Weird Magic: Nondeterministic Combos
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- Опубліковано 19 жов 2024
- Everyone's favourite frog player might have attempted to explain nondeterministic combos and mathematical convergence at your kitchen table, but what do the rules actually say about it?
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If you're wondering how this all applies to The Frog, In case the video wasn't clear, the rules say you can't actually do that whole combo in a tournament. But if you're an American, the third amendment says you don't have to let any level 1 judges into your house, so go nuts.
this is the best follow up you could have given lol
It doesn't apply to Gitrog though.
"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."
Since during the gitrog's combo you are always milling or adding draw triggers. and if you mill a shuffle effect right away, you just respond by dredging and getting more draw triggers, the loop is constantly changing the game state, and therefore not slow play.
@@deathkitty23 Gitrog loop can end in a slow play violation, however, as gitrog loop into shuffle titan twice *is* considered slowplay, as dredging the Titan into the yard twice is basically doing nothing to advance the gamestate- If,you haven't milled any lands in either dredges. Gitrog is definitely the one least at-risk, though, so that's good
@@manarager3413 that only happens if your list runs more than one eldrazi titan.
Eh, that Third Amendment is very literally understood. Otherwise American taxpayers having to fund not only the world's richest armed force but also the world's _third_ richest armed force in American Policing, an armed force which is designed to target _only_ the very same people funding them, would probably be considered a violation of the ban against forced Quartering
The moral is, if you have infinite mana and can untap artifacts infinitly, you should have a better way of winning than wirefly hive
It DOEs matter if the mana you can produce an infinite amount of is colorless, a specific color or of any color. The example given does allow for any combination of colors, since you net an "extra" Blue every time you activate for Blue, so you just need to remember to call "Blue" every other activation until you have enough of the other colors you need.
But even with colorless mana there are more consistent options than Wirefly Hive, even Serpent Generator would be a better choice here because Poisson laughs at Life Total. However I don't think any other option would be as strong of a flex as winning via Wirefly Hive and sometimes all a Johnny wants to do is flex on the Spikes.
@@vladspellbinder Flexing on the Spikes is the best part. Like Donating a Blood Funnel to a Landstill player. BWAHAHA
@@vladspellbinder Myr Matrix for example. Indestructible, gives all myr +1/+1 and produces myr for generic mana without tap.
Yeah, like Aladdin's Ring!
just get rocket launcher lol
One thing most people aren't mentioning is that the first heads flip is functionally guaranteed as you can flip any number of consecutive tails with no negative impact. This means that for any required number of 2/2s, reduce the "heads" flips needed by one. That's all!
But that assumes that they flip heads at all - and there is a chance they might never flip heads
@@petermaguire8139 sure in theory, but if I make an infinite combo and say "I'll flip until I flip heads" which is an accepted until condition on a combo, provided that the tails don't impact the game state (which they don't until you flip your first heads).
@@petermaguire8139 unless the coin is tails on both sides the probability of "never flippi g heads" is 0.
I'm not that familiar with magic but wouldn't the wirefly hive destroy itself if you flip incorrectly? It says "destroy all wireflies" and it is a wirefly since it has wirefly in the name, no?
@@johnnickfanaccount3492 It wouldn't destroy itself. The text on this card has been oracled to be clearer about this. Now it says "destroy all permanents named Wirefly" in magic this always means that they need to be named exactly "wirefly".
Also it would be a pretty terrible card if it did destroy itself.
"Whenever an artifact comes into play, you MAY gain 1 life"
Leonin player could just choose not to and lose staunchly on their own terms, a respectable course
And THAT'S why you always put Krark's Thumb in a deck where you're flipping coins. Turn that 50% heads into a 75% heads!
What about Krark's other thumb?
@@stopmotionstupidity2440 Nah, that's only for dice. Krark is not a clever goblin.
I am a mathematician and spent an hour researching what the probability of this problem was. it is more difficult than what it seems. The sum of these events converge to a small number. The best way to proceed is therefore to make a single launch but the probability of happening is 6-7%
6.4 better
Every Krark Skashima player ever "Fine I'll play it out. Time to play solitaire for an hour or two"
At 1:12 "But loops unfortunately cannot be used when the outcome is random"
Loops can still be used, shortcut for loops that can't. Small but relevant difference
If the outcome is random then it isn’t a loop. A loop in magic is a sequence of infinitely repeatable actions that doesn’t repeat effectively identical game states. I can tap a basalt monolith for 3 mana and use 3 mana to untap basalt monolith infinitely, but because it doesn’t do anything it doesn’t qualify as a loop. If however I generate more than 3, pay fewer than 3, or trigger an ability when the basalt monolith activates, becomes tapped, or becomes untapped, then the game state after each iteration is different and non-repeating which can be counted as a loop.
@@anthonycannet1305 the game state here gets permanently altered by the other player's health going up, no?
@@LyonsTheMad but it goes up by a random amount every attempt, so you cant just say, i do this X times, because there is not way to predict what will happen
@@devforfun5618 well yes, this is true of any finite number. But if we say "I do this infinite times" then you can calculate what the odds converge to, since it's just an infinite sum. If the odds don't converge to one, you just use a random number generator to determine success according to the chance of success over infinite trials.
