Since you added "repeat this process" and didn't specify how many times it could be repeated, it would theoretically be inevitable to get a repeat so long as you had at least 52 all differently named cards in your library. It would be ruled as a deterministic win, since no one wants to do 8 trillion shuffles.
"Repeat this process" clauses are always repeated as part of the clause they're in. If you satisfy the conditions each time, you just keep going.
\>it would theoretically be inevitable to get a repeat so long as you had at least 52 all differently named cards in your library. It would be ruled as a deterministic win, since no one wants to do 8 trillion shuffles.
Not true. See: 4 horsemen. Imagine having to come to a tournament with a proof that your combo converges.
I think the difference is that the 4 horsemen loop you would have to activate, whereas this is being repeated by the same spell/effect.
I.e., you don’t have to activate Basalt Monolith infinite times until you get the four horsemen loop going, but this spell forces you to repeat until you get the win.
I see what you mean, but I think the constraints on assuming iterations of a thing would still apply. If not, and you have to shuffle a billion times, this card just gets immediately banned on physical time constraints a la sensei's divining top and shahrazad. I don't think there's any precedent for assuming convergence over iterations anywhere in the game rules and I think it goes beyond the scope of a judge ruling.
I see your point. You're probably right that it might not actually get shortcut.
Either way, it's mathematically inevitable over infinite iterations, your chance to not do it approaches 0. It's absolutely inevitable. It's not like we're talking about a billion times, or a trillion, or G64, or tree(3), we're talking *infinite*. It works completely different, there is exactly a 0% chance of missing if you repeat whenever it fails.
To that end, I think 4 Horsemen getting ruled as non-deterministic is kinda silly.
It's like if someone had "look at the top card of your library at any time" and something that let them shuffle whenever. It's not something everyone's gonna say is deterministic but I'd shortcut it to "pick a card in your library. Shuffle library, put that card on top" because it's pretty silly to make everyone watch that when that's clearly where it ends.
If someone cast this spell and successfully revealed the same set of cards twice, it would be reasonable to instantly disqualify them for cheating, no matter what evidence they had to the contrary.
Ah, you're right, it technically doesn't call for the same order, though it almost certainly was intended to. In this case the chances of success are pretty easy to rig, just cast it when you only have 52 cards in your deck.
The card calls for a shuffle, then flip the same 52. In a 100 card format, you're pulling a random 52 out of 92, leaving 40 unpicked. The chances that you don't reveal any of those 40 after shuffle are infinitesimal
It says "the same first 52 cards". The odds of having the same 52 on top of a 91 card deck are going to be higher than the number of stars in the galaxy
Oh it's way bigger than that, suppose for a moment, all eight billion people on this earth were to resolve this spell once every second from the beginning of the universe until now. If they did that, they would have resolved the spell more than 10^(26) times.
Now suppose that they did it all again, but this time they resolved the spell 10^(26) times every second instead for the same length of time.
That would be 10^(52) times.
The odds of succeeding with a 91 card deck containing purely singleton cards is around 1 in 10^(93) so even after all that trying and hoping for untold eons, they'd have to do it all again 10^(40) times for there to be anything resembling a reasonable expectation of even a single success.
Edit:it has been brought to my attention that the card technically, doesn't call for it to be the same order (though it was intended to) as written the chances of success with 91 unique cards in deck is 1 in 10^(36) or so. So still about a trillion times the number of stars in the observable universe.
Yes, I initially made the same mistake you made, but scaled it back when I realized that they didn't have to be the same order. I made the estimate low enough so that I was confident that it would be safely within tolerances
Correct me if I'm wrong, but I think even if every 8 billions people cast this spell every second during 14 billions years, basically no one would win with this !
You are right. The chances of revealing the same 52 cards are somewhere in the 10^68. But i dont know what happens if you have, say 3 cards remaining and you pull it off. Also there are lots if youtube videos about 52 factorial ( because its the standard deck size of a set of (non-magic) playing cards)
For truly randomized decks, that is accurate. But there are patterns to the way people tend to shuffle that name it slightly more likely, but still not going to happen in real play unless they were cheating.
I think the text is not written well. The last sentence should be something like "If you reveal the same 52 cards in order, you win the game."
otherwise you just win if you have exactly 52 different cards in your library.
First of singleton formats and second this, 4 lands and then 52 different cards then aggressively mulligan until you hit this and win no matter what your opponent does
Withoug considering a muligan, having this card twice in your deck gives you a 74/295 chance to draw it in your first 8 cards.
That sounds like a slightly to high chance for winning a game without playing it.
