne7723If you were going out of your way to make things as difficult as possible for the judges to resolve, wouldn't that just get you dinged for Unsporting Conduct?

    • @TheJacklikesvideos
      @TheJacklikesvideos 4 หลายเดือนก่อน

      @@Idran that's not unsporting, that's funmaxing.

  • @olivercarle
    @olivercarle 4 หลายเดือนก่อน +3

    I actually had the Scrambleverse one come up in EDH once but with Sai thopters instead of squirrels. Before casting Scrambleverse my friend (on a Grixis Chaos deck) asked me where I would have stopped making thopters (uh oh!) and I just said 10,000, expecting a Massacre Wurm. He drops The Scram, we all laugh, then agree to just take 2,500 thopters each before divvying the rest of the permanents as per card rules. Next turn we all mercifully die to a Craterhoof.

  • @bluerendar2194
    @bluerendar2194 2 ปีที่แล้ว +7

    For a time-limited tourney, the *eventual* "effective" answer to the infinite-squirrels is:
    Flip a coin. If the opponent calls it correct, they have arbitrarily more squirrels. Otherwise, the player has arbitrarily many squirrels, and can choose to attack to attempt victory with still arbitrarily many squirrels left behind. Next turn, then the opponent can choose to attack to attempt victory with the same conditions if they have more squirrels.
    The "functionally infinite" is since with a sufficiently large number, the chance of any fixed finite number advantage or less is vanishingly small, so need not be considered.

    • @ich3730
      @ich3730 ปีที่แล้ว +1

      Doesnt work because you cant have an arbitrary number of things. The rules need you to specify a number.

    • @Aaron0000014
      @Aaron0000014 4 หลายเดือนก่อน

      That does not make a lick of sense. What if you randomly divided the squirrels evenly enough that you wouldnt have lethal? It is unlikely but a possibility

    • @bluerendar2194
      @bluerendar2194 4 หลายเดือนก่อน

      @@Aaron0000014 The chance of that with "enough" squirrels is "effectively zero"
      That is, no matter how small of a probability you want of not evenly dividing, you can create enough squirrels to be under that probability.
      Certainly less than the chance of the coin landing on the edge, or even spontaneously combusting if you really wanted to

    • @Aaron0000014
      @Aaron0000014 4 หลายเดือนก่อน

      @@bluerendar2194 It depends on how many squirrels the person feels like summoning. Catch someone only saying 300 squirrels and the odds of it mattering are a lot different

    • @bluerendar2194
      @bluerendar2194 4 หลายเดือนก่อน

      @@Aaron0000014 oh 300 ain't *nearly* enough - that's only a standard deviation of 8.66 from the average, or like ~20% to have an imbalance of 20 or more
      But even for a relatively modest 300 million, on average one person has 273.8 extra from an even split, or that is, 500+ extra squirrels, and the chance for a

  • @Thatwasademo
    @Thatwasademo 2 ปีที่แล้ว +2

    I'm surprised nobody in the comments 7 months ago gave the simple answer to the (two-player case of the) scrambleverse question of "the binomial distribution is right there (and understood by every undergrad student, not a PhD level thing at all), this can be solved by generating a single random real", even the ones that were already reaching for computer assistance (why would you waste cycles simulating all million coin flips).

  • @jundachen9518
    @jundachen9518 5 หลายเดือนก่อน +1

    For scrambleverse on a large number of identical objects, use a random number between 0 and 1 and map it to a binomial distribution cdf and you're done.

  • @calliopehu1924
    @calliopehu1924 2 ปีที่แล้ว +4

    the soft sigh into a second take at 9:55 is such a mood

    • @JudgingFtW
      @JudgingFtW  2 ปีที่แล้ว +2

      If I ever get big enough to hire an editor and a voice actor, we will have multiple DDRs released every day.

    • @KVWI
      @KVWI 2 ปีที่แล้ว

      @@JudgingFtW I will, without a hint of irony, voice one of these for free if you ever decide to outsource any of the work, I think my friend might get a kick out of it.