@@LyonsTheMad but you can't say i do this an infinite amount of times because then the games doesn't end, there needs to be a finite condition
The algorithm decided to bless me with this video. Man your content is nicely edited and structured, hope to see more in the future.
Same. Let´s see if our paths cross again in the infinite future of the algorithm.
The algorithm lile syou back. Do you want a replila??
One of those rare cases where it's a really complex board state but the ruling is simply "Don't be a dick"
I think the comments here prove why these rules exist. Most people, including most MTG players, don't understand math well enough to realize that having a sum of infinitesimal chances to win does not equal a 100% chance to win, not even with infine time.
great video on a neat mtg thing I love. I've heard of this little scenario before but your presentation was super concise, clear and simple which is awesome, this vid is my new way to introduce people to this funny concept! good stuff!
I like this, currently designing my first paper modern deck and requires a non deterministic infinite combo so its nice to know that my opponents will have to wait for me to dig into my library and hit them in the face with comet storm
Technically they can concede at any point and don't _have_ to wait for you to finish your combo. Most players _will_ wait though to ensure you _actually have it_ as some people have been known to start taking the "I'm finding my combo" steps and _not actually have the win condition_ (for whatever reason, such as having sided it out after game one) hoping their opponent would concede for them.
@@vladspellbinder the infamous Mike Long bluff!
@@vladspellbinder isn't that illegal though? like if the judge found out about them not having the win con they would be DQ'ed?
@@JohnnyYeTaecanUktena No, it is not. The whole thing is that in a previous game you show your opponent that you have the win con by using it to win, then in a later game you play a search spell that would allow you to find the win con and the opponent concedes instead of waiting for you to search. At no point are you performing an illegal operation, all plays are legal.
Search effects have a built in "failed to find" clause, that if you can't find what you are searching for the spell still resolves fine. So if a spell would cause you to "search your library for an Enchantment card, reveal it and put it into your hand then shuffle" but you have no Enchantments in your deck you search your library, find nothing and then shuffle.
Since a player can concede at any point in the game no matter of what is being resolved it is their fault for not waiting to see if you actually find your win condition instead of just assuming you have it and giving up.
@@vladspellbinder if i recall, people would build Dragonstorm and not put any dragons in. When they hit their combo and played it people would generally scoop anyway.
Subscribed for this one. Interesting scenario, clear rules explanation, no wasted time in video. Good stuff
The answer is simply no, you shouldn't win. This is a parallel to the paradox of Achilles and the Tortoise. Infinitely many chances decreasing in proppability can combine into a finite chance less than 100%.
I would disagree. The parallel is quite nice, but the conclusion that you shouldn't win is false. Just because the probability that you win after infinite tries does not sum up to 1 does not mean that you can under no circumstances win. There is clearly a chance larger than 1/8 for you to win.
I guess in a real turnament you could probably try a few times before the judge would have you either stop or loose the game for stalling
@@jakobmoderl3331 Yeah, you are right. I might have not expressed myself well. I meant it as "you shouldn't win on the spot as you aren't guarenteed to win".
So I messed around a little bit with this thing and found that out:
P(x) = (2P(x+1)-P(x+2)/2^x-P(x+3)/2^(x+1) +3/2^(x+1))/2
where P(x) is the probability of winning when your opponnent has x life points, assuming you never stop trying to summon the wireflies.
so i wrote a program that could compute backward using this formula, that is from P(x),P(x+1) and P(x+2) for large x down to P(4)
since the probability of winning when your opponnent has a lot of life gets very low when x is big, setting P(x), P(x+1) and P(x+2) to 0 can yield good approximations.
by setting P(101), P(102) and P(103) to 0, i wrote a program that found this approximation of P(4): 0.09253236056061417
then, I set P(1001), P(1002) and P(1003) to 0 to see if i would get a better approximation, but I just got the same number.
my result fits nicely between the obvious lower bound of 1/16 and the 1/8 upper bound found by Benjamin Buchan.
why would you never stop trying to summon once you have lethal? interesting though
Hell yeah. I have no idea what the hell you're talking about but I like it.
@@Kakerate2 He meant that you never stop no matter how big the op life total is. If you get lethal, you win xD
@@FishMTG Roughly translated:
Assume that you cannot win, at all, once your opponent reaches 101 life or above. Calculating the odds of winning before 101 life is reached, is 9.25%.
Then we calculate the odds of winning if we assume that you cannot win if your opponent has 1001 life or above. This gives the same result (9.25% but with slightly different decimals WAY down the line).
This means that the actual odds of winning, in ANY (big) number of finite attempts, is 9.25%
As the first few attempts have way better odds, it's optimal for the player to manually do the first 3-5 loops or so, and then quit, rather than saying "I use this loop 1000 times, let's roll a dice to see if I win."
This is a nice attempt, but your equation looks incorrect. I can give an alternative formulation.
You can model this as a Markov process with transition probabilities given by
p(n->m) = 1/2^n (if m = 0)
p(n->m) = 1/2^(k+1) (if m = n + k with 0 m),
with a_0 = 1. Notice that a_n is just what you called P(n).
In our case, since n = 4, the equation looks like this:
a_4 = 1/16 + a_4/2 + a_5/4 + a_6/8 + a_7/16.