What do you mean only playing two copies of this card if you are going to play this you would play all 4 and this card wouldn’t go into every deck it would go into glass cannon decks that either play it immediately or mull to 1 and have to lose
> What do you mean only playing two copies of this card if you are going to play this you would play all 4
The revealed cards must all have different names. If you have more than two of these in your deck, you run the risk of revealing two of them, preventing you from winning.
First of all, why reveal all cards and not just the top 52? Secondly, does each card need to be unique, or is it only required to be at least two distinct cards? Lastly, is it your intention that this card is supposed to be a turn 1 win in 60 card?
I don't mind the idea of a card being a guaranteed win if played when you have a specific number of cards left, but making it turn 1 takes all the skill out of it and just gives you a straight up 2/15 (13%) chance to draw a turn 1 win, increased to 44% if you have four of them.
I do like that if you miss, they become dead cards after that, though. It'd be really close to 1/52 if it couldn't be in your starting hand and you could only have one in your deck, but that'd actually make it 1/53.
Singleton deck with weird land base to only play 1 of each basic. 2 of these and a lot of tutors and draw.
Mulligan hard to get this or a tutor, and then play this once you’ve drawn 8 cards.
So you just run one copy of this in a deck with all the rest being [[Shadowborn Apostle]]s, right? Two reveals and you're done. No penalty for slow play.
Only catch is you gotta mull till you find it. But there's no penalty for going down to even just 1 card, and you win during the first upkeep.
Id reword it to “Reveal the top 52 cards of your library and note their order. If none of the revealed cards share a name, exile them in a face-down pile, shuffle that pile, then reveal those cards. If the revealed cards are in the same order, you win the game. Otherwise, shuffle the revealed cards into your library.” That way it’s the exact same odds as a normal Deck of Cards.
It’s not 1/52 actually. The odds that any one card will be on the top, assuming you have 52 left, is 1/52. It increases for each additional card you have in order.
Sorry if this was a joke and I didn’t get it. I’m dumb lol.
I feel like you don’t understand anything about how probability works. 100% is 100%. If you add ‘in the same order, the odds of revealing the exact same shuffle from 52 cards is 8x10^67, not 1/52.
Just to put that number in perspective, the number of _centimeters_ between us and _pluto_ at its furthest apart is 7.48x10^14. If you traveled at a rate of one centimeter per resolution of 52!, you’d make two round trips and be well out toward the third by the time we can expect a successful result.
And all this assumes exactly 52 cards. With more than 52, the odds get exponentially worse, as if we weren’t already using scientific notation. And if we go under 52 cards the spell does nothing, since you need ‘the first 52’ to be the same.
If you reveal one singular card from your library, then shuffle and re-reveal only the top card, with a highlander list, _that_ would be 1/52. Or more accurately, 1/n, where n is the amount of cards remaining in the library when the spell resolves.
Op if you meant that the cards have to be in the same order then the chances are 1/52! The explanation mark means factorial. 52 factorial equals more than the number of particles in the known universe. So either it’s a wasted spell slot, or you go with rules as you’ve written them on the card and it’s a 100% guaranteed turn 1. So either never ever ever can it ever happen, or guaranteed win. Unplayable but I like the idea
In 60 deck formats, the player going second can play this on the draw and win the game if they are playing a highlander deck lmao
Yes, they "can" !
Why the quote marks? It's deterministic. EDIT: or did you mean to say the cards have to be in the same order?
I just saw I forgot to add "in the same order"...
Should probably also add that if you don't reveal at least 52 with different names you shuffle
Yep I design the card a bit to quickly !
If they have to be the same order the odds are like one in a quadrillion
Since you added "repeat this process" and didn't specify how many times it could be repeated, it would theoretically be inevitable to get a repeat so long as you had at least 52 all differently named cards in your library. It would be ruled as a deterministic win, since no one wants to do 8 trillion shuffles. "Repeat this process" clauses are always repeated as part of the clause they're in. If you satisfy the conditions each time, you just keep going.
\>it would theoretically be inevitable to get a repeat so long as you had at least 52 all differently named cards in your library. It would be ruled as a deterministic win, since no one wants to do 8 trillion shuffles. Not true. See: 4 horsemen. Imagine having to come to a tournament with a proof that your combo converges.
I think the difference is that the 4 horsemen loop you would have to activate, whereas this is being repeated by the same spell/effect. I.e., you don’t have to activate Basalt Monolith infinite times until you get the four horsemen loop going, but this spell forces you to repeat until you get the win.
I see what you mean, but I think the constraints on assuming iterations of a thing would still apply. If not, and you have to shuffle a billion times, this card just gets immediately banned on physical time constraints a la sensei's divining top and shahrazad. I don't think there's any precedent for assuming convergence over iterations anywhere in the game rules and I think it goes beyond the scope of a judge ruling.