  • @miaschwartz1074
    @miaschwartz1074 2 ปีที่แล้ว +6

    In regards to that last story about the abzan control deck, my L2 was telling a story about a tournament Where the finals were won with sultai food mirror was won because over the course of a 3 hour game because he had to give out 3 GRV to one player

  • @TheExFloridaMan
    @TheExFloridaMan 4 หลายเดือนก่อน +1

    The Scrambleverse example sounds like Pole Position in Yugioh (a long an convoluted combo that made it illegal for the opponent to put a card on the field [doing so creates an infinite loop which at the time you werent allowed to do]), my solution would be to coin flip for any unique creatures, and then apply the chance of any one coin toss to the lump of tokens (being 50 50) and splitting them in half

  • @Martini_1911
    @Martini_1911 ปีที่แล้ว +1

    This reminds me of the time I was playing my chaos EDH deck and cast a scrambleverse with a Hive Mind in play when one of my opponents had 15-20 token copies of an equipment that duplicates itself upon dealing damage. After I realized what I had done I scooped at instant speed to remove my effects from the stack since it would have taken probably at least 30 minutes to resolve that stack

  • @garscow
    @garscow 2 ปีที่แล้ว +3

    That's a great answer! Thank you for putting the effort in on a difficult question.

  • @zebmaxwell7979
    @zebmaxwell7979 2 ปีที่แล้ว +6

    Me and a few friends play an old dead format called Prismatic. This type of thing happens all the time. We usually take it to a vote if we can't figure out on the correct interaction is.

    • @Adrianovaz2007
      @Adrianovaz2007 2 ปีที่แล้ว +3

      Flipping a coin about a rules dispute used to be the end all be all decision back in the Alpha rulebook so you guys are still technically correct :P

  • @theemathas
    @theemathas 2 ปีที่แล้ว +5

    If I were to face the coin flip scenario, I would just write a quick short computer program that does a million coin flips lol

    • @DerpyLaron
      @DerpyLaron 2 ปีที่แล้ว

      That still take a while. That's a lot of operations. Randomness for PC is doing some math. Doing that a million times is gonna take a moment

    • @theemathas
      @theemathas 2 ปีที่แล้ว +2

      @@DerpyLaron Not if you use a fast (but not suitable for cryptography) pseudorandom number generator.

    • @antitheta777
      @antitheta777 2 ปีที่แล้ว

      Might be a good idea to already have it before the question comes up. In an earlier question I tried to write one on a calculator emulator, because that is all I would be likely to have for a judge call. Could not exceed 2,000 flips a minute.

  • @Blackstar-rf9yp
    @Blackstar-rf9yp 5 หลายเดือนก่อน

    The best bet to force the interaction is to put a silver bullet Scrambleverse in Peregrine Took food combo. It makes infinite game objects, Mana, and cards, which lets you dig for and cast your Scrambleverse.

  • @DtLS
    @DtLS 4 หลายเดือนก่อน

    3:55 A better scenario is for the opposing player to be playing Scrambleverse on their turn after someone has done an infinite creature generating combo. Now it's not just complex for the sake of making a judge's head hurt but instead is a possible out for them.

  • @elikorn8777
    @elikorn8777 2 ปีที่แล้ว +2

    So I was pointed to this video today and I now see why. I once had an infinite creatures with scapeshift. Somone combed of with krenko in edh but a player had played out moat and had some (4ish) blockers. The moat player also had a win on the untap (unblockable infect) The krenko player could only dig for about 5 or 6 cards and found scapeshift (he did have infinite mana). This was a edh tournament with a box on the line so we where trying to be strict. The judge in the end ruled that we should only scapeshift moat. If the moat and collosus ended up on one side. That player won. If alternate sides the one with the moat won. If all 1 million tokens ended up on one side, and the other had the moat, then moat won.

  • @Playingwithproxies
    @Playingwithproxies 2 ปีที่แล้ว +2

    You could roll a dice to decide the winner of the majority of the squirrels then roll the dice again to determine if they get more than a single squirrel as a majority. If they win a second roll they get two more squirrels a third 3 additional as so on. Until they lose a roll. Then do the regular resolve of the spell for all other permits.

  • @jaredcody6869
    @jaredcody6869 ปีที่แล้ว

    Thanks very much for the upload. I have really enjoyed it especially since I am building up my Chaos Commander deck again and seeing how best to tackle tricky questions and interactions in such things.