Sadly this is not a closed system because it propagates upwards indefinitely, but it might be possible to solve it in Fourier space. I'll have to look into it some more.
Numerically, a Python-coded algorithm that emulates the game seems to return a probability of about 18-19%, which sounds more reasonable than ~9.25% since at the very beginning you have at least 1/8 = 12.5% chance of winning by flipping three heads in a row (since you can re-flip the first coin indefinitely, so flipping the 'zero'-th heads has 100% success rate).
This is why the 4 horseman combo doesn’t work in legacy anymore
I mean if youre lucky it does XD VERY VERY LUCKY
Also it does in the only relevant variant of legacy (MODO legacy)
4 horseman can be proven to be deterministic given an infinite number of attempts. It's basically the same as this combo but without the opponent gaining life, so EVENTUALLY you'll flip 10 in a row and win.
@@opo33333 true, although the problem is that you can't shortcut that loop, because the opponent could have a response to the loop when the loop is in a certain state (like maybe they want to exile your graveyard, but only once a certain configuration of cards is in it).
Maybe someday, someone will find a way to make it impossible for their opponent to do anything before executing the loop, at which point there's no reason they shouldn't be able to execute it. In theory at least
@@jinxed7915 there is a beholder that prevents the opponent from playing anything on your turn
If your turn is gonna be over an hour long, we're gonna have problems. *Cracks knuckles*
This type of slow play is much more pronounced in four horsemen where you have to mill exactly four cards before hitting emurkral when you have a way of infinitely milling yourself. So while you can shortcut it to just mill and not perform the tap/untap necessary for the loop you still leads to the same exact issue cause once emurkral shuffles back into the deck the game state has not progressed.
The situation you described is more of a mathematical problem approach to the issue but still leads to the same ruling issue.
Though for that you could argue that if you do it enough times, you will eventually succeed.
Four horsemen isn't as precise as you describe, you just need to have the cards get milled at any point before the emrakul and there are 4 of them, you don't need to get exactly 4 and upon reset via emrakul the ones you did mill stay in play thus making each of them a coin flip of either being above or below the emrakul in the deck.
0:48 I think you misplaced the decimal point. That's a lot of useless leading zeros...
You are correct and now I'm mad about it
I love the idea of a research video trying to lie that a statistic is super small by using 0s in this manner. We're trained to never do this so it's not something we pay attention to when someone else does it.
Happy to see more content from you! Your editing is getting so much better
So basically, if you can’t say with 100% certainty what the game state will be after X iterations, an “infinite” combo cannot be shortcut. And if it can’t be shortcut then you only have a small finite number of iterations to achieve the desired game state because each iteration requires some amount of the limited time you have
100% is necessary but not sufficient, even if set of expections has probability 0, but it is non-empty loop is not allowed.
Simple, any infinites, that require chance, are assigned as slowplay in competitive format, it depends on the judge, some might allow it, some might deny it. This conlusion came from few judges, that I personally know... trust me, there was tons of yelling :D
this was a treat, I also like your editing
I wrote this code which outputs 0.186464 with an input of 4 life and depth of 30. As BobFaceASDF pointed out, we can ignore all series of tails and begin with one guaranteed head for each iteration. If there is a mistake in this code please let me know.
double function(int n, int maxDepth)
{
double output = pow(2, -(n-1));
if (n >= maxDepth) return output;
for (int i = 1; i < n; ++i)
{
if (n + i
Updated accuracy: 4 30
0.186453699627530330
double function(int n, int maxDepth)
{
double output = pow(2, -(n-1));
if (n >= maxDepth) return output;
for (int i = 1; i < n; ++i)
{
if (n + i
Really excellent job accounting for everything here. I briefly thought you were wrong about needing to succeed on four consecutive flips; The Sages can attack as well meaning you can attack for lethal with only three Wireflies right? But that’s wrong because the our opponent’s Elder can block the Sages meaning we do need four. Well done spotting that! Great video!
I believe in this situation the probability of success given infinite attempts does not approach 1, which is possible in situations where the odds of an attempt approach 0. First of all, the length of runs of tails does not matter and we can ignore them. Every 'attempt' starts with one head and your opponent gaining one life. They could gain more, but it's definitely at least 1 no matter what. So to take an upper bound, let's assume that it's always only one life. The first attempt has a 1/16 chance of success. If it fails, in the best case your opponent gained one life so you now have at most a 1/32 chance of success. The next try has at most a 1/64 chance of success, then 1/128 and so on. The correct thing to do with these probabilities is to multiply their inverses, but I don't want to deal with an infinite product and the chance is definitely less than the sum of the individual chances of success. The sum of a geometric series like this is easy, in this case 1/8. We now have an upper bound on the chance of success after an arbitrary number of attempts, and it's less than 1, so you are not guaranteed to succeed.
This is interesting. So you're saying that although you can try an infinite amount of times to succeed, each attempt is so smaller that the last that it doesn't matter? Would it be correct to say that this infinite sum is converging to a value less than 1?
Makes sense actually. When I first saw this problem, I went by the assumption that because you can try indefinitely, you would succeed eventually. But you convinced me that it is wrong.