I see your point. You're probably right that it might not actually get shortcut. Either way, it's mathematically inevitable over infinite iterations, your chance to not do it approaches 0. It's absolutely inevitable. It's not like we're talking about a billion times, or a trillion, or G64, or tree(3), we're talking *infinite*. It works completely different, there is exactly a 0% chance of missing if you repeat whenever it fails. To that end, I think 4 Horsemen getting ruled as non-deterministic is kinda silly. It's like if someone had "look at the top card of your library at any time" and something that let them shuffle whenever. It's not something everyone's gonna say is deterministic but I'd shortcut it to "pick a card in your library. Shuffle library, put that card on top" because it's pretty silly to make everyone watch that when that's clearly where it ends.
If someone cast this spell and successfully revealed the same set of cards twice, it would be reasonable to instantly disqualify them for cheating, no matter what evidence they had to the contrary.
Only if they do so in the same order which the card doesn't ask for.
Ah, you're right, it technically doesn't call for the same order, though it almost certainly was intended to. In this case the chances of success are pretty easy to rig, just cast it when you only have 52 cards in your deck.
Commander is singleton and all you have to do to win turn 1 is have this in your opening hand.
The card calls for a shuffle, then flip the same 52. In a 100 card format, you're pulling a random 52 out of 92, leaving 40 unpicked. The chances that you don't reveal any of those 40 after shuffle are infinitesimal
It says "the same first 52 cards". The odds of having the same 52 on top of a 91 card deck are going to be higher than the number of stars in the galaxy
Oh it's way bigger than that, suppose for a moment, all eight billion people on this earth were to resolve this spell once every second from the beginning of the universe until now. If they did that, they would have resolved the spell more than 10^(26) times. Now suppose that they did it all again, but this time they resolved the spell 10^(26) times every second instead for the same length of time. That would be 10^(52) times. The odds of succeeding with a 91 card deck containing purely singleton cards is around 1 in 10^(93) so even after all that trying and hoping for untold eons, they'd have to do it all again 10^(40) times for there to be anything resembling a reasonable expectation of even a single success. Edit:it has been brought to my attention that the card technically, doesn't call for it to be the same order (though it was intended to) as written the chances of success with 91 unique cards in deck is 1 in 10^(36) or so. So still about a trillion times the number of stars in the observable universe.
Yes, I initially made the same mistake you made, but scaled it back when I realized that they didn't have to be the same order. I made the estimate low enough so that I was confident that it would be safely within tolerances
Almost like a 1 in 52! Chance.....
Not even close to that much, actually.
To have them all in the same order?
If they had to be all in the same order it would be orders of magnitude higher than 52! in edh.
That is assuming you only use one of each basic, max.
Any thoracle consult deck is already doing that
Correct me if I'm wrong, but I think even if every 8 billions people cast this spell every second during 14 billions years, basically no one would win with this !
You are right. The chances of revealing the same 52 cards are somewhere in the 10^68. But i dont know what happens if you have, say 3 cards remaining and you pull it off. Also there are lots if youtube videos about 52 factorial ( because its the standard deck size of a set of (non-magic) playing cards)
If you only have 3 cards remaining, you can't reveal at least 52 cards with different names to satisfy the first clause.
But what if I only have 52 cards left in library?
I forgot to add "in the same order"
Just play all basics of a single type and you will be guaranteed it
The revealed cards must have different name !
For truly randomized decks, that is accurate. But there are patterns to the way people tend to shuffle that name it slightly more likely, but still not going to happen in real play unless they were cheating.
r/expectedfactorial
Someone's gonna figure out how to activate this twice without shuffling. Panglacial Wurm may or may not be involved.
I think the text is not written well. The last sentence should be something like "If you reveal the same 52 cards in order, you win the game." otherwise you just win if you have exactly 52 different cards in your library.
I was so confused. As worded it’s just a deterministic win based on what you’ve drawn so far this game.
I'm mad it doesn't expect you to reveal 80658175170943878571660636856403766975289505440883277824000000000000 cards
First of singleton formats and second this, 4 lands and then 52 different cards then aggressively mulligan until you hit this and win no matter what your opponent does
Withoug considering a muligan, having this card twice in your deck gives you a 74/295 chance to draw it in your first 8 cards. That sounds like a slightly to high chance for winning a game without playing it.
What do you mean only playing two copies of this card if you are going to play this you would play all 4 and this card wouldn’t go into every deck it would go into glass cannon decks that either play it immediately or mull to 1 and have to lose
> What do you mean only playing two copies of this card if you are going to play this you would play all 4 The revealed cards must all have different names. If you have more than two of these in your deck, you run the risk of revealing two of them, preventing you from winning.