  • @jonhu4127
    @jonhu4127 2 ปีที่แล้ว +2

    In the scrambleverse example, I'd say that the match is immediately a draw as the million coin flips would take the entire round to time and well beyond. That's me, though
    Edit: wow, that resolution came up at the end of the video. I think I did pretty good

  • @mateuszboczkiewicz5170
    @mateuszboczkiewicz5170 ปีที่แล้ว

    WIth the scrambleverse, if it would've happened in tournament game, I'd determine random controler for 100 tokens, then multiply that outcome by 10 000 and that how many tokens every player gets. The question about Replenish is pretty simple - you resolve the spell. If you have multiple enters the battlefield trigger, you put them on the stack. Deal with it, its not that hard to do. With the Mastery of the Unseen - you help players determine the life total if needed, and that's it. If they play mirror match up with that card and cant kill each other, let the timer run out and that's it

  • @kylejohnson4662
    @kylejohnson4662 2 ปีที่แล้ว +4

    About the best answer you can give to the question that has no good answer.

  • @galelululu
    @galelululu 4 หลายเดือนก่อน

    “What would that do?”
    “I don’t know. Why did you do that. What’s wrong with you”

  • @Ragnarok633
    @Ragnarok633 ปีที่แล้ว

    I've been getting into magic lately and your videos have been a wonderful tool with making me familiar with the rules processes of magic. This is probably my favorite one though because it reminds me to not treat my kitchen table games with my friends like tournaments.

  • @theetiologist9539
    @theetiologist9539 2 ปีที่แล้ว +5

    Might not happen in modern but this kind of scrambleverse thing happens in commander all the time.

  • @tehmadflip
    @tehmadflip 2 ปีที่แล้ว +3

    I remember a commander game, with warp world, and big board states. The ETB triggers took us forever to stack and resolve.

    • @mn6334
      @mn6334 2 ปีที่แล้ว

      I was in a Commander game a month or two ago where someone Radiated a Chaos Warp. After taking about 5-10 minutes to work out the best way to even start resolving it I eventually just conceded and left to start a new game. I think the two people who actually tried to play it out took at least half an hour to resolve things.

    • @kyubikirby
      @kyubikirby 2 ปีที่แล้ว +1

      one of the few times I scooped at instant speed in a commander game was when a player had out Hive Mind, the cast a Warp World... I'm not sitting through 4 resolutions of warp world.

  • @colacadstink
    @colacadstink 2 ปีที่แล้ว +2

    GOOD GRIEF Abzan Control was the WORST!! My first large event as an L1 was Khans standard, and everyone - judges, players, didn't matter - hated how long matches were going! I thought I'd blocked those memories out... And now they come rushing back... :P

  • @kimcoleeppling8177
    @kimcoleeppling8177 2 ปีที่แล้ว +17

    “If it could actually come up in a magic tournament, there’s no reason why you as a judge couldn’t go through all the steps necessary to take the answer down.”
    I know you’re _really_ going to hate this question, but there is a legacy-legal decklist that can assemble a game state on turn 1 where no players can make any meaningful decisions for the rest of the game, which is forced to play out as a universal Turing machine running a program of the player’s design. This program may run forever, in which case the game is a forced draw, or it may end, in which case the player who started the combo wins. However, it is possible to set up the program in such a way that the answer to “does the program ever finish” is provably independent of the standard axioms of mathematics. In other words, you can’t tell if the player wins or if it’s a draw, and if you bring in a mathematician they’ll tell you it’s impossible to know.
    If it’s mathematically impossible to determine the outcome of a sequence of game actions in finite time, where do you go from there?

    • @tamsinm
      @tamsinm 2 ปีที่แล้ว +9

      If you're at a tournament, there's a limit of 50 minutes (plus 5 turns) on the match. Get the mathematician to work out if the program would end *within the allotted time* assuming a reasonable rate of play (1 action per second, maybe? to be generous) - if not, it's a draw.
      Also suggest to the Turing machine player that they buy a lottery ticket, because they must have phenomenal luck to draw the right opening hand to set this up.

    • @888ian2
      @888ian2 2 ปีที่แล้ว

      doesnt the Turing machine only work if the opp always chooses yes?