I wonder if we can create a MTG situation where the sum is diverging or converging toward 1, and it would take so much time to go there that it would be illegal to both shortcut the loop (because of random) and play it out (because it would take too long, maybe years). So in this situation the player should win in theory but by these rules they wouldn't.
It would work with a different amount of life right? Or a different amount of cats that gain life
@@Programme021 Others have mentioned the four horsemen combo, which requires milling cards in a specific order. Since the last card is Emrakul you can keep reshuffling your graveyard into your library to try again. The shuffling is random so you can't shortcut it, but the probability of getting the desired order doesn't decrease which is a much simpler case where you are guaranteed to get it eventually. To make the loop take longer you can put extra graveyard-shuffling Eldrazi in the deck.
Actually, the first attempt has a 1/8 chance of success, because you can flip as many tails as you want at the beginning without your opponent gaining anything.
So your upper bound should be 1/4, not 1/8.
nice math. quick video. well done.
Neat analysis video! Thanks for uploading!
Rip Four Horsemen. Loved that Legacy deck.
What happens if you create a Turing machine for which it isn't known whether it halts or not? All of the actions are automatic. If it halts you win, if it loops it's a draw.
The game keeps on giving
Yes, you never have exactly a 0% chance of winning, so you could keep going since you can try indefinitely.
But your opponent is definitely approaching 100% chance of winning.
There comes a point where you just say "fuck it, round up"
As I understand even physics does this, at the Planck scale.
I'm sure you can figure out when that moment is, some time before dying while flipping coins.
If you keep playing it indefinitely there is a 100% chance of winning.
Your opponent doesn’t win. They gain life. That’s different.
This is an interesting case compared to the classic four horsemen example. In the four horsemen, the probability that you win in a finite number of turns is 1, meaning you could argue that even though it's nondeterministic you still are certain to win. (Also, let's be honest, even having to play it out a four horseman deck would kill faster than High Tide.) Here the probability is something like 20%, meaning you can't even argue there's no need to play it out.
High tide still one of my favorites to bust out. (Four Horsemen of the Apocalypse, on the other hand, usually confuses the hell out of my opponents, yet so fun to play).
The chance of this working converges to 0 the more failures you go through
Probability wise it's really weird, but it's pretty easy to simulate just so that you're not robbed of your sub 1% chances of this succeeding
The chance of this working is 1 or 100%. Since the nuber of possible tries is infinite, and it resolves in a finite number of iterations of the loop. The chance that converges to 0 is the chance of succeeding in one iteration of the loop.
@@jayaragon2787 You are wrong, if the opponent didn't gain any life when you fail the chance to win with this would be 1, but since every failure drastically decreases the chance for other iterations to succeed it turns out that the chance to win at all is actually pretty small. If you calculate the odds of winning it's actually less than 1/8.
@@jayaragon2787 that's not how convergent probability works
If the chance doesn't change with the number of iterations, you'd be right, but it does, meaning the probability becomes infinitely small as you reach an infinite number of tries
It’s so simple! Give your creatures indestructible.
Would a similar situation be say my opponent has a way to infinitely mill me but I have a card in my deck like an eldeazi that shuffles my graveyard into my deck. Can my opponent shortcut so that that Eldrazi is the last card in my deck so that I'm forced to draw it?
no, its the reason the 4 horseman isn't tournament legal. (Combo that infinitelly mill yourself with a way to shuffle your graveyard into your deck, but at each time it does some damage by milling certain cards. The problem is you don't know how much damage you do every loop before you have to shuffle your graveyard into your deck. theoretically you can be unlucky and always hit the card that reshuffle your deck before you do damage, even if with infinite loop you are garanteed to kill your opponent)
Ah yes, Wirefly Hive, good old wirefly hive.
In this scenario they also have a 2/3 doing the untapping, so they'd only need 3 successful flips in a row.
nope, the 2/3 can be blocked by the opponents 1/1 ;)
And thats why krarks thumb comes into play
My man is back
If you lose first flip game state did not change, you are not allowed to continue. Otherwise you are only allowed to continue after losing flip if opponent gained life between that and previous losing flip as otherwise game state is same
I had a combo like this then I realized that lab man and spectral sailor could be more consistent. I draw my deck oh they counter lab man? Well I have every counterspell from my deck in my hand. Draw one more card and win. Only time it fails is if they counter an ability which I can just activate again cause infinite mana.
Laughs in Disciple of the Vault topdeck
I've tried calculating this out manually to three iterations (each one adds a new dimension to the math, so I'm stopping there) to try and see if the chance of survival approaches 50%.
The first iteration it is an obvious 7/8 or 87.5% chance of surviving and growing life (either by one two or three depending on if they lose the flip on the second, third or fourth). Discounting the first flip as no change, they have a 50/50 of losing on the second flip for +1, 25% of losing on the third flip for +2 and 1/8 of losing on the fourth for +3.
Continuing to iteration 2, off of those three scenarios and weighted accordingly, they have an 83.4% total chance of survival - dropping by 4.1%.
But when I continue to the Third iteration, survival drops to 78.2%, dropping by 5.1%.
The win chance appears to be accelerating.
I'd love for someone else to check my math on this.