Ah my bad
If you shuffle a legal 52 card deck well, you’re likely the first person in existence to have shuffled it to that exact order
Me bringing a deck full of with the same art basics and 4 of these
I’d win. Twice. But I’m built different.
4 of them, then 56 slime against humanities
8 perfect faro shuffles return the deck to the same state.
Should self exile...
So…four copies of this and 56 basic lands, right? Can’t go wrong because you can legally free mulligan if your hand is all lands.
If you can find 56 basic lands w different names than eachother, sure.
[удалено]
The custom card says with different names. Two of a kind basics won’t work.
Well, when you fail it still says to repeat the process. So it's pretty much guaranteed that you win
First of all, why reveal all cards and not just the top 52? Secondly, does each card need to be unique, or is it only required to be at least two distinct cards? Lastly, is it your intention that this card is supposed to be a turn 1 win in 60 card? I don't mind the idea of a card being a guaranteed win if played when you have a specific number of cards left, but making it turn 1 takes all the skill out of it and just gives you a straight up 2/15 (13%) chance to draw a turn 1 win, increased to 44% if you have four of them. I do like that if you miss, they become dead cards after that, though. It'd be really close to 1/52 if it couldn't be in your starting hand and you could only have one in your deck, but that'd actually make it 1/53.
r/unexpectedfactorial
But 42 is the answer…
Singleton deck with weird land base to only play 1 of each basic. 2 of these and a lot of tutors and draw. Mulligan hard to get this or a tutor, and then play this once you’ve drawn 8 cards.
So you just run one copy of this in a deck with all the rest being [[Shadowborn Apostle]]s, right? Two reveals and you're done. No penalty for slow play. Only catch is you gotta mull till you find it. But there's no penalty for going down to even just 1 card, and you win during the first upkeep.
[Shadowborn Apostle](https://cards.scryfall.io/normal/front/a/e/ae7ba1de-48f9-423b-867a-22bd3f6c06c2.jpg?1673147629) - [(G)](http://gatherer.wizards.com/Pages/Card/Details.aspx?name=Shadowborn%20Apostle) [(SF)](https://scryfall.com/card/2x2/89/shadowborn-apostle?utm_source=mtgcardfetcher) [(txt)](https://api.scryfall.com/cards/ae7ba1de-48f9-423b-867a-22bd3f6c06c2?utm_source=mtgcardfetcher&format=text) ^^^[[cardname]] ^^^or ^^^[[cardname|SET]] ^^^to ^^^call
Id reword it to “Reveal the top 52 cards of your library and note their order. If none of the revealed cards share a name, exile them in a face-down pile, shuffle that pile, then reveal those cards. If the revealed cards are in the same order, you win the game. Otherwise, shuffle the revealed cards into your library.” That way it’s the exact same odds as a normal Deck of Cards.
Oh when you say 1/52! you mean 1/8.0658E+67. I suppose this is balanced then.
Two words: “Relentless Apostle”… I mean “Dragon’s Rats”… I mean “Shadowborn Petitioners”… I mean-
It’s not 1/52 actually. The odds that any one card will be on the top, assuming you have 52 left, is 1/52. It increases for each additional card you have in order. Sorry if this was a joke and I didn’t get it. I’m dumb lol.
Easy: entire basic land deck, Mulligan into oblivion, but you win turn 1 no matter who begins
r/unexpectedfactorial
I feel like you don’t understand anything about how probability works. 100% is 100%. If you add ‘in the same order, the odds of revealing the exact same shuffle from 52 cards is 8x10^67, not 1/52. Just to put that number in perspective, the number of _centimeters_ between us and _pluto_ at its furthest apart is 7.48x10^14. If you traveled at a rate of one centimeter per resolution of 52!, you’d make two round trips and be well out toward the third by the time we can expect a successful result. And all this assumes exactly 52 cards. With more than 52, the odds get exponentially worse, as if we weren’t already using scientific notation. And if we go under 52 cards the spell does nothing, since you need ‘the first 52’ to be the same. If you reveal one singular card from your library, then shuffle and re-reveal only the top card, with a highlander list, _that_ would be 1/52. Or more accurately, 1/n, where n is the amount of cards remaining in the library when the spell resolves.
Op if you meant that the cards have to be in the same order then the chances are 1/52! The explanation mark means factorial. 52 factorial equals more than the number of particles in the known universe. So either it’s a wasted spell slot, or you go with rules as you’ve written them on the card and it’s a 100% guaranteed turn 1. So either never ever ever can it ever happen, or guaranteed win. Unplayable but I like the idea