    • @kimcoleeppling8177
      @kimcoleeppling8177 2 ปีที่แล้ว +1

      @@888ian2 there was a paper in 2019 with a decklist where your opponent can't do anything because everything they own is exiled and they don't draw, and the closest thing to a choice is when you're forced to cast a spell that asks a target but there's only one legal target

    • @spongemanhere
      @spongemanhere 2 ปีที่แล้ว +1

      pretty sure at that point you just go home

    • @electra_
      @electra_ ปีที่แล้ว

      I was thinking the same thing. It's weird to have a situation where it's impossible to know whether it's a win or a draw.
      Realistically though, you probably just call it a draw on time.

  • @The_Arisen
    @The_Arisen 5 หลายเดือนก่อน

    These hypotheticals strike me as the magic equivalent of zen kōans. And this is a channel of academia. At least, so far as I'm aware.

  • @SpitefulAZ
    @SpitefulAZ 2 ปีที่แล้ว +1

    Thanks for the long video. I think this was a Great response to such a question.

  • @ManSubhu
    @ManSubhu 2 ปีที่แล้ว +1

    Scambleverse squirrel nest statistics say that somebody will get a lot more squirrels than the other side. Toss a coin and whoever wins is assumed to have enough squirrels to overwhelm the blockers on the otherside and win.

  • @Roan_Harris
    @Roan_Harris 4 หลายเดือนก่อน

    Times like these it is important to remember that conceding is instant speed

  • @espicelmecanicodecombustio1632
    @espicelmecanicodecombustio1632 2 ปีที่แล้ว +2

    Imagine being a judge and getting called over what RNG used should be used to simulate distributing millions of squirrel tokens. MTG can be a silly game sometimes

  • @korytoombs886
    @korytoombs886 2 ปีที่แล้ว

    Scrambleverse = My solution = Everyone gets half.

  • @mstieler8480
    @mstieler8480 ปีที่แล้ว

    I'm guessing the "Replenish" question falls down to "OK, how many of these have continuous effects that work above the layer of the Humility & Opalescense that are for sure in this 75-enchantment pile because what enchantment-shenanigan player is without them" and ignore everything that happens under :D

  • @RandonActs5
    @RandonActs5 2 ปีที่แล้ว +2

    If there were such a game where one person made a million tokens and the opponent then played scrambleverse I think I would boil the question down to a statistical analysis question and figure out the likely hood of each distribution of the million tokens and then have one of the players roll percentages to figure out where on the scale they landed, as for the rest of the permanents on the field they could just flip as normal. If it were a commander game instead of a 1v1, I have no idea then the different distributions get even crazier then.

    • @deathinabox9100
      @deathinabox9100 2 ปีที่แล้ว

      If I remember my stats class correctly, random events get significantly less "random" the more times they are repeated. 1 mil is an insanely large number, to the point where you'd just have the math out the average distribution and it'd be functionally guaranteed to result in that exact distribution.

  • @nicholasiverson9784
    @nicholasiverson9784 2 ปีที่แล้ว +3

    The best answer I feel for scrambleverse with arbitrary large number of tokens is - assume it's Very unlikely the the difference between the number of tokens each player gets Isn't going to be small. One person will have a Lot more tokens than the other almost every time. Even if they're "close" close in this case is within several hundred of each other, which will still lead to a players imminent demise. Do the normal procedure for Scrambleverse on the board except for the tokens, flip a coin to see who "wins" in the token distribution, and if the game makes it to a relevant combat phase before someone blows up everything - they can win or lose from there. So there's no ambiguity they can agree on a finite split before the coin flip, say 600,000 to the winner 400,000 to the loser or something like that.

  • @DragonPriem
    @DragonPriem ปีที่แล้ว

    The large the number being randomly divided the closer the distribution would be to the statistical average. So splitting the squirrels in half would be the simplest and most likely result anyway

  • @xd_lead
    @xd_lead 2 ปีที่แล้ว +2

    Thank you! This has been enlightening as well as compelling to get an infinite tokens deck with scramble verse player haha

  • @colebear
    @colebear 4 หลายเดือนก่อน

    for the scramble verse id say if you have any unique creatures flip for them then divde the rest evenly as that's how itd end up
    if odd numbers flip for them

  • @tubenewsintern6446
    @tubenewsintern6446 5 หลายเดือนก่อน

    For the Million Squirrel problem, I'd say that since someone will be extremely likely to end up with 40+ more creatures than the other, that you could just flip a single coin

  • @markmccauley8460
    @markmccauley8460 2 ปีที่แล้ว +1

    You know what Dave, I will have an awesome day. Better than great.