I don't understand what you are saying, each failure drops win odds precipitously. First chain needs 3 heads, and is the best odds the attacker will get. However winning 3 50/50s in a row is a chance of 12.5%, with a 87.5% chance of the other player surviving. If you only lose 1 flip of the 3, the odds of flipping 5-to-7 more heads is 3.125% to 0.78125%.
The point is that each iteration has dramatically worse odds, and those odds get worse exponentially. If you fail 2 flips in your first 6-to-10 tries infinite more tries just results in your opponent having infinite health.
@@MonochromaticPrism That is for an individual flip. When we instead look at the chances of the opponent surviving the entire process up to that point we can get a more holistic picture of what may actually occur. First, don't forget we drop the first flip so our chances of winning are 6.25%, 3.125%, 1.5625% - so the chances of survival are 93.75%, 96.875%, and 98.4375% in those brackets, and we land in those (if we survive round one) 50%, 25%, and 12.5% of the time, respectively which comes out to 0.46875 + 0.242188 + 0.123047 or 83.3984% chance of surviving round 2. The math for round 3 is similar, but 3 dimensional: four results under 5, five results under 6, six results under 7.
i don't play this game and i have no idea what he's talking about, but i'm going to eat it anyway
I'd just mix it with darksteel forge so that destruction effect doesn't matter
Or you could play dimir and use meathook massacre.
It would probably be faster to do a mathematical proof showing that you are more likely to win than lose (or vis versa) than to actually perform the loop
The worst part about this whole shitty situation is that given the chance to perform infinite iterations, you would 100% find the winning string. That’s because every possible combinations of H and T flips exist in the sequence, including an infinite string of H.
Rule 107.1. The only numbers the Magic game uses are integers.
Infinities are non-integer numbers so performing an infinite number of iterations would violate the rules and thus isn't permitted.
You would not have a 100% chance of winning, the chance to win each iteration goes down so quickly that even with infinite tries you are not guaranteed to succeed.
@@AoE_Freak That's absolutely not how infinity works. It doesn't matter how low the chance is. If you can try infinite times you will for certain get it eventually.
@@Cman21921 Nope, you are the one who is wrong about how infinity works. Let's say that every failed flip until the winning chain of succesful flips only created 1 wirefly, under this assumption our chance of success is even greater than the actual scenario, Now this assumption makes the calculations very easy. Our chance to win in our n+1 attempt is (1/2)^n+7. So the overall chance to win is 2^(-7)*(sum (1/2)^n with n=1 to infinity). The value of the infinite series here is famously 1 so overall we have 2^(-7) chance to succeed. But in the original scenario some failed attempts will give more than 1 life to our opponent so the actually probability of winning is even less than 2^(-7) which is already a fairly small number.
The game rules may allow it, but would these two planeswalkers really be sitting around long enough to produce the countless wireflies necessary to achieve victory?
what's NOT being discussed is how often you'll have that combo out while your opponent has a leonin elder...... HMMMMM
isn't the point that if it wasn't the leonin you could repeat the loop until you have lethal and win, but the leonin being in the game causes the lethal threshold to change at every attempt meaning you don't win automatically anymore
ln arena it happened to me that the opponent had a an infinite combo that would win the game, but i had soul warden so the combo lasted until his turn ended while my life was increased and decreased
@@devforfun5618 it was mostly a joke about what are the chances of seeing this interaction lol
But what do I do with my Gitrog CEDH deck? I mean... i could sit here shuffeling up for millennia....
How to turn Friday Night Magic into Saturday Morning Slog.
Especially if you're playing against that ONE MTG guy who says, "I can do this ALL night"
And then the shop you guys are in kicks you out. SOmetimes permanently. Sometimes for a week.
One thing you could do, is argue that on fair flips you have half chance. So you only get half as many tokens. Doesn't matter much anyway. Just make half of infinite tokens.
Sure they gain half of infinite life, but you will have half infinite blockers. So unless they manage to win by a means other than combat... "For all intents and purposes" you've won.
Four Horsemen is the actual deck that this is talking about.
Seems like the best bet would be to go for 2 tokens or at most for 3.
I don't get why the second half of MTR 4.4: LOOPS is being glossed over "a player must stop if at any point in the process a previous game state (or one identical in relevant ways) is reached again" I'd say you get one shot then your done and since the other player is only gaining more life making your plan more difficult but still being in the same board state otherwise
Because the other player gains life and thus an identical gamestate is never reached. It's impossible to argue that the other player gaining life is irrelevant since it directly changes your win condition by increasing the necessary number of heads.
@@Mr.ToadJanfu your opponent gaining life is not relevant to your goal or loop you need n-1 consecutive coin flip wins n= your opponents life total at the start. that never changes
@@ClaimingKarma I think it should be obvious that any situation you have to use a VARIABLE to express is one that is subject to changes.
I was summoned from Noah the Gathering's youtubeversal game.
Lots of math
I feel like if this ever actually happened in a reasonably important tournament, they’d probably make more solid rules about it; something that would prevent such infinite recursion
There was a legacy combo deck called 4 horsemen that won with a nondeterministic combo, basically you wanted to mill narcomebas from your graveyard to bring them back into play, sac them with grinding station to deal damage to your opponent and then reshuffle your graveyard into your deck by milling emrakul, the aeons torn.