  • @MrBotofdoom
    @MrBotofdoom 2 ปีที่แล้ว +6

    I think in the scrambleverse scenario it's more reasonable that two different players have the two halves, one has a million tokens and the other wants to scrambleverse, neither person is doing something "unkind" to the other like when one person would do both. Following the rules exactly will always end in time running out and shortcutting by splitting the tokens half-and-half is against gamerules. I will be stepping on the toes of many here but i still think splitting half and half leads to the best and fairest result.
    Yeah this is a tough one though, understandable frustration there.

    • @matthewspear2053
      @matthewspear2053 2 ปีที่แล้ว +2

      Splitting half and half is not a very good representation of the end result though. There is only a 1.6% chance that one player doesn't get at least 20 more squirrel tokens then the other player which would give them lethal. So 49.2% player one wins with an attack, 49.2% player two wins with an attack, and 1.6% it takes more than one attack. The "fairest" way to resolve this if you don't use a program to flip a million coins is to just flip a coin to determine who wins. Assuming that no players have anything to change the board state.

    • @JuniperHatesTwitterlikeHandles
      @JuniperHatesTwitterlikeHandles 2 ปีที่แล้ว +2

      This exactly, especially in commander, where infinite token generation and scrambleverse are both key components of different very popular deck archetypes. I mean, idk how likely it is for someone to bring a chaos deck to a cEDH table, but if they could get this to happen at a sanctioned tournament... well that's the kind of thing that a person who plays chaos would find very funny, I wouldn't be surprised if this has happened somewhere.

  • @harriyama
    @harriyama 2 ปีที่แล้ว

    I don't care about rules and interactions that influence my games. I want crazy and whacky and convoluted rules. That's why I watch your channel other than just googling the interaction. So yes, please do a lot more "I bring back these 75 enchantments from my graveyard, what happens?" videos.

  • @Enja_Near
    @Enja_Near 5 หลายเดือนก่อน

    I have successfully put Scrambleverse under an Eye of the Storm against someone that made infinite squirrel tokens in the early days of commander.
    Fun stuff.

  • @only1sn1not1taken
    @only1sn1not1taken 5 หลายเดือนก่อน

    I had an actual game come up where it was Scrambleverse being copied several, several times. We just scooped because there were multiple activated abilities that would drastically change the boardstate between each Scrambleverse. It was not worth playing through the sequence of events.

  • @rylandmalcolm3825
    @rylandmalcolm3825 2 ปีที่แล้ว +1

    In the spirit of asking for nteractions with fewer peices.
    Thawing Glaciers and Amulet of Vigor and Patron of the moon.
    When thawing glaciers is returned to the hand at the end of turn, would the player get priority at any point to bring the land back down?
    Would they be able to activate it's abillity again? Would there even be a seccond endstep to repeat it all?
    These vidos are straight to the point and super clear!

    • @herbert164
      @herbert164 2 ปีที่แล้ว

      from what i read about the cleanup step, there will be a round of priority to respond to the trigger of thawing glaciers. Once the trigger resolves, however, there will be no priority to activate patron. If you had 2 thawing glaciers or a vesuva, you could respond to the trigger of thawing glaciers to put in the other land. However, since it triggers at the beginning of the cleanup step the 2nd will not trigger until your opponents cleanup.

  • @pa7764
    @pa7764 5 หลายเดือนก่อน

    I thought that with effects like Replenish or Rise of the dark realms, the player casting the spell chooses in which order the cards go on the battlefield? This lets them start with all the "whenever an enchantment etb" cards, then double triggers, then ETB enchantments, then all the remaining ones, one by one as though they all went on the stack together. Which they probably do and the opponent could respond to individual enchantments entering (like blowing up the blood moon immediately before the next enchantment enters).

  • @psymar
    @psymar ปีที่แล้ว

    I feel the solution to million squirrel scrambleverse is get a computer to flip coins and tabulate the results.

  • @mallie2057
    @mallie2057 2 ปีที่แล้ว

    I believe Frank Karstens PhD dissertation was on a problem related to magic the gathering

  • @SuPeRMeGaJoE0
    @SuPeRMeGaJoE0 ปีที่แล้ว

    Okay. If I were judging a tournament and someone generated a million tokens and played Scrambleverse, I know what I would do. Unsporting Conduct--Stalling. IPG says its a DQ-able offense, but I would probably just issue a game loss.

  • @linkXness
    @linkXness 2 ปีที่แล้ว

    About that Scrambleverse-Thing.
    Player A makes a million Creature Tokens, then attacks the whole EDH Table.
    Player C tells the others to not scoop and plays a fog, announcinghe has a way to cleaar the board.
    Player Bs turn starts, and he is a mono red Chaos player, so he plays brambleverse
    Heres how it can logically happen in a game of magic.

  • @Mechsrule1
    @Mechsrule1 5 หลายเดือนก่อน

    How to respond to scrambleverse with a million tokens? Well there are coin flip programs that can simulate any number and give you the results (and if you can't find one, they're very easy to code). The judge asks one player to pick heads or tails and runs the program for the number of identical tokens, and divide them up that way, then handle all the normal permanents with physical coins or dice or whatever

  • @apocalypticoreos6460
    @apocalypticoreos6460 2 ปีที่แล้ว

    for the scrambleverse question, i would probably just randomly assign everything that was not a squirrel, and then give out an even number of squirrels to each player (even with respect to how the assignment is being done, if its gonna be uneven)

  • @telph3223
    @telph3223 2 ปีที่แล้ว +2

    "Alexa, flip 1 million coins"

  • @HavoCentral
    @HavoCentral 2 ปีที่แล้ว +3

    The answer, in casual commander at least, is move on to the next game.
    Had someone with Thousand year storm, storm count of 57, cast warp world. That would have took hours to resolve considering all of the ETB

    • @dstreetz91
      @dstreetz91 2 ปีที่แล้ว +1

      This goes beyond warp world into warp the multiverse. I love warp world, but 58 copies of it? Nah.

  • @creepercraft1107
    @creepercraft1107 2 ปีที่แล้ว +4

    Why didnt you answer the scrambleverse question? You posed it to yourself only to not answer it, instead saying IF you are playing it at low level to answer it yourself, why not just answer or not make a video on a question you can't answer?

  • @rowanedmunds7574
    @rowanedmunds7574 2 ปีที่แล้ว

    I actually was in a situation where a CMC matters spell deck cast a scrambleverse on a board state with something like 200 tokens (not infinate but a bunch of token doublers) to resolve it there was rolling for sets of 10 tokens to save time.

  • @DragoSmash
    @DragoSmash 4 หลายเดือนก่อน

    "all right sirs, we are doing this, both of you are getting 500,000 squirrels, because that's the statistic half, you sir, are getting a warning for playing Scrambleverse intentionally in your own board state and will be take a game loss due to slow play if you continue doing it in further games"

  • @WilhelmScreamer
    @WilhelmScreamer 2 ปีที่แล้ว

    I appreciate the sensible responsse

  • @JD-eq1gk
    @JD-eq1gk 5 หลายเดือนก่อน

    Funny enough if Nick really wanted to push the issue and you had a computer nearby you could write a C++ or Matlab script that performs the coin flip using a uniformly distributed random variable and the Rand function and then plot the histogram histogram and actually get the hundred something thousand results

  • @SethPentolope
    @SethPentolope 5 หลายเดือนก่อน

    It it were my group, I would vote that everything is finite and we use the python interpreter app on my smartphone to do this:
    import random
    print(sum([random.randint(0,1) for _ in range(numOfTokens)]))
    And that’s how many tokens one of those players gets. The other gets the rest. It isn’t too hard to extend if there are >2 players.
    Like you said, the rules of the game are whatever you can convince the rest of table to agree with. Like placing a +2 on another +2 in UNO.

  • @MrIluvbutts
    @MrIluvbutts 4 หลายเดือนก่อน

    I can understand why you are a judge. Because the way you speak is clear concise and intelligent

  • @Deh9o11en8or
    @Deh9o11en8or 5 หลายเดือนก่อน

    for the second one I'd probably go with having a computer simulate the coinflips (via something like anydice) and then assign the tokens in bulk

  • @BrotherAlpha
    @BrotherAlpha 2 ปีที่แล้ว

    Cast Scrambleverse and then Massacre Worm. This would kill all of the opponent's 1/1 and make them lose 2 life for each.