However this was nondeterministic because you could mill emrakul before milling any narcomeba, therefore resetting the loop.
This deck was geniuenly very good, however it was pretty much soft banned, because tournament rules didn't allow you to shortcut nondeterministic loops, so it was unfeasible to win before being disqualified for slow play
0:52 that percentage chance you have there is the same as 9% lmao. A lot of zeros means nothing if you put the decimal in the wrong place
No, your success is not guaranteed even if you try unlimited times. It’s complicated math but the overall probability might not converge to one if the individual probability converges to zero
All tournaments have a REL.
i'm pretty sure even with limits this specific combo doesn't work, like I think it's just not in your favor
Is it actually nondeterministic? If you keep going forever, you will win, right?
If a player starts a creature token combo that they can’t stop from triggering infinity, and the word “may” isn’t involved in said combo, …..wouldn’t the game end in a draw because the player with said combo can’t end the phase it’s happening in and therefore can’t end their turn?
Yep. If an infinite loop that only involves mandatory actions begins, the game is a draw.
Could you write a program that simulates a series of coinflips and tells you if you’ve won? So all you would have to do is pick a number and run the program.
I am sad about these rules and would find it much more fun if you can play a loop as long as you can show that your probability of success = 1, i.e. you’ll *almost certainly* succeed. And as long as you can calculate the final probability distribution of the different end states quickly without having to play out a long loop manually.
four horsemen did nothing wrong
Realy like the video, but surely its then possible to determine what the odds are for winning if you go to infinity. say its 1/6 then your roll a d6 once. no?
The problem is you can't declare "infinte" as a number of iterations to shortcut. You must specify shortcuts that end so you need to declare a number. No matter how big you make that number you cannot say whether or not you have won when it's over.
Like someone mentioned you need only you need only 1/3 to hit becouse the first one doesnt mattter.
So the chance you win in infinit amount of try is:
Sum of (1/3)^x where x goes from 1 to inf
that convergence to 0.5 so you have 50 percent chance to win in infinit amount of time
If you keep playing it indefinitely there is a 100% chance of winning.
It doesnt really matter how high the life total gets. Because there is always a chance for you to win.
Infinite fluctuation is reached and you only need to hit zero once. Think of it like a cosine wave.
@@pieter8481 actually it is not. I dont know if you learned summation. Just becouse you are adding numbers infinitely doesnt mean you will end on infity.
For example: adding (1/10)^x= 1+0,1+0,01+0,001+0,0001+0,00001 ... = 1,111111.....
As you can see even if you add this number infinity amount of time you will never get more than 1,2
Same applies here. You are adding smaller and smaller number.
First time you have 1/3 chance. Second time 1/9. Third time 1/27.
So 1/3+1/9+1/27+....=0,5
@@lukasblaha2274
It doesnt actually really matter what the chance of winning is. With infinite tries there is infinite variance in possible outcomes.
Furthermore achieving a game-position where the chance of winning is exactly equal to zero is impossible.
There exists always a future ahead where you will be able to win given enough luck.
@@pieter8481 there is always a future where you can win, but it gets so low so quickly that even with an infinite number of tries, you are not guaranteed to win. Lukas' equation is correct.
@@pieter8481 a shortcut is a shortcut, if you actually try this combo it wouldn't go on forever because the judge would interrupt the match, maybe a less problematic rule would be to only allow a loop to go on an arbitrary amount of times, lets say 1000, then you could calculate the real probability and roll a dice, if you can prove that you can win in 1000 times, you win, otherwise you dont and cant try the loop until your next turn
Thats weird. As the leonin holder i would argue my opponent wins. Gaining infinite+4 life is not as good as having infinite*2 damage
I think the Wirefly Hive breaks down from abuse before reaching infinite wireflies, though.
Mathematics says otherwise. Those are both functionally the same on the scale of infinites.
Ok so this is the type writer effect in magic the only problem being that you can't win unless you show your work witch is impossible unless you got infinite trying to work with.
You could show your work easily. The only way to prove he wouldn’t win is to show some connection proving that each attempt sabotages itself. But they don’t; instead they simply make future attempts less likely which is not the same thing.
EDIT: He still can’t win this way, per the rule book tho.
I can’t remember the exact name, but this mathematical idea is related to a type of gambling fallacy. (Not THE gamblers fallacy though)
The idea is that if you play a fair game, or even a biased game, against an opponent with an arbitrarily large amount of money you will always eventually lose if you keep playing because eventually you’ll run out and have to stop and they won’t.
Here the player being attacked can gain life but never win, and so they must eventually lose.
St. Petersburg problem
I seen something simular happen once
Player 1: had infinite 1 at a time mill
Player 2: had a kozilek butcher of truth in deck
because the milling can be stopped and started at any time. Player 1 could mill 1 card at a time until either kozilek resets the deck and start again or kozilek is the last card in the deck
Since it was in a casual game we just cut to kozilek being the 1 card in deck.
Mine's more...Erm...Extreme.
Melira, Sylvok Outcast
Kitchen Finks
Bloodthrone Vampire
Add in:
Pawn of Ulamog and/or Sifter of Skulls
Butcher of Malakir
Field becomes "Yes" Eldrazi Spawn/Scions, I have a Yes/Yes Bloodthrone Vampire, and I have Yes Life while being able to borderline control my opponent's field.