  • @jmr5125
    @jmr5125 ปีที่แล้ว

    Well, I think a new rule is needed to address these issues. I suggest:
    If a player has a board state where victory is inevitable casts a spell or activates an effect (or any combination of spells and effects) which renders the board state indeterminate in a way not otherwise resolvable by these rules, that player loses the game.
    It could be called the "Don't screw with the judges" rule. :)

  • @drycereal1
    @drycereal1 5 หลายเดือนก่อน

    I think you should take a normal distribution segmented into 20 parts, roll a 20 sided die, and distribute the tokens based on the point on the distribution curve at that point. Easy

  • @guybuckridge7326
    @guybuckridge7326 2 ปีที่แล้ว

    There are two main solutions to the Scrambleverse problem, based on circumstance. The first being the simplest as it assumes that the Token player has active infinite combo and all they need is a turn to attack. In this case the Red mage is just out of luck as the Token player can deploy an extra million 'squirrels' after the sorcery resolves and we have a dead mage soon after.
    If the combo is stopped somehow (i.e. it's single turn effect or the Token player loses a required permanent) then you have the quandary presented to us. My solution for that is divide the non-squirrel creatures as normal for Scrambleverse then divide the squirrels evenly. Mathmatically as the number of tokens approaches 1 million the variation will tend very strongly towards even, leaving just a fraction of a percentage of difference. I think is the best answer if the same player has done both things since they wanted to stall the game but you have effectively neutered the obscene Token part of the game (by letting them neutralize each other) but letting the swapping potential of the other creatures be relevant. If it was two different players where they each wanted to win you could offer them a die roll to simulate the inevitable tiny percentage of variation between the squirrel numbers. Once they chose who had more I would use a D20 to determine the difference between the two players which would have a high chance to quickly result in a winner.
    Bidet

    • @Zeekfox
      @Zeekfox 2 ปีที่แล้ว +1

      I would almost say the end result is practically flipping a coin twice to decide the game. The first flip is to decide if the token player gets to keep Earthcraft. If they do, they can just make more tokens and they win. The second flip is essentially who gets the greater share of tokens. With a million 1/1's, while it's practically a statistical guarantee that each player will get 499,000 of them, what of the other 2,000? It's quite likely one player would get at least 20 more tokens than the other, and without further intereference, whoever gets that many more tokens wins on their next attack. And while there is technically at least some chance of the million tokens going either exactly 50/50 or within, say, 5 tokens one way or the other, I believe that's unlikely enough to be considered.

  • @zukiginagato2215
    @zukiginagato2215 2 ปีที่แล้ว

    These happens a lot on commander to be fair... usually we simply see 'will u win the game?'
    If not, just average results.

  • @Aaackermann
    @Aaackermann 5 หลายเดือนก่อน

    The "solution" could be an online tool like a coin flip simulator, where you can enter a number (like 1,000,000) and look at the result, if the opponent is addamant on getting a final result.

  • @magemanne7723
    @magemanne7723 ปีที่แล้ว

    How I see scrambleverse being played is that squirrel guy doesn't win on turn the rodents hit the field, opposing burn player have sided scrambleverse for game 2 and plays it.

  • @ademisc
    @ademisc 2 ปีที่แล้ว

    And the turn prior to Scrambleverse somebody had the nerve to put down an Eye of the Storm...

  • @spazmatCc
    @spazmatCc 2 ปีที่แล้ว

    I can say the few times this has happened in commander, I will usually quit the game entirely, remove the problem pieces from my deck, and never play that combo ever again. I highly dislike having infinite Revillark combos so I have to be careful when using it in a deck or strategy.

  • @WingLMui
    @WingLMui 5 หลายเดือนก่อน

    fine i'm gonna make one million creatures and scrambleverse in a game now

  • @yuyu63
    @yuyu63 2 ปีที่แล้ว

    The scramble verse issue is came up a few weeks ago with a barid stuard of Argive under opponents control

  • @jtvanilla1776
    @jtvanilla1776 5 หลายเดือนก่อน

    in the case of the scrambleverse with a million creatures, i would roll d100 x1000 for each player and give that many to each player, then d100x100, then d100x10 etc., until all have been distributed

  • @JervisGermane
    @JervisGermane 5 หลายเดือนก่อน

    In summary, judges answer rules questions, not thought experiments.