…Just realized Wirefly Hive is a Changeling hate card.
Oh... oh God.
Please just Reanimate an Avacyn, Angel o Hope already
Or.. you could prove the exact probability of the strategy actually working, I didn't run the math but I expect it to converge to a pretty small finite number
Did you watch the full video? It doesn't, that's the point. The probability does in fact become infinitely small without ever converging.
@@johngalt200 dude read my comment again, i said the probability of "the strategy" working, i.e., the result of the infinite sum of the probabilities of each attempt
I love magic.
Theoretically you might make a rule to resolve this as the chance to win from this. Having the chance as one out of the number heads needed times the number of chances needed to get that at 4HP thats 4*4=16 so 1/16 roll a D16b and pick one number form that as a winner, when/if that fails, give the winner half the benefit of a full failure on every roll, so gain 2 health in this case, and forbid further tries until rollers next turn. but that wouldn't work for every case of a mechanic like this I would imagine. That and as 20hp it would take a d400 to resolve, tho one might in that case might cut their losses anyway. as 1/400 is horrid odds. and the equation breaks at 1hp as that requires a... a d1... yeah, no... just flip the coin really for that how ever.
It's somewhat disturbing that you're not allowed to just roll for the converged value in such cases
You expect a judge after a long day of arguing with grifters to calculate the precise probability, translate it into an equivalent dice roll, then clearly communicate what's happening to both players? Your optimism is inspiring!
@@nomukun1138 the only way to do it would be sending the presentation in advance to the judges in case you use it in game so they can prepare to roll in advance, and i dont think they should be this nice
What about summoning sickness? 🤔
Maybe it’s the Op’s end step
Why didn't you show the description of training grounds. I spent so long trying to figure out how this is an infinite...................................................................................................................................
Is wirefly hive technically a wirefly? The card says destroy all "wireflies" and not "wirefly" or wirefly hive".
It tells you to destroy all of a creature type. Wirefly hive is not a creature.
What's with the yugioh tier art?
Alt portrait?
But what is an actual judge's ruling?
"I just sat down, who starts a conversation like that!?"
Does MTG have a distinction between “restore 1 health up to a maximum of current max health” vs “increase current health (no matter what the current health is)” ?
Because, if the other card only does the former, and e.g. opponent won’t ever get over 40 health, then they would definitely lose eventually (if you played it out indefinitely).
If the card does the second thing, and can increase their health arbitrarily high, then nvm, the chance that you eventually win is strictly less than 1, even with no limit on play time
There is no limit to the amount of life you can gain, so it could truly go on forever
@@ScrubLordCasual thanks
even with mtg having no lifepoint maximum, this was a really good point for you to bring up. well done.
To add on top of this, kinda tangentially, unlike some other tcgs magic also allows for infinite creatures. Mtg just loves infinites tbh
It also worth noting that the starting health is 20, unless you playing edh/commander in which case it 40.
0:53 0000000000000000000000000000000000000000000000009.332636% is still 9.332636% regardless how many zeroes you put infront of 9
so the answer is N'yesn't... good to know.
How about this, You are in an paper tournament. You have this almost exact combo but the opponent does not have the Leonin elder.
The opponent has 30 life. So you need 15 Wireflies.
You have infinite mana so given enough time you should be able to end up with 15 eventually. But in tournament time out is a factor. Can the opponent force you to go through it manually because of timing?
Since you can't say I flip 1000 times and kill you. It's possible you need to flip 3000st times if you're unlucky.
-Second question:
Also what happens if you manage your inf combo and you say a number. One trillion mana. Then the inf mana combo goes away and all you're left with is an untap effect and wirefly Hive.
I guess you're no longer allowed to short cut because in theory you could end up with less than 15 wireflies if you tried to flip however many times you can flip with one trillion mana.
The only way to know for sure is to do it manually right?
I know almost nothing about magic
Attacking player would win in this call because of the nature of the game itself and how infinite combos work. By saying that the only obstacle to the combo are the arbitrary time limits enforced by the event or the subjective lens of "slow play", that means that the only thing you would need to say is "I complete this loop until the eventual winning of 4 coin flips in a row". It's a part of casual play that hasnt been integrated into the rules for what i can only imagine is some weird attempt not to shift meta. You're opponent doesn't win if a judge rules against it, you are just made to lose because of some silly rule.
In a coin flip you'd say you have a 50% chance to win, and your opponent does as well.
You wouldn't say that there are an infinite number of instances in the multiverse where you win the flip so you win without needing to flip.
Because your opponent wins just as often.
So why would you do that here, when your opponent's chance to win approaches 100% while yours drops?
Amongst that multiverse there would be infinite situations where you win, but a significantly larger infinite situations where your opponent wins.
Just because you're the attacking player?