  • @lukasm5254
    @lukasm5254 5 หลายเดือนก่อน

    And there I was thinking this is my solution to Riemann Hypothesis, MtG is Turing complete, so set up a state that stops if it is true, and then go ask a judge and ask if this is a draw or if we would go on.

    • @JudgingFtW
      @JudgingFtW  5 หลายเดือนก่อน

      Only problem is that the judge would just claim the $1million prize in response

  • @martinmackey7191
    @martinmackey7191 11 หลายเดือนก่อน +1

    Basketball rules guru here with a similar rules situation that can arise. I've been asked "what happens if someone inbounds the ball while the clock is stopped, but nobody pursues it and it comes to rest on the floor, couldn't we be there forever since the clock won't start until a player in bounds touches the ball?"
    Yes, but we don't need a rule for that, since both teams will want the ball and someone will go after it.
    So sometimes rule questions which appear to have no governing rule don't actually need a true answer.

  • @robertbcardoza
    @robertbcardoza ปีที่แล้ว

    Sounds like there should be some standard calculators to quickly resolve arbitrarily large interactions like this.

  • @matterhorn731
    @matterhorn731 2 ปีที่แล้ว

    The million tokens plus Scrambleverse thing seems like it might be a little more likely to come up if coming from two different players. Like if Amy has a two card combo that makes a million tokens but doesn't have haste, and then *_Nick_* untaps and casts Scrambleverse. Nick has a 25% chance of taking Amy's combo and winning himself, plus a 50% chance of disrupting the combo by taking one piece, at which point he'd have a roughly 50% chance of taking enough tokens to survive Amy's attack, maybe even winning on the crack back.
    (Why is Nick running Scrambleverse in the first place? No clue. Maybe he's just a troll.)

  • @TheYoutublover96
    @TheYoutublover96 5 หลายเดือนก่อน +2

    A questions about the first situation with Frenetic Efreet, since efreet phases out or buries of the first flip, does efreet not become an illegal target causing the remaining flips to fizzle? new to magic and thought when a card leaves play any remaining instances of its ability or upcoming effect on the stack goes away, is this wrong? e.g. ligtning bolt targeting a 1/1 in response to the 1/1 tapping for an effect/activating a triggered ability; Does the tap effect/ability of the 1/1 resolve after the LB?

    • @nathanweise8160
      @nathanweise8160 5 หลายเดือนก่อน +1

      Frantics Effret's effect does not target frantic efreet because it does not use the word "target", so it does not fizzle due to all the targets being illegal. Abilities on the stacker independent of their source. If you lightning bolt 1/1 in response to its effect the effect still resolves and does as much as possible.

    • @TheJacklikesvideos
      @TheJacklikesvideos 4 หลายเดือนก่อน

      if you're new to M:tG, you should read the rules. 113.7a clarifies your misunderstanding. Once activated or triggered, an ability exists on the stack independently of its source. Destruction or removal of the source after that time won’t affect the ability. [...] if the source is no longer in the zone it’s expected to be in at that time, its last known information is used. The source can still perform the action even though it no longer exists.

  • @untappedlands9853
    @untappedlands9853 2 ปีที่แล้ว +1

    honestly Scrambleverse seems like an interesting sideboard against infinite creature combos, also Scute swarm with Mutate has been crazy for me to figure out in paper magic because each token becomes unique to every other token. Just won't play it in Paper magic.

  • @phatkin
    @phatkin 2 ปีที่แล้ว

    solution for the scrambleverse problem: whip out your laptop and write a quick python script

  • @feyfiren
    @feyfiren 2 ปีที่แล้ว

    8:15 i cracked up rofl when i saw the list of enchantments xD

  • @brianarsuaga5008
    @brianarsuaga5008 5 หลายเดือนก่อน

    I would TOTALLY leyline of anticipation out a scrambleverse lol

  • @bebits3173
    @bebits3173 2 ปีที่แล้ว

    "this type of things never happens..."
    *laughs in chaos*

  • @Closer2Zero
    @Closer2Zero 2 ปีที่แล้ว

    Some people want to watch the world burn... and some people just wanna watch the Squirrels dance

  • @themostdresden
    @themostdresden ปีที่แล้ว

    You make me want to be a judge. Well answered