@Jordan Rodrigues I wouldn't apply the gamblers paradox to this because you are effectively the house and opposing player in a game of magic. In poker or black jack your opponent has the same odds as you within the operations of the game and maintain those odds throughout. In this specific example the combo cannot fall off in a way that would be able to give the opposing player an opportunity at winning without the attacking player consciously doing so. By building a magic deck you are already doing your best to spit in the face of Lady Luck. You specifically add in specific numbers of cards to have a higher probability of appearing when you need them. Even in draft you do the same thing. When you're dealing with a Turing complete game you need to view a combo as completing a function and not as an ad infinitum system.
If you keep playing it indefinitely there is a 100% chance of winning.
It doesnt really matter how high the life total gets. Because there is always a chance for you to win.
Infinite fluctuation is reached and you only need to hit zero once. Think of it like a cosine wave.
You can't shortcut it that way though, because of the life your opponent gains each time a wirefly etbs. You need to know how exactly how much life they gain for it to be fair, so you would need to resolve everything manually.
@@AoE_Freak it wouldn't be a problem though. If you have infinite opportunities to win and an avenue to do so, the gain in life becomes wholly unimportant. Life totals stop being a scale of 1-20(1-100 for the gentlemen(and ladies) among us) and become a binary. With this specific combo, the game goes from a very complicated logic web/ algorithm ending in a yet to be defined victor, to a predefined outcome-machine.
TLDR in this specific scenario, and given the defending player cannot act in a meaningful way, life gained and triggers resolved become unimportant since the outcome is "Either I win or I let the combo fall and lose".
You do not win. I cannot tell you what the chance that you win is, but I can tell you that there is less than 12.5% chance that you will win, even with infinite flips.
Math time.
LP is the opponent's current life points. Each individual flip has a 50% chance to succeed. So each attempt has (0.5)^LP chance to succeed.
Assuming you already failed on your attempt, the best case scenario gives your opponent only 1 additional LP.
So to get a probability we can sum the probabilities for all our flips, since if we succeed in any of them we win. In this case we add (0.5)^LP + (0.5)^(LP+1) + (0.5)^(LP+2) + (0.5)^(LP+3) + ...
Luckily this infinite sum is well known. It actually is equal to (0.5)^LP / (1 - 0.5). If our LP starts at 4 then the sequence comes out to 0.125 or 12.5%. But this is not the percent chance to succeed. Because we could fail by more than 1 flip, skipping the chance to succeed on that number.
So in the situation present that person has great than 87.5% chance of giving their opponent infinite mana and less than 12.5% chance of winning.
For time sake, I propose we roll a d8.
disagree
@@Kakerate2 there are lots of assumptions made here to make the math solvable.
Technically you should be calculating the inverse probably. And need to add a factor in for the is off that term bring hit.
Offering to roll a d8 is an offer to gamble on the outcome of the game and your opponent is obligated to report it to a judge and you get a game loss. Better luck next time pal 😉
@@MK-13337 that's not gambling on the outcome of the game. that's using gambling to determine the outcome of the game.
@@yargolocus4853 True. I brainfarted.
The title "Schrödinger Combos" was right there
Ok. It was just 1 combo.
Why is the title in Plural then?
This has nothing to do with Schroedinger as the outcome doesn't change depending on if you observe it or not
@@oelboy Well, the outcome depends on the judge and how he is feeling that day
Uh… the wire flies don’t have haste. So you lose either way if you were about to lose anyways, or if the opponent just clears.
He stated that the scenario is taking place in the opponents end step. So he would untap with the right amount of Flies to kill his opponent. No haste necessary.
SUBBED!
1:03 timestone for cardgame statistic
coin flips and dice rolls have no place in magic
0:52 oh nice it's only a 9.332636% chance
Almost 1/10 LMAO
I wonder if this video was recommended to me because we're using the same Magic art as an avatar.
can somebody explain me why the enemy should lose when you have 4 wireflies? i don't see him losing any life, nor any other win condition... i'm really confused ^^
He is doing this at the opponents end step so he will be able to swing with them right away.
But if you can try an infinite amount of times, you will succeed eventually. It is deterministic. Onpymhow many tries and thisnartifacts you create is nondeterministic.
The head judges don't like game that are decided by having to know calculus. I had this very discussion a number of years ago with the powers that be and was told, "We're not deciding games on calculus."
this is actually a counterintuitive probability case where despite infinite tries and at no point reaching 0 probability to win, the odds of winning drop off so fast that they counteract even the power of an infinite series, and you are not guaranteed to win. It's counterintuitive which is why your gut reaction is to disbelieve me, and admittedly I'm not strong enough at probability to give you a proper mathematical proof, but it's true, I've seen some other comments explaining it better than I could. Have a browse.
Basically you are falling for the misconception that infinity works how you think it should, rather than infinity working how it actually does, which is often not what we expect.
@@empty5013
If you keep playing it indefinitely there is a 100% chance of winning.
It doesnt really matter how high the life total gets. Because there is always a chance for you to win.
Infinite fluctuation is reached and you only need to hit zero once. Think of it like a cosine wave.
@@pieter8481 no, there isn't a 100% chance of winning. If the probability drops quickly enough, it will converge towards a certain value instead of being 100%. Kinda like how you can add up an infinite amount of numbers, and they will converge towards a certain number of they are decreasing rapidly enough. 0.1 + 0.01 + 0.001... converges to 0.11111..., not infinity.
@@pieter8481 but you cant play indefinitelly, can you